RD Sharma Class 12 Exercise 11.1 Higher Order Derivatives Solutions Maths - Download PDF Free Online

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# RD Sharma Class 12 Exercise 11.1 Higher Order Derivatives Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 27, 2022 02:55 PM IST

The RD Sharma Class 12th Exercise 11.1 have become an indispensable part of the class 12 students’ exam preparation routine. They use these books to complete their homework, assignments and study for the exams. These books are available for all the subjects and chapters.

Also Read - RD Sharma Solution for Class 9 to 12 Maths

## Higher Order Derivatives Excercise:11.1

Higher Order Derivatives exercise 11.1 question 1(i)

$6x+2sec^{2}x\: tan\, x$
Hint:
You must know about derivative of tan x
Given:
$x^{3}+tan\: x$
Solution:
$Let\: \: y=x^{3}+tan\: x$
\begin{aligned} &\frac{d y}{d x}=3 x^{2}+\sec ^{2} x \quad \quad\left[\frac{d(\tan x)}{d x}=\sec ^{2} x \text { and } \frac{d x^{3}}{d x}=3 x^{2}\right] \\ &\frac{d^{2} y}{d x^{2}}=6 x+2 \sec x \cdot \sec x \tan x \frac{d \sec x}{d x}=\sec x \tan x \\ &\frac{d^{2} y}{d x^{2}}=6 x+2 \sec ^{2} x \tan x \end{aligned}

Higher Order Derivatives exercise 11.1 question 1(ii)

$\frac{-[\sin (\log x)+\cos (\log x)]}{x^{2}}$
Hint:
You must know about derivative of sin x & log x
Given:
$sin(log\: x)$
Solution:
$Let\: \: y=sin(log\: x)$
\begin{aligned} &\left.\frac{d y}{d x}=\cos (\log x) \times \frac{d}{d x} \log x \quad \text { ( } \frac{d \sin x}{d x}=\cos x \right) \\ &\left.\frac{d y}{d x}=\cos (\log x) \times \frac{1}{x} \quad \text { ( } \frac{d \log x}{d x}=\frac{1}{x}\right) \\ &\frac{d y}{d x}=\frac{\cos (\log x)}{x} \end{aligned}
Use quotient rule
$As\: \: \frac{u}{v}=\frac{u'v-v'u}{v^{2}}$
$W\! here\: \: v=x\: \, and \: \: u=cos(log\: x)$
\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{\left.x \frac{d}{d x}(\cos (\log x))-\cos (\log x)\right) \frac{d}{d x}(x)}{x^{2}} \\ &\frac{d^{2} y}{d x^{2}}=\frac{-x \sin (\log x) \cdot \frac{d(\log x)}{d x}-1 \cdot \cos (\log x)}{x^{2}}\left(\frac{d}{d x} x=1\right) \\ &\frac{d^{2} y}{d x^{2}}=\frac{-\sin (\log x) \cdot \frac{1}{x} x-\cos (\log x)}{x^{2}} \quad\left(\frac{d \log x}{d x}=\frac{1}{x}\right) \\ &\frac{d^{2} y}{d x^{2}}=\frac{-\sin (\log x)-\cos (\log x)}{x^{2}} \\ &\frac{d^{2} y}{d x^{2}}=\frac{-[\sin (\log x)+\cos (\log x)]}{x^{2}} \end{aligned}

Higher Order Derivatives exercise 11.1 question 1(iii)

$-cosec^{2}x$
Hint:
You must know about derivative of sin x & log x
Given:
$log(sin\, x)$
Solution:
$Let\: \: y=log(sin\, x)$
$\begin{array}{ll} \frac{d y}{d x}=\frac{1}{\sin x} \frac{d \sin x}{d x} & \left(\frac{d \log x}{d x}=\frac{1}{x}\right) \\\\ \frac{d y}{d x}=\frac{1}{\sin x} \cos x & \left(\frac{d \sin x}{d x}=\cos x\right) \\\\ \frac{d y}{d x}=\cot x & \left(\frac{\cos x}{\sin x}=\cot x\right) \\\\ \frac{d^{2} y}{d x^{2}}=\cot x & \\\\ \frac{d^{2} y}{d x^{2}}=-cosec^{2} x & \left(\frac{d}{d x} \cot x=-cosec^{2} x\right) \end{array}$

Higher Order Derivatives exercise 11.1 question 1(iv)

$-24e^{x}sin\: 5x+10e^{x}cos\: 5x$
Hint:
You must know about derivative of sin x & ex
Given:
$e^{x}sin\: 5x$
Solution:
$Let\: \: y=e^{x}sin\: 5x$
Use multiplicative rule
$As\: \: UV=UV^{1}+U^{1}V$
Where U=ex & V=sin 5x
\begin{aligned} &\frac{d y}{d x}=e^{x} \frac{d}{d x} \sin 5 x+\frac{d}{d x} e^{x} \cdot \sin 5 x \\ &\frac{d y}{d x}=e^{x} \cdot \cos 5 x \frac{d}{d x} 5 x+e^{x} \cdot \sin 5 x \quad\left(\frac{d}{d x} e^{x}=e^{x}\right) \quad\left(\frac{d \sin 5 x}{d x}=\cos 5 x\right) \\ &\frac{d y}{d x}=e^{x} \cdot \cos 5 x \cdot 5+e^{x} \cdot \sin 5 x \quad \quad\left(\frac{d}{d x} 5 x=5\right) \\ &\frac{d y}{d x}=5 e^{x} \cos 5 x+e^{x} \sin 5 x \end{aligned}
\begin{aligned} &\frac{d}{d x}(\frac{d y}{d x})=\frac{d}{d x}(5 e^{x} \cos 5 x+e^{x} \sin 5 x) \end{aligned}
Again use multiplication rule
$As\: \: UV=UV^{1}+U^{1}V$
Where U=ex & V=sin 5x
U=ex & V=cos 5x
\begin{aligned} &\frac{d^{2} y}{d x^{2}}=5\left[e^{x} \frac{d}{d x} \cos 5 x+\frac{d}{d x} e^{x} \cdot \cos 5 x\right]+\left[e^{x} \frac{d}{d x} \sin 5 x+\frac{d}{d x} e^{x} \cdot \sin 5 x\right] \\ &\frac{d^{2} y}{d x^{2}}=5\left[e^{x} \cdot(-\sin 5 x) \frac{d}{d x} 5 x+e^{x} \cdot \cos 5 x\right]+\left[e^{x} \cos 5 x \frac{d}{d x} 5 x+e^{x} \cdot \sin 5 x\right] \\ &\left(\frac{d \sin x}{d x}=\cos x, \frac{d \cos x}{d x}=-\sin x, \frac{d 5 x}{d x}=5\right) \end{aligned}
\begin{aligned} &\frac{d^{2} y}{d x^{2}}=5\left[e^{x} \cdot(-\sin 5 x) 5+e^{x} \cdot \cos 5 x\right]+\left[e^{x} \cos 5 x .5+e^{x} \cdot \sin 5 x\right] \\ &\frac{d^{2} y}{d x^{2}}=\left[25 e^{x} \sin 5 x+5 e^{x} \cdot \cos 5 x\right]+\left[5 e^{x} \cos 5 x +e^{x} \cdot \sin 5 x\right] \\ &\frac{d^{2} y}{d x^{2}}=-24 e^{x} \sin 5 x+10 e^{x} \cdot \cos 5 x \end{aligned}

Higher Order Derivatives exercise 11.1 question 1(v)

$27e^{6x}cos3x-36e^{6x}sin3x$
Hint:
You must know about derivative of cos3x & e6x
Given:
$e^{6x}cos3x$
Solution:
$Let\: \: y=e^{6x}cos3x$
Use multiplicative rule
$As\: \: UV=UV^{1}+U^{1}V$
Where U=e6x and V=cos3x
\begin{aligned} &\frac{d y}{d x}=e^{6 x} \frac{d}{d x} \cos 3 x+\frac{d}{d x} e^{6 x} \cdot \cos 3 x \\ &\frac{d y}{d x}=e^{6 x}-\sin 3 x \frac{d}{d x} 3 x+6 e^{6 x} \cos 3 x \quad\left(\frac{d \cos x}{d x}=-\sin x, \frac{d}{d x} e^{6 x}=e^{6 x} .6\right) \\ &\frac{d y}{d x}=-3 e^{6 x} \sin 3 x+6 e^{6 x} \cos 3 x \\ &\frac{d^{2} y}{d x^{2}}=-3 e^{6 x} \sin 3 x+6 e^{6 x} \cos 3 x \end{aligned}
$As\: \: UV=UV^{1}+U^{1}V$
Where U=e6x and V=sin3x
\begin{aligned} &\frac{d^{2} y}{d x^{2}}=-3\left[e^{6 x} \cos 3 x(3)+e^{6 x} \cdot \sin 3 x .6\right]+6\left[e^{6 x} \cos 3 x \cdot 6+e^{6 x} \cdot(-\sin 3 x)\right] \\ &\left(\frac{d \sin 3 x}{d x}=3 \cos 3 x, \frac{d \cos 3 x}{d x}=3(-\sin 3 x), \frac{d e^{6 x}}{d x}=6 e^{6 x}\right) \\ &\frac{d^{2} y}{d x^{2}}=-3\left[3 e^{6 x} \cos 3 x+6 e^{6 x} \cdot \sin 3 x\right]+6\left[e^{6 x} \cos 3 x \cdot 6-3 e^{6 x} \cdot \sin 3 x\right] \\ &\frac{d^{2} y}{d x^{2}}=27 e^{6 x} \cos 3 x-36 e^{6 x} \sin 3 x \end{aligned}

Higher Order Derivatives exercise 11.1 question 1(vi)

$5x+6x\: log\: x$
HInt:
You must know about derivative of log x & x3
Given:
$x^{3}\: log\: x$
Solution:
$Let\: \: y=x^{3}\: log\: x$
Use multiplicative rule
As UV=UV1+U1V
Where U=x3 & V=log x
\begin{aligned} &\frac{d y}{d x}=x^{3} \frac{d}{d x} \log x+\frac{d}{d x} x^{3} \log x \\ &\frac{d y}{d x}=x^{3} \frac{1}{x}+3 x^{2} \log x \quad\left(\frac{d}{d x} \log x=\frac{1}{x}, \frac{d}{d x} x^{3}=3 x^{2}\right) \\ &\ \end{aligned}
Use multiplicative rule
As UV=UV1+U1V
Where U=x2 & V=log x
\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{d}{d x} x^{2}+3\left[x^{2} \frac{d}{d x} \log x+\frac{d}{d x} x^{2} \cdot \log x\right] \\ &\left.\frac{d^{2} y}{d x^{2}}=2 x+3[x+2 x \cdot \log x] \quad \text { ( } \frac{d}{d x} x^{2}=2 x, \frac{d}{d x} \log x=\frac{1}{x}\right) \\ &\frac{d^{2} y}{d x^{2}}=2 x+3 x+6 x \cdot \log x \\ &\frac{d^{2} y}{d x^{2}}=5 x+6 x \log x \end{aligned}

Higher Order Derivatives exercise 11.1 question 1(vii)

$\frac{-2x}{(1+x^{2})^{2}}$
Hint:
You must know about derivative of tan-1x
Given:
tan-1x
Solution:
$Let\: \: y=tan^{-1}x$
\begin{aligned} &\frac{d y}{d x}=\frac{1}{1+x^{2}} \quad\left(\frac{d \tan ^{-1} x}{d x}=\frac{1}{1+x^{2}}\right) \\ & \end{aligned}
Use Quotient rule
\begin{aligned} &\text { As } \frac{u}{v}=\frac{u^{1} v-v^{1} u}{v^{2}} \\ &\text { Where } v=1+x^{2} \& u=1 \\ &\frac{d^{2} y}{d x^{2}}=\frac{\left(1+x^{2}\right) \frac{d}{d x} 1-1 \cdot \frac{d}{d x}\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{2}} \\ &\left.\frac{d^{2} y}{d x^{2}}=\frac{0\left(1+x^{2}\right)-1.2 x}{\left(1+x^{2}\right)^{2}} \quad \text { ( } \frac{d}{d x} 1=0, \frac{d}{d x}\left(1+x^{2}\right)=2 x\right) \\ &\frac{d^{2} y}{d x^{2}}=\frac{0-2 x}{\left(1+x^{2}\right)^{2}} \\ &\frac{d^{2} y}{d x^{2}}=\frac{-2 x}{\left(1+x^{2}\right)^{2}} \end{aligned}

Higher Order Derivatives exercise 11.1 question 1(viii) maths

$-xcos\: x-2sin\: x$
Hint:
You must know about derivative of x cos x
Given:
x cos x
Solution:
$Let\: \: xcos\: x$
Use multiplicative rule
As UV=UV1+U1V
Where U=x & V=cos x
\begin{aligned} &\frac{d y}{d x}=x \frac{d}{d x} \cos x+\frac{d}{d x} x \cos x \quad\left(\frac{d}{d x} x=1, \frac{d}{d x} \cos x=-\sin x\right) \\ &\frac{d y}{d x}=-x \sin x+\cos x \end{aligned}
Use multiplicative rule
As UV=UV1+U1V
Where U=x & V=sin x
\begin{aligned} &\frac{d^{2} y}{d x^{2}}=-\left(x \frac{d}{d x} \sin x+\frac{d}{d x} x \cdot \sin x\right)+\frac{d \cos x}{d x} \\ &\frac{d^{2} y}{d x^{2}}=-(x \cos x+\sin x)-\sin x \quad\left(\frac{d}{d x} \cos x=-\sin x\right) \\ &\frac{d^{2} y}{d x^{2}}=-x \cos x-\sin x-\sin x \\ &\frac{d^{2} y}{d x^{2}}=-x \cos x-2 \sin x \end{aligned}

Higher Order Derivatives exercise 11.1 question 1(ix)

$\frac{-(1+log\: x)}{x^{2}(log\: x)^{2}}$
Hint:
You must know about derivative of log(log x)
Given:
$log(log\: x)$
Solution:
$Let\: \:y= log(log\: x)$
\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x} \log (\log x) \quad\left(\frac{d \log x}{d x}=\frac{1}{x}\right) \\ &\frac{d y}{d x}=\frac{1}{\log x} \frac{d}{d x}(\log x) \\ &\frac{d y}{d x}=\frac{1}{\log x} \cdot \frac{1}{x} \\ &\frac{d y}{d x}=\frac{1}{x \log x} \end{aligned}
Use Quotient rule
\begin{aligned} &\text { As } \frac{u}{v}=\frac{u^{1} v-v^{1} u}{v^{2}}\\ &\text { Where } u=1 \& v=x \log x\\ &\text { Use multiplicative rule }\\ &\text { As } U V=U V^{1}+U^{1} V\\ &\text { Where } U=1 \& V=\log x \end{aligned}
\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{\frac{d}{d x} 1 x \log x-1 \frac{d}{d x} x \log x}{(x \log x)^{2}} \\ &\frac{d^{2} y}{d x^{2}}=\frac{0 . x \log x-x \log x}{x^{2}(\log x)^{2}} \quad\left(\frac{d}{d x} \log x=\frac{1}{x}, \frac{d}{d x} 1=0, \frac{d}{d x} x=1\right) \\ &\frac{d^{2} y}{d x^{2}}=\frac{-\log x-1}{x^{2}(\log x)^{2}} \\ &\frac{d^{2} y}{d x^{2}}=\frac{-(1+\log x)}{x^{2}(\log x)^{2}} \end{aligned}

Higher Order Derivatives exercise 11.1 question 2

$2e^{-x}sin\: x$
Hint:
You have to show
$\frac{d^{2}y}{dx^{2}}=2e^{-x}sin\: x$
Given:
If y=e-xcos x, show that
$\frac{d^{2}y}{dx^{2}}=2e^{-x}sin\: x$
Solution:
Let y=e-xcos x
Use multiplicative rule
\begin{aligned}\text { As } U V=U V^{1}+U^{1} V \\ \text { Where } U=e^{-x} \& V=\cos x \\\frac{d y}{d x}=e^{-x} \frac{d}{d x} \cos x+\cos x \frac{d}{d x} e^{-x}\\ \frac{d y}{d x}=-e^{-x} \sin x-\cos x e^{-x} & \left(\frac{d \cos x}{d x}=-\sin x, \frac{d e^{-x}}{d x}=-1 e^{-x}\right) \end{aligned}
Again differentiating w.r.t.x we get
Use multiplicative rule
\begin{aligned} &\text { As } U V=U V^{1}+U^{1} V \\ &\text { Where } U=\cos x\: \&\: V=e^{-x} \\ &\frac{d^{2} y}{d x^{2}}=-\left[\sin x \cdot \frac{d}{d x} e^{-x}+e^{-x} \cdot \frac{d}{d x} \sin x+\cos x \frac{d}{d x} e^{-x}+e^{-x} \frac{d}{d x} \cos x\right] \\ &\frac{d^{2} y}{d x^{2}}=-\left[\sin x \cdot\left(-e^{-x}\right)+e^{-x} \cos x+\cos x\left(-e^{-x}\right)+e^{-x}(-\sin x)\right] \end{aligned}
\begin{aligned} &\frac{d^{2} y}{d x^{2}}=-\left[-e^{-x} \sin x+e^{-x} \cos x-e^{-x} \cos x-e^{-x} \sin x\right] \\ &\left(\frac{d}{d x} \sin x=\cos x, \frac{d}{d x} \cos x=-\sin x, \frac{d}{d x} e^{-x}=-1 e^{-x}\right) \\ &\frac{d^{2} y}{d x^{2}}=-\left[-2 \sin x e^{-x}\right] \\ &\frac{d^{2} y}{d x^{2}}=2 e^{-x} \sin x \end{aligned}
Hence proved

Higher Order Derivatives exercise 11.1 question 3

$\cos ^{2} x \frac{d^{2} y}{d x^{2}}-2 y+2 x=0$
Hint:
You have to show
$\cos ^{2} x \frac{d^{2} y}{d x^{2}}-2 y+2 x=0$
Given:
If y=x+tan x show that
$\cos ^{2} x \frac{d^{2} y}{d x^{2}}-2 y+2 x=0$
Solution:
Let y=x+tan x
$y=x+tan x\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; (\frac{d\: tan\: x}{dx}=sec^{2}x,\: \frac{\mathrm{d} }{\mathrm{d} x}x=1)$
\begin{aligned} &\frac{d y}{d x}=1+\sec ^{2} x \\ &\frac{d^{2} y}{d x^{2}}=2 \sec x(\sec x \tan x) \quad\left(\frac{d}{d x} \sec ^{2} x=2 \sec x(\sec x \tan x)\right. \\ &\frac{d^{2} y}{d x^{2}}=2 \sec ^{2} x \tan x \quad \quad\left(\sec ^{2} x=\frac{1}{\cos ^{2} x}, \tan x=\frac{\sin x}{\cos x}\right) \\ &\frac{d^{2} y}{d x^{2}}=\frac{2 \sin x}{\cos ^{3} x} \end{aligned}
According to question
$\cos ^{2} x \frac{d^{2} y}{d x^{2}}-2 y+2 x=0$
substituting these values we get
\begin{aligned} &\left(\frac{2 \sin x}{\cos x}\right)-2(x+\tan x)+2 x \\ &\frac{2 \sin x}{\cos x}-2 x-2 \tan x+2 x \\ &2 \tan x-2 x-2 \tan x+2 x=0 \end{aligned}
Hence proved

Higher Order Derivatives exercise 11.1 question 4 maths

$\frac{6}{x}$
HInt:
You must know about derivative of x3 & log x
Given:
If y=x3log x, Prove that
$\frac{d^{4}y}{dx^{4}}=\frac{6}{x}$
Solution:
Let y=x3log x
Use multiplicative rule
$As \: \: UV=UV^{1}+U^{1}V$
\begin{aligned} &\text { Where } U=x^{3} \& V=\log x \\ &\frac{d y}{d x}=x^{3} \frac{d}{d x} \log x+\frac{d}{d x} x^{3} \log x \\ &\frac{d y}{d x}=x^{3} \frac{1}{x}+3 x^{2} \log x \quad\left(\frac{d \log x}{d x}=\frac{1}{x}, \frac{d}{d x} x^{3}=3 x^{2}\right) \\ &\frac{d y}{d x}=x^{2}+3 x^{2} \log x \end{aligned}
Use multiplicative rule
$As \: \: UV=UV^{1}+U^{1}V$
\begin{aligned} &\text { Where } U=x^{2} \& V=\log x \end{aligned}
\begin{aligned} &\frac{d^{2} y}{d x^{2}}=3\left(x^{2} \frac{d}{d x} \log x+\frac{d}{d x} x^{2} \log x\right)+\frac{d}{d x} x^{2} \\ &\frac{d^{2} y}{d x^{2}}=3\left(x^{2} \frac{1}{x}+2 x \log x\right)+2 x \quad\left(\frac{d \log x}{d x}=\frac{1}{x}, \frac{d}{d x} x^{2}=2 x\right) \\ &\frac{d^{2} y}{d x^{2}}=3 x+6 x \log x+2 x \\ &\frac{d^{2} y}{d x^{2}}=5 x+6 x \log x \end{aligned}
Use multiplicative rule
$As \: \: UV=UV^{1}+U^{1}V$
\begin{aligned} &\text { Where } U=x\: \&\: V=\log x \end{aligned}
\begin{aligned} &\frac{d^{3} y}{d x^{3}}=\frac{d}{d x} 5 x+6\left(x \frac{d}{d x} \log x+\frac{d}{d x} x \log x\right) \\ &\frac{d^{3} y}{d x^{3}}=5+6\left(x \frac{1}{x}+\log x\right) \quad\left(\frac{d \log x}{d x}=\frac{1}{x}, \frac{d}{d x} 5 x=5\right) \\ &\frac{d^{3} y}{d x^{3}}=5+6+6 \log x \end{aligned}
\begin{aligned} &\frac{d^{3} y}{d x^{3}}=11+6 \log x \\ &\left.\frac{d^{4} y}{d x^{4}}=0+\frac{6}{x} \quad \text { ( } \frac{d \log x}{d x}=\frac{1}{x}, \frac{d}{d x} 11=0\right) \\ &\frac{d^{4} y}{d x^{4}}=\frac{6}{x} \end{aligned}
Hence proved

Higher Order Derivatives exercise 11.1 question 5

$2cos\: x\: cosec^{3}x$
Hint:
You must know about how third derivatives be find
Given:
If y=log x(sin x). Prove that
$\frac{d^{3}y}{dx^{3}}=2cos\: x\: cosec^{3}x$
Solution:
$Let\: \: y=log (sin x)\; \; \; \; \; \; \; \; (\frac{d\: log\: x}{dx}=\frac{1}{x})$
\begin{aligned} &\frac{d y}{d x}=\frac{1}{\sin x} \frac{d}{d x} \sin x \\ &\frac{d y}{d x}=\frac{1}{\sin x} \cos x \\ &\frac{d y}{d x}=\cot x \end{aligned}
again differentiating we get
$\frac{d^{2} y}{d x^{2}}=-cosec^{2} x$ $(\frac{d\: cot\: x}{dx}=-cosec^{2}x)$
Again differentiating w.r.t. x we get
\begin{aligned} &\frac{d^{3} y}{d x^{3}}=-2 \operatorname{cosec} x\cdot(-\cos ec x\cdot \cot x) \quad\left(\frac{d \operatorname{cosec}^{2} x}{d x}=-cosec\: x \cot x\right) \\ &\frac{d^{3} y}{d x^{3}}=-2 \cos x \operatorname{cosec}^{3} x \end{aligned}

Higher Order Derivatives exercise 11.1 question 6

$\frac{d^{2}y}{dx^{2}}+y=0$
Hint:
You have to know about how to find derivative of second order
Given:
If y=2sin x+3cos x,show that
$\frac{d^{2}y}{dx^{2}}+y=0$
Solution:
Let y=2sin x+3cos x
\begin{aligned} &\frac{d y}{d x}=2 \cos x-3 \sin x \quad \quad\left(\frac{d \cos x}{d x}=-\sin x, \frac{d \sin x}{d x}=\cos x\right) \\ &\frac{d^{2} y}{d x^{2}}=-2 \sin x-3 \cos x \\ &\frac{d^{2} y}{d x^{2}}=-y \\ &\frac{d^{2} y}{d x^{2}}+y=0 \end{aligned}
Hence proved

Higher Order Derivatives exercise 11.1 question 7

$\frac{d^{2} y}{d x^{2}}=\frac{2 \log x-3}{x^{3}}$
Hint:
You have to know about derivative of
$\frac{ \log x}{x}$
Given:
$I\! f\: \:y= \frac{ \log x}{x}, show\: \: that$
$\frac{d^{2} y}{d x^{2}}=\frac{2 \log x-3}{x^{3}}$
Solution:
$Let \:y= \frac{ \log x}{x}$
Use quotient rule
\begin{aligned} &\frac{u}{v}=\frac{u^{1} v-v^{1} u}{v^{2}} \\ &u=\log x\: \&\: v=x \\ &\frac{d y}{d x}=\frac{\log x}{x} \\ &\frac{d y}{d x}=\frac{x \frac{d}{d x} \log x-\log x \frac{d}{d x} x}{x^{2}} \quad\left(\frac{d \log x}{d x}=\frac{1}{x}\right) \\ &\frac{d y}{d x}=\frac{x \frac{1}{x}-\log x \cdot 1}{x^{2}} \\ &\frac{d y}{d x}=\frac{1-\log x}{x^{2}} \end{aligned}
Use Quotient rule again
\begin{aligned} &\frac{u}{v}=\frac{u^{1} v-v^{1} u}{v^{2}} \quad \\ &u=1-\log x \: \&\: v=x^{2} \\ &\frac{d^{2} y}{d x^{2}}=\frac{x^{2} \frac{d}{d x}(1-\log x)-\frac{d}{d x} x^{2}(1-\log x)}{\left(x^{2}\right)^{2}} \\ &\frac{d^{2} y}{d x^{2}}=\frac{x^{2}\left(\frac{-1}{x}\right)-(1-\log x) 2 x}{x^{4}} \quad\left(\frac{d-\log x}{d x}=\frac{-1}{x}, \frac{d}{d x} x^{2}=2 x\right) \end{aligned}
\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{-x-2 x+2 x \log x}{x^{4}} \\ &\frac{d^{2} y}{d x^{2}}=\frac{2 x \log x-3 x}{x^{4}} \\ &\frac{d^{2} y}{d x^{2}}=\frac{2 x \log x-3}{x^{3}} \end{aligned}
Hence proved

Higher Order Derivatives exercise 11.1 question 8 maths

$\frac{-b^{4}}{a^{2}y^{3}}$
Hint:
You must know about derivative of
$tan\: \theta \: \: and\: \: sec\: \theta$
Given:
$I\! f\: \: x=a\: sec\: \theta ,y=b\: tan\: \theta$
Prove that
$\frac{d^{2}y}{dx^{2}}=\frac{-b^{4}}{a^{2}y^{3}}$
Solution:
$Let\: \: x=a\: sec\: \theta \\y=b\: tan\: \theta$
\begin{aligned} &\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}} \\ &\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{d y}{d x}\right) \\ &x=a \sec \theta, y=b \tan \theta \end{aligned}
\begin{aligned} &\left.\frac{d x}{d \theta}=a \sec \theta \tan \theta \quad \text { ( } \frac{d \tan \theta}{d x}=\sec ^{2} x, \frac{d \sec \theta}{d x}=\sec \theta \tan \theta\right) \\ &\frac{d y}{d \theta}=b \sec ^{2} \theta \\ &\frac{d y}{d x}=\frac{b}{a} \frac{\sec \theta}{\tan \theta}=\frac{b}{a \sin \theta} \\ &\frac{d}{d \theta}\left(\frac{d y}{d x}\right)=-\frac{b}{a} \operatorname{cosec} \theta \cot \theta \end{aligned}
\begin{aligned} &=\frac{-\frac{b}{a} \operatorname{cosec} \theta \cot \theta}{a \sec \theta \tan \theta} \\ &\frac{d^{2} y}{d x^{2}}=\frac{-b}{a^{2} \tan ^{3} \theta} y=b \tan \theta \\ &\frac{d^{2} y}{d x^{2}}=\frac{-b}{a^{2}\left(\frac{y}{b}\right)^{3} \theta} y=\frac{-b^{4}}{a^{2} y^{3}} \end{aligned}

Higher Order Derivatives exercise 11.1 question 9

$\frac{sec^{3}\theta }{a\theta }$
Hint:
You must know about derivative of
$cos\: \theta \: \: and\: \: sin\: \theta$
Given:
\begin{aligned} &\text { If } x=a(\cos \theta+\theta \sin \theta), y=a(\sin \theta-\theta \cos \theta) \text { Prove that }\\ &\frac{d^{2} x}{d \theta^{2}}=a(\cos \theta-\theta \sin \theta), \frac{d^{2} y}{d \theta^{2}}=a(\sin \theta-\theta \cos \theta), \frac{d^{2} y}{d x^{2}}=\frac{\sec ^{3} \theta}{a \theta} \end{aligned}
Solution:
\begin{aligned} &\text { Let } x=a(\cos \theta+\theta \sin \theta), y=a(\sin \theta-\theta \cos \theta) \end{aligned}
Use multiplicative rule
\begin{aligned} &\text { As } U V=U V^{1}+U^{1} V \\ &\text { Where } U=\theta \& V=\sin \theta \\ &\begin{array}{l} \frac{d x}{d \theta}=a[-\sin \theta+\theta \cos \theta+\sin \theta] \quad\left[\frac{d \sin \theta}{d x}=\cos \theta, \frac{d \cos \theta}{d x}=-\sin \theta\right] \\ \\ \frac{d x}{d \theta}=a \theta \cos \theta \end{array} \end{aligned}
again
Use multiplicative rule
\begin{aligned} &\text { As } U V=U V^{1}+U^{1} V \\ &\text { Where } U=\theta\: \&\: V=\cos \theta \\ &\frac{d y}{d \theta}=a[\cos \theta+\theta \sin \theta-\cos \theta] \\ &\frac{d y}{d \theta}=a \theta \sin \theta \end{aligned}
Again Use multiplicative rule
\begin{aligned} &\text { As } U V=U V^{1}+U^{1} V \\ &\text { Where } U=\theta \& V=\cos \theta \\ &\frac{d^{2} x}{d \theta^{2}}=a[-\theta \sin \theta+\cos \theta] \\ &\frac{d^{2} x}{d \theta^{2}}=a[\cos \theta-\theta \sin \theta] \end{aligned}
Again Use multiplicative rule
\begin{aligned} &\text { As } U V=U V^{1}+U^{1} V \\ &\text { Where } U=\theta \& V=\sin \theta \\ &\frac{d^{2} y}{d \theta^{2}}=a\{\theta \cos \theta+\sin \theta\} \\ &\frac{d^{2} x}{d \theta^{2}}=a(\cos \theta-\theta \sin \theta) \end{aligned}
\begin{aligned} &\frac{d^{2} y}{d \theta^{2}}=a(\theta \cos \theta+\sin \theta) \\ &\frac{d y}{d x}=\frac{a \theta \sin \theta}{a \theta \cos \theta}=\tan \theta \\ &\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d \theta}\left(\frac{d y}{d x}\right) \frac{d \theta}{d x} \end{aligned}
\begin{aligned} &\frac{d}{d \theta}(\tan \theta) \times \frac{1}{a \theta \cos \theta} \\ &\frac{\sec ^{2} \theta}{a \theta \cos \theta} \\ &\frac{d^{2} y}{d x^{2}}=\frac{\sec ^{3} \theta}{a \theta} \end{aligned}

Higher Order Derivatives exercise 11.1 question 10

$2e^{x}cos(x+\frac{\pi }{2})$
Hint:
You must know about derivative of ex cos x
Given:
If y = ex cos x, prove that
$\frac{d^{2}y}{dx^{2}}=2e^{x}cos(x+\frac{\pi }{2})$
Solution:
Let y = ex cos x
Use multiplicative rule
\begin{aligned} &\text { As } U V=U V^{1}+U^{1} V \\ &\text { Where } U=e^{x}\: \&\: V=\cos x \\ &\frac{d y}{d x}=e^{x}(-\sin x)+e^{x} \cos x \quad \quad\left(\frac{d}{d x} \cos x=\sin x\right) \end{aligned}
Again Use multiplicative rule
Differentiating again
\begin{aligned} &\frac{d^{2} y}{d x^{2}}=-\left[e^{x} \cos x+\sin x e^{x}\right]+\left[-e^{x} \sin x+e^{x} \cos x\right] \\ &\frac{d^{2} y}{d x^{2}}=-2 \sin x e^{x} \\ &\frac{d^{2} y}{d x^{2}}=2 e^{x} \cos \left(x+\frac{\pi}{2}\right) \end{aligned}

Higher Order Derivatives exercise 11.1 question 11

$\frac{-b}{a^{2}y^{3}}$
Hint:
You must know about derivative of second order
Given:
\begin{aligned} &{\text { If } x}=a \cos \theta, y=b \sin \theta \\ &\text { Show that } \frac{d^{2} y}{d x^{2}}=\frac{-b}{a^{2} y^{3}} \end{aligned}
Solution:
\begin{aligned} &{\text {Let}\: \: x}=a \cos \theta, y=b \sin \theta \\ \end{aligned}
\begin{aligned} &\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}} \\ &\frac{d^{2} y}{d x^{2}}=\frac{\frac{d}{d \theta}\left(\frac{d y}{d x}\right)}{\frac{d x}{d \theta}} \\ &x=a \cos \theta, y=b \sin \theta \\ &\frac{d x}{d \theta}=-a \sin \theta \frac{d y}{d \theta}=b \cos \theta \\ &\frac{d y}{d x}=\frac{-b}{a} \cot \theta \end{aligned}
\begin{aligned} &\frac{\frac{\mathrm{d} }{\mathrm{d} \theta }(\frac{dy}{dx})}{\frac{dx} {d\theta }}=\frac{\frac{b}{a} \cos e c^{2} \theta}{-a \sin \theta} \quad\left(\frac{d}{d \theta} \cot \theta=\cos e c^{2} \theta\right) \\ &\frac{-b \times b^{3}}{a^{2} \sin ^{3} \theta \times b^{3}} \\ &\frac{d^{2} y}{d x^{2}}=\frac{-b}{a^{2} y^{3}} \end{aligned}
Hence proved

Higher Order Derivatives exercise 11.1 question 12 maths

$\frac{32}{27a}$
Hint:
You must know about derivative of
$\frac{cos^{3}\theta}{sin^{3}\theta }$
Given:
\begin{aligned} &{\text { If } x=a\left(1-\cos ^{3} \theta\right), y=a \sin ^{3} \theta} \\ &\text { Prove that } \frac{d^{2} y}{d x^{2}}=\frac{32}{27 a} \; a t\; \theta=\frac{\pi}{6} \end{aligned}
Solution:
\begin{aligned} &{\text { Let } x=a\left(1-\cos ^{3} \theta\right), y=a \sin ^{3} \theta} \end{aligned}
\begin{aligned} &\frac{d y}{d \theta}=3 a \sin ^{2} \theta \cos \theta\left(\frac{d}{d \theta} \sin ^{3} \theta=3 a \sin ^{2} \theta \cos \theta\right) \\ &\frac{d y}{d \theta}=3 \cos ^{2} \theta \sin \theta \ \left(\frac{d}{d \theta} \cos ^{3} \theta=3 \cos ^{2} \theta \sin \theta\right) \\ &\frac{d y}{d x}=\tan \theta \quad \quad\left(\frac{d}{d \theta} \tan \theta=\sec ^{2} \theta\right) \end{aligned}
\begin{aligned} &\frac{\frac{d}{d \theta}\left(\frac{d y}{d x}\right)}{\frac{d x}{d \theta}}=\frac{\sec ^{2} \theta}{3 a \cos ^{2} \theta \sin \theta}\\ &\frac{d^{2} y}{d x^{2}}=\frac{\sec ^{4} \theta}{3 a \sin \theta} \quad\left(\theta=\frac{\pi}{6}\right)\\ &\frac{d^{2} y}{d x^{2}}=\frac{\sec ^{4}\left(\frac{\pi}{6}\right)}{3 a \sin \left(\frac{\pi}{6}\right)}=\frac{32}{27 a} \end{aligned}
Hence proved

Higher Order Derivatives exercise 11.1 question 13

$\frac{-a}{y^{2}}$
HInt:
You must know about derivative of second order
Given:
\begin{aligned} &{\text { If } x=a(\theta+\sin \theta), y=a(1+\cos \theta)} \\ &\text { Prove that } \frac{d^{2} y}{d x^{2}}=\frac{-a}{y^{2}} \end{aligned}
Solution:
\begin{aligned} &{\text { Let } x=a(\theta+\sin \theta), y=a(1+\cos \theta)} \\ \end{aligned}
\begin{aligned} &\left.\frac{d x}{d \theta}=a[1+\cos \theta] \quad \text { ( } \frac{d}{d \theta} \theta=1, \frac{d}{d \theta} \sin \theta=\cos \theta\right) \\ &\frac{d y}{d \theta}=a(-\sin \theta) \\ &\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{-\sin \theta}{(1+\cos \theta)} \end{aligned}
\begin{aligned} &\sin 2 \theta=2 \sin \theta \cos \theta \\ &\cos 2 \theta=2 \cos ^{2} \theta-1 \\ &\frac{d y}{d x}=\frac{-2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}} \end{aligned}
\begin{aligned} &\frac{d y}{d x}=-\tan \frac{\theta}{2} \\ &\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d \theta}\left(\frac{d y}{d x}\right) \frac{d \theta}{d x} \\ &\frac{d}{d \theta}\left(-\tan \frac{\theta}{2}\right) \cdot \frac{1}{a(1+\cos \theta)} \end{aligned}
\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{-1}{2} \sec ^{2} \frac{\theta}{2} \cdot \frac{1}{a(1+\cos \theta)} \\ &=\frac{-\sec ^{2} \frac{\theta}{2}}{2 a(1+\cos \theta)} \\ &\frac{-a}{y^{2}}=\frac{-a}{a(1+\cos \theta)^{2}}=\frac{-a}{a^{2}(1+\cos \theta)(1+\cos \theta)} \end{aligned}
\begin{aligned} &\frac{-1}{a\left(1+2 \cos ^{2} \frac{\theta}{2}-1\right)(1-\cos \theta)} \\ &=\frac{-1}{2 a \cos ^{2} \frac{\theta}{2}(1+\cos \theta)}=\frac{-\sec ^{2} \frac{\theta}{2}}{2 a(1+\cos \theta)} \end{aligned}
\begin{aligned} &R H L:-\frac{-a}{y^{2}}=\frac{-a}{a(1+\cos \theta)^{2}}=\frac{-1}{a\left(1+2 \cos ^{2} \frac{\theta}{2}-1\right)(1-\cos \theta)} \\ &\frac{d^{2} y}{d x^{2}}=\frac{-\sec ^{2} \frac{\theta}{2}}{2 a(1+\cos \theta)}=\frac{-a}{y^{2}} \\ &\frac{d^{2} y}{d x^{2}}=\frac{-a}{y^{2}} \end{aligned}
Hence proved

Higher Order Derivatives exercise 11.1 question 14

$\frac{-cosec\: x\frac{4\theta }{2}}{4a}$
Hint:
You must know about derivative of
$cos\: \theta \: and\: sin\: \theta$
Given:
\begin{aligned} &\text { If } x=a(\theta-\sin \theta), y=a(1+\cos \theta) \\ &\text { Find } \frac{d^{2} y}{d x^{2}} \end{aligned}
Solution:
\begin{aligned} &\text { Let } x=a(\theta-\sin \theta), y=a(1+\cos \theta) \end{aligned}
\begin{aligned} &\frac{d x}{d \theta}=a[1-\cos \theta] \; \; \; \; \; ......(1)\\ &\frac{d y}{d \theta}=a(-\sin \theta)\; \; \; \; \; ......(2) \\ &\text { Now } \frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{\sin \theta}{(1-\cos \theta)} \end{aligned}
Using identity
\begin{aligned} &\sin 2 \theta=2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \\ &2 \sin ^{2} \frac{\theta}{2}=1-\cos \theta \\ &\cos 2 \theta=1-2 \sin ^{2} \theta \\ &2 \sin ^{2} \theta=1-\cos 2 \theta \end{aligned}
\begin{aligned} &\frac{d y}{d x}=\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin ^{2} \frac{\theta}{2}} \\ &\frac{d y}{d x}=\cot \frac{\theta}{2} \end{aligned}
Now diff on both sides
\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d \theta}\left(\frac{d y}{d x}\right) \frac{d \theta}{d x} \\ &\frac{d^{2} y}{d x^{2}}=-\operatorname{cosec}^{2} \theta \cdot \frac{1}{2} \bullet \frac{1}{a(1-\cos \theta)} \\ &\frac{d^{2} y}{d x^{2}}=\operatorname{cosec}^{2} \frac{\theta}{2} \bullet \frac{1}{2} \bullet \frac{1}{a 2 \sin ^{2} \frac{\theta}{2}} \end{aligned}
\begin{aligned} &=\frac{-\cos e c^{2} \frac{\theta}{2} \times \cos e c^{2} \frac{\theta}{2}}{4 a} \\&\frac{d^{2} y}{d x^{2}}=\frac{\cos e c^{4} \frac{\theta}{2}}{4 a} \end{aligned}
Hence proved

Higher Order Derivatives exercise 11.1 question 15

$\frac{-1}{a}$
Hint:
\begin{aligned} &{{\text { }}\text { You must know about derivative of } \cos \theta\: \&\: \sin \theta}\\ \end{aligned}
Given:
\begin{aligned} &\text { If } x=a(1-\cos \theta), y=a(\theta+\sin \theta)\\ &\text { Prove that } \frac{d^{2} y}{d x^{2}}=\frac{-1}{a} \: and\: \theta=\frac{\pi}{4}\\ \end{aligned}
Solution:
\begin{aligned} &{{\text { Let }} x=a(1-\cos \theta), y=a(\theta+\sin \theta)} \end{aligned}
\begin{aligned} &\frac{d x}{d \theta}=a \sin \theta \\ &\frac{d y}{d \theta}=a(1+\cos \theta) \\ &\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{1+\cos \theta}{\sin \theta} \end{aligned}
Use Quotient rule
\begin{aligned} &\text { As } \frac{u}{v}=\frac{u^{1} v-v^{1} u}{v^{2}} \\ &\text { Where } u=1+\cos \theta \: \&\: v=\sin \theta \\ &\frac{d}{d \theta}\left(\frac{d y}{d x}\right)=\frac{\sin \theta(-\sin \theta)-(1+\cos \theta) \cos \theta}{\sin ^{2} \theta} \end{aligned}
\begin{aligned} &\frac{d}{d \theta}\left(\frac{d y}{d x}\right)=\frac{-1+\cos \theta}{(1-\cos \theta)(1+\cos \theta)} \\ &\frac{d^{2} y}{d x^{2}}=\frac{\frac{-1}{1-\cos \theta}}{a \sin ^{2} \theta} \\ &\frac{d^{2} y}{d x^{2}}=\frac{-1}{a \sin \theta(1-\cos \theta)} \quad\left(\theta=\frac{\pi}{2}\right) \end{aligned}
\begin{aligned} \frac{d^{2} y}{d x^{2}} &=\frac{-1}{a \sin \frac{\pi}{2}\left(1-\cos \frac{\pi}{2}\right)} \\ \frac{d^{2} y}{d x^{2}} &=\frac{-1}{a} \end{aligned}
Hence proved

Higher Order Derivatives exercise 11.1 question 16 maths

$\frac{-1}{a}$
Hint:
\begin{aligned} &{{\text { }}\text { You must know about derivative of } \cos \theta\: \&\: \sin \theta}\\ \end{aligned}
Given:
\begin{aligned} &\text { If } x=a(1+\cos \theta), y=a(\theta+\sin \theta)\\ &\text { Prove that } \frac{d^{2} y}{d x^{2}}=\frac{-1}{a}\: and\: \theta=\frac{\pi}{4}\\ \end{aligned}
Solution:
\begin{aligned} &{{\text { Let }} x=a(1+\cos \theta), y=a(\theta+\sin \theta)} \end{aligned}
\begin{aligned} &\frac{d x}{d \theta}=-a \sin \theta \\ &\frac{d y}{d \theta}=a(1+\cos \theta) \\ &\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{1+\cos \theta}{-\sin \theta} \end{aligned}
Use Quotient rule
\begin{aligned} &\text { As } \frac{u}{v}=\frac{u^{1} v-v^{1} u}{v^{2}} \\ &\text { Where } u=1+\cos \theta \& v=\sin \theta \\ &\frac{d}{d \theta}\left(\frac{d y}{d x}\right)=\frac{-\sin \theta(-\sin \theta)+(1+\cos \theta) \cos \theta}{\sin ^{2} \theta} \end{aligned}
\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{1+\cos \theta}{\sin ^{2} \theta} \cdot \frac{-1}{a \sin \theta} \\ &\frac{d^{2} y}{d x^{2}}=\frac{-(1+\cos \theta)}{a \sin ^{2} \theta} \\ &\left.\frac{d^{2} y}{d x^{2}}=\frac{-\left(1+\cos \frac{\pi}{2}\right)}{a \sin ^{2} \frac{\pi}{2}} \quad (\theta =\frac{\pi}{2}\right) \\ &\frac{d^{2} y}{d x^{2}}=\frac{-1}{a} \end{aligned}
Hence proved

Higher Order Derivatives exercise 11.1 question 17

\begin{aligned} &\text { } 3 \sin ^{2} \theta\left[5 \cos ^{2} \theta-1\right]\\ \end{aligned}
Hint:
\begin{aligned} &\text { You must know about derivative of } \cos \theta\: \&\: \sin \theta\\ \end{aligned}
Given:
\begin{aligned} &\text { If } x=\cos \theta, y=\sin ^{3} \theta\\ &\text { Prove that } y \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=3 \sin ^{2} \theta\left[5 \cos ^{2} \theta-1\right] \end{aligned}
Solution:
\begin{aligned} &\text { Let } x=\cos \theta, y=\sin ^{3} \theta\\ \end{aligned}
\begin{aligned} &\frac{d x}{d \theta}=-\sin \theta \\ &\frac{d y}{d \theta}=3 \sin ^{2} \theta \cos \theta \\ &\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{3 \sin ^{2} \theta \cos \theta}{-\sin \theta}=3 \sin \theta \cos \theta \end{aligned}
\begin{aligned} &\frac{d^{2} y}{d x^{2}}=d\left(\frac{\frac{d y}{d x}}{d x}\right)=\frac{d\left(\frac{d y}{d \theta}\right) d \theta}{\frac{d x}{d \theta}} \\ &=\frac{-\cos ^{2} \theta \cdot 3+3 \sin ^{2} \theta}{-\sin \theta} \end{aligned}
\begin{aligned} &y \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=\frac{\sin ^{3} \theta\left(3 \cos ^{2} \theta-3 \sin ^{2} \theta\right)}{\sin \theta} \\ &y \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=3 \sin ^{2} \theta\left[5 \cos ^{2} \theta-1\right] \end{aligned}
Hence proved

Higher Order Derivatives exercise 11.1 question 18

0
Hint:
You must know about derivative of
$cos\: \theta\: and\: tan\: \theta$
Given:
$I\! f\: y=sin(sin\: x)$
\begin{aligned} &Prove\: \: that\: \: \frac{d^{2} y}{d x^{2}}+\tan x \frac{d y}{d x}+y \cos ^{2} x =0\\ \end{aligned}
Solution:
$Let\: y=sin(sin\: x)$
\begin{aligned} &\frac{d y}{d x}=\cos (\sin x) \cdot \cos x \\ &\frac{d^{2} y}{d x^{2}}=-\cos (\sin x)(\sin x)+\cos x(-\sin (\sin x) \cos x) \\ &\frac{d^{2} y}{d x^{2}}=-\sin x \cos (\sin x)-\cos ^{2} x(\sin (\sin x)) \end{aligned}
\begin{aligned} &L H S:-\frac{d^{2} y}{d x^{2}}+\tan x \frac{d y}{d x}+y \cos ^{2} x \\ &-\sin x \cos (\sin x)-\cos ^{2} x(\sin (\sin x))+\tan x \cos x \cos (\sin x)+\sin (\sin x) \cos ^{2} x \\ &-\sin x(\cos (\sin x))+\frac{\sin x}{\cos x} \cdot \cos x \cdot \cos (\sin x) \\ &-\sin x \cos x \sin x+\sin x \cos x \sin x \end{aligned}
Hence proved

Higher Order Derivatives exercise 11.1 question 19

0
HInt:
You must know about derivative of sin pt
Given:
$I\! f\; x=sin\; t,\; y=sin\; pt$
$\text { Prove that }\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}+p^{2} y=0$
Solution:
$Let\; x=sin\; t,\; y=sin\; pt$
\begin{aligned} &\frac{d x}{d t}=\cos t \text { and } \frac{d y}{d t}=p \cos p t \\ &\text { Now } \frac{d y}{d x}=\frac{p \cos p t}{\cos t} \\ &\frac{d y}{d x} \cos t=p \cos p t \end{aligned}
\begin{aligned} &\left(\frac{d y}{d x}\right)^{2}\left(1-\sin ^{2} t\right)=p^{2}\left(1-\sin ^{2} p t\right) \\ &\left(\frac{d y}{d x}\right)^{2}\left(1-x^{2}\right)=p^{2}\left(1-y^{2}\right) \end{aligned}
Differentiating with x
\begin{aligned} &\left(1-x^{2}\right) 2 \frac{d y}{d x} \times \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}(-2 x) \\ &=-p^{2} \times 2 y \frac{d y}{d x} \\ &2 \frac{d y}{d x}\left[\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}\right]=p^{2} y 2 \frac{d y}{d x} \\ &\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}+p^{2} y=0 \end{aligned}
Hence proved

Higher Order Derivatives exercise 11.1 question 20 maths

0
Hint:
You must know about derivative of sin-1 x
Given:
$I\! f y=(sin^{-1}x)^{2} \; \; Prove\: that\: \: (1-x^{2})y_{2}-xy_{1}-2=0$
Solution:
$Let\; \; y=(sin^{-1}x)^{2}$
\begin{aligned} &\frac{d y}{d x}=2 \sin ^{-1} x\left(\frac{1}{\sqrt{1-x^{2}}}\right) \quad\left(\frac{d}{d x} \sin ^{-1} x=\left(\frac{1}{\sqrt{1-x^{2}}}\right)\right) \\ &\frac{d^{2} y}{d x^{2}}=\frac{2 \sqrt{1-x^{2}}\left(\frac{1}{\sqrt{1-x^{2}}}\right)-\sin ^{-1} x \times \frac{1}{2}\left(\frac{-2 x}{\sqrt{1-x^{2}}}\right)}{1-x^{2}} \end{aligned}
\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{2\left(1+\frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}}\right)}{1-x^{2}} \\ &\left(1-x^{2}\right) y_{2}=\frac{2+2 x \sin ^{-1} x}{\sqrt{1-x^{2}}} \\ &\left(1-x^{2}\right) y_{2}-y_{1} x-2=0 \end{aligned}
Hence proved

Higher Order Derivatives exercise 11.1 question 21

Proved
Hint:
You must know about the derivative of exponential function and tangent inverse $x$
Given:
$y=e^{tan^{-1}x},\; \; Prove\: (1+x^{2})y_{2}+(2x-1)y_{1}=0$
Solution:
$Let\; \; y=e^{tan^{-1}x}$
\begin{aligned} &\frac{d y}{d x}=e^{\tan ^{-1} x} \times \frac{1}{\left(1+x^{2}\right)} \\ &\left(x^{2}+1\right) \frac{d y}{d x}=e^{\operatorname{tan}^{-1} x} \\ &\left(x^{2}+1\right) \frac{d^{2} y}{d x^{2}}+2 x \frac{d y}{d x}=e^{\tan ^{-1} x} \times \frac{1}{\left(1+x^{2}\right)} \end{aligned}
\begin{aligned} &\left(x^{2}+1\right) \frac{d^{2} y}{d x^{2}}+2 x \frac{d y}{d x}=\frac{d y}{d x}\\ &\left(x^{2}+1\right) \frac{d^{2} y}{d x^{2}}+(2 x-1) \frac{d y}{d x}=0\\ &\text { or }\\ &\left(x^{2}+1\right) y_{2}+(2 x-1) y_{1}=0 \end{aligned}

Higher Order Derivatives exercise 11.1 question 22

Proved
Hint:
You must know about the derivative of logarithm function and cos x and sin x
Given:
$y=3cos(log\: x)+4sin(log\: x)$
Solution:
$Let\: \: y=3cos(log\: x)+4sin(log\: x)$
\begin{aligned} &\text { Differentiating both sides w.r.t } x\\ &\frac{d y}{d x}=-3 \sin (\log x) \frac{d(\log x)}{d x}+4 \cos (\log x) \frac{d(\log x)}{d x} \end{aligned}
\begin{aligned} &\text { By using product rule of derivation }\\ &\frac{d y}{d x}=\frac{-3 \sin (\log x)}{x}+\frac{4 \cos (\log x)}{x}\\ &x \frac{d y}{d x}=-3 \sin (\log x)+4 \cos (\log x) \end{aligned}
\begin{aligned} &\text { Again differentiating both sides w.r.t } x \text { , }\\ &x \frac{d}{d x}\left(\frac{d y}{d x}\right)+\frac{d y}{d x} \frac{d}{d x}(x)=\frac{d}{d x}[-3 \sin (\log x)+4 \cos (\log x)] \end{aligned}
\begin{aligned} &\text { By using product rule of derivation, }\\ &x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}(1)=-3 \cos (\log x) \frac{d}{d x}(\log x)-4 \sin (\log x) \frac{d}{d x}(\log x)\\ &x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}(1)=\frac{-3 \cos (\log x)}{x}-\frac{4 \sin (\log x)}{x} \end{aligned}
\begin{aligned} &x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=\frac{-[3 \cos (\log x)+4 \sin (\log x)]}{x} \\ &x^{2} \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}=-[3 \cos (\log x)+4 \sin (\log x)] \\ &x^{2} \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}=-y \end{aligned}
\begin{aligned} &\therefore x^{2} \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}+y=0\\ &\text { or }\\ &x^{2} y_{2}+x y_{1}+y=0 \end{aligned}

Higher Order Derivatives exercise 11.1 question 23

Proved
Hint:
You must know about the derivative of exponential function
Given:
$y=e^{2x}(ax+b), \: \: show\: \: that\: \: y_{2}-4y_{1}+4y=0$
Solution:
$y=e^{2x}(ax+b)\; \; \; \; \; \; \; \; \; ......(1)$
\begin{aligned} &\text { By using product rule of derivation }\\ &\frac{d y}{d x}=e^{2 x} \frac{d y}{d x}(a x+b)+(a x+b) \frac{d}{d x} e^{2 x} \\ &\frac{d y}{d x}=a e^{2 x}+2(a x+b) e^{2 x}\\ &\frac{d y}{d x}=e^{2 x}(a+2 a x+2 b) \end{aligned}$.....(2)$
\begin{aligned} &\text { Again differentiating both sides w.r.t } x, \text { using product rule }\\ &\frac{d^{2} y}{d x^{2}}=e^{2 x} \frac{d y}{d x}(a+2 a x+2 b)+(a+2 a x+2 b) \frac{d}{d x} e^{2 x}\\ &\frac{d^{2} y}{d x^{2}}=2 e^{2 x}+2(a+2 a x+2 b) e^{2 x} \quad \ldots .(3) \end{aligned}
$In\; order\; to\; prove\; the\; expression\; try\; to\; get\; the\; required\; f\! orm \\Subtracting\; 4 \times equation\; (2)\; f\! rom\; equation (3)\\ \frac{d^{2} y}{d x^{2}}-4 \frac{d y}{d x}=2 e^{2 x}+2(a+2 a x+2 b) e^{2 x}-4 e^{2 x}(a+2 a x+2 b)\\ \frac{d^{2} y}{d x^{2}}-4 \frac{d y}{d x}=2 a e^{2 x}-2 e^{2 x}(a+2 a x+2 b)\\ \frac{d^{2} y}{d x^{2}}-4 \frac{d y}{d x}=-4 e^{2 x}(a x+b)$
\begin{aligned} &\text { Using equation }(1)\\ &\frac{d^{2} y}{d x^{2}}-4 \frac{d y}{d x}=-4 y\\ &\frac{d^{2} y}{d x^{2}}-4 \frac{d y}{d x}+4 y=0\\ &y_{2}-4 y_{1}+4 y=0 \end{aligned}

Higher Order Derivatives exercise 11.1 question 24 maths

Proved
Hint:
You must know about the derivative of sin function and logarithm function
Given:
$x=\sin \left(\frac{1}{a} \log y\right) \text { show that }\left(1-x^{2}\right) y_{2}-x y_{1}-a^{2} y=0$
Solution:
\begin{aligned} &x=\sin \left(\frac{1}{a} \log y\right)\\ &\log y=a \sin ^{-1} x\\ &y=e^{a \sin ^{-1} x}\; \; \; \; \; \; \; .....(1) \end{aligned}
\begin{aligned} &\text { Let } t=a \sin ^{-1} x\\ &\frac{d t}{d x}=\frac{a}{\sqrt{1-x^{2}}} \quad\left[\frac{d}{d x} \sin ^{-1} x=\frac{1}{\sqrt{1-x^{2}}}\right]\\ &\text { and } y=e^{t}\\ &\frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}\\ &\frac{d y}{d x}=e^{t} \frac{a}{\sqrt{1-x^{2}}}=\frac{a e^{a \sin ^{-1} x}}{\sqrt{1-x^{2}}}\; \; \; \; \; \; \; \; .......(2) \end{aligned}
\begin{aligned} &\text { Again differentiating both sides }\\ &\frac{d^{2} y}{d x^{2}}=a e^{a \sin ^{-1} x} \frac{d}{d x}\left(\frac{1}{\sqrt{1-x^{2}}}\right)+\frac{a}{\sqrt{1-x^{2}}} \frac{d}{d x} e^{a \sin ^{-1} x} \end{aligned}
\begin{aligned} &\text { Using chain rule and equation }\\ &\frac{d^{2} y}{d x^{2}}=\frac{-a e^{a \sin ^{-1} x}}{2\left(1-x^{2}\right) \sqrt{1-x^{2}}}(-2 x)+\frac{a^{2} e^{a \sin ^{-1} x}}{\left(1-x^{2}\right)}\\ &\frac{d^{2} y}{d x^{2}}=\frac{x a e^{a \sin ^{-1} x}}{\left(1-x^{2}\right) \sqrt{1-x^{2}}}+\frac{a^{2} e^{a \sin ^{-1} x}}{\left(1-x^{2}\right)}\\ &\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}=a^{2} e^{a \sin ^{-1} x}+\frac{x a e^{a \sin ^{-1} x}}{\sqrt{1-x^{2}}} \end{aligned}
\begin{aligned} &\text { Using equation }(1) \text { and }(2)\\ &\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}=a^{2} y+x \frac{d y}{d x}\\ &\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-a^{2} y-x \frac{d y}{d x}=0\\ &\left(1-x^{2}\right) y_{2}-x y_{1}-a^{2} y=0 \end{aligned}

Higher Order Derivatives exercise 11.1 question 25

Proved
HInt:
You must know the derivative of logarithm and tangent inverse $x$
Given:
$\log y=\tan ^{-1} x, \text { show }\left(1+x^{2}\right) y_{2}+(2 x-1) y_{1}=0$
Solution:
\begin{aligned} &\log y=\tan ^{-1} x\\ &\text { Differentiate the equation w.r.t } x\\ &\frac{1}{y} \frac{d y}{d x}=\frac{1}{1+x^{2}}\\ &1+x^{2} \frac{d y}{d x}=y \end{aligned}
$Di\! f\! ferentiate\; again\\ \left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}(2 x)=\frac{d y}{d x}\\ \left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}+(2 x-1) \frac{d y}{d x}=0\\ or\\ \left(1+x^{2}\right) y_{2}+(2 x-1) y_{1}=0$

Higher Order Derivatives exercise 11.1 question 26

Proved
Hint:
You must know the derivative of logarithm and tangent inverse $x$
Given:
$y=\tan ^{-1} x, \text { show }\left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}+(2 x) \frac{d y}{d x}=\frac{d y}{d x}$
Solution:
$y=\tan ^{-1} x$
$Di\! f\! f\! erentiate\; the\; equation\; w.r.t \; x \\ \frac{d y}{d x}=\frac{1}{1+x^{2}}\\ \left(1+x^{2}\right) \frac{d y}{d x}=1 \\ Di\! f\! f\! erentiate\; again \\ \left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}+(2 x) \frac{d y}{d x}=0 \\ or \\ \left(1+x^{2}\right) y_{2}+(2 x) y_{1}=0$

Higher Order Derivatives exercise 11.1 question 27

Proved
Hint:
You must know the derivative of logarithm and tangent inverse $x$
Given:
$y=\left\{\log \left(x+\sqrt{x^{2}+1}\right)\right\}^{2}, \text { show }\left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}=2$
Solution:
$y=\left\{\log \left(x+\sqrt{x^{2}+1}\right)\right\}^{2}$
\begin{aligned} &\text { Differentiate the equation w.r.t } x\\ &\frac{d y}{d x}=2 \log \left(x+\sqrt{x^{2}+1}\right) \cdot \frac{1}{x+\sqrt{x^{2}+1}} \times\left(1+\frac{2 x}{2 \sqrt{x^{2}+1}}\right)\\ &=\frac{2 \log \left(x+\sqrt{x^{2}+1}\right)}{x+\sqrt{x^{2}+1}} \times \frac{x+\sqrt{x^{2}+1}}{\sqrt{x^{2}+1}}\\ &=\sqrt{x^{2}+1} \cdot \frac{d y}{d x}=2 \log \left(x+\sqrt{x^{2}+1}\right) \end{aligned}
\begin{aligned} &\text { Differentiate again }\\ &\frac{d^{2} y}{d x^{2}} \sqrt{1+x^{2}}+\frac{2 x}{2 \sqrt{1+x^{2}}} \frac{d y}{d x}=\frac{2}{x+\sqrt{x^{2}+1}} \times\left(1+\frac{2 x}{2 \sqrt{x^{2}+1}}\right)\\ &\frac{\left(x^{2}+1\right) \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}}{\sqrt{x^{2}+1}}=\frac{2}{x+\sqrt{x^{2}+1}} \times \frac{x+\sqrt{x^{2}+1}}{\sqrt{x^{2}+1}}\\ &\left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}=2 \end{aligned}

Higher Order Derivatives exercise 11.1 question 28 maths

Proved
Hint:
You must know the derivative of logarithm and tangent inverse $x$
Given:
$y=\left(\tan ^{-1} x\right)^{2}, \text { show }\left(1+x^{2}\right)^{2} y_{2}+2 x\left(1+x^{2}\right) y_{1}=2$
Solution:
\begin{aligned} &y=\left(\tan ^{-1} x\right)^{2}\\ &\text { Differentiate the equation w.r.t } x\\ &\frac{d y}{d x}=2 \tan ^{-1} x \cdot \frac{1}{1+x^{2}}\\ &\left(1+x^{2}\right) \frac{d y}{d x}=2 \tan ^{-1} x \end{aligned}
\begin{aligned} &\text { Differentiate again }\\ &\left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}+(2 x) \frac{d y}{d x}=\frac{2}{1+x^{2}}\\ &\left(1+x^{2}\right)^{2} \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}\left(1+x^{2}\right)(2 x)=2\\ or\\ &\left(1+x^{2}\right)^{2} y_{2}+2 x\left(1+x^{2}\right) y_{1}=2 \end{aligned}

Higher Order Derivatives exercise 11.1 question 29

Proved
Hint:
You must know the derivative of cot x function
Given:
$y=\cot x, \text { show } \frac{d^{2} y}{d x^{2}}+(2 y) \frac{d y}{d x}=0$
Solution:
$Let\; \; y=\cot x$
\begin{aligned} &\text { Differentiate the equation w.r.t } x\\ &\frac{d y}{d x}=\frac{d(\cot x)}{d x}\\ &\frac{d y}{d x}=-\operatorname{cosec}^{2} x \end{aligned}
\begin{aligned} &\text { Differentiate again }\\ &\frac{d^{2} y}{d x^{2}}=-[2 cosec\: x(-cosec\: x \cot x)]\\ &\frac{d^{2} y}{d x^{2}}=-2 \operatorname{cosec}^{2} x \cot x \end{aligned}
\begin{aligned} &\frac{d^{2} y}{d x^{2}}=-2 \frac{d y}{d x} y \\ &\frac{d^{2} y}{d x^{2}}+(2 y) \frac{d y}{d x}=0 \end{aligned}

Higher Order Derivatives exercise 11.1 question 30

$\frac{-2}{x^{2}}$
Hint:
You must know the derivative of logarithm function
Given:
$y=\log \left(\frac{x^{2}}{e^{2}}\right), \text { find } \frac{d^{2} y}{d x^{2}}$
Solution:
$Let\: \: y=\log \left(\frac{x^{2}}{e^{2}}\right)$
$Di\! f\! f\! erentiate\; the\; equation\; w.r.t\; x \\ \frac{d y}{d x}=\frac{1}{\frac{x^{2}}{e^{2}}} \cdot \frac{1}{e^{2}} \cdot 2 x=\frac{2}{x}\\ Di\! f\! f\! erentiate\; again\\\\ \frac{d^{2} y}{d x^{2}}=-2\left[\frac{1}{x^{2}}\right]=\frac{-2}{x^{2}}\\ Hence, \frac{d^{2} y}{d x^{2}}=\frac{-2}{x^{2}}$

Higher Order Derivatives exercise 11.1 question 31

Proved
Hint:
You must know the derivative of exponential function
Given:
$y=a e^{2 x}+b e^{-x}, \text { show } \frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}-2 y=0$
Solution:
$y=a e^{2 x}+b e^{-x}$
\begin{aligned} &\text { Differentiate the equation w.r.t } x\\ &\frac{d y}{d x}=2 a e^{2 x}+b e^{-x}\\ &\frac{d^{2} y}{d x^{2}}=4 a e^{2 x}+b e^{-x} \end{aligned}
\begin{aligned} &\text { LHS }=\frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}-2 y \\ &=4 a e^{2 x}+b e^{-x}-2 a e^{2 x}+b e^{-x} \\ &=-2 a e^{2 x}-2 b e^{2 x} \\ &=0=\text { RHS } \end{aligned}

Higher Order Derivatives exercise 11.1 question 31

Proved
Hint:
You must know the derivative of exponential function
Given:
$y=a e^{2 x}+b e^{-x}, \text { show } \frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}-2 y=0$
Solution:
$y=a e^{2 x}+b e^{-x}$
\begin{aligned} &\text { Differentiate the equation w.r.t } x\\ &\frac{d y}{d x}=2 a e^{2 x}+b e^{-x}\\ &\frac{d^{2} y}{d x^{2}}=4 a e^{2 x}+b e^{-x} \end{aligned}
\begin{aligned} &\text { LHS }=\frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}-2 y \\ &=4 a e^{2 x}+b e^{-x}-2 a e^{2 x}+b e^{-x} \\ &=-2 a e^{2 x}-2 b e^{2 x} \\ &=0=\text { RHS } \end{aligned}

Higher Order Derivatives exercise 11.1 question 32 maths

Proved
Hint:
You must know the derivative of exponential sin and cos functions
Given:
$y=e^{x}(\sin x+\cos x), \text { show } \frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0$
Solution:
$y=e^{x}(\sin x+\cos x)$
\begin{aligned} &\text { Differentiate the equation w.r.t } x\\ &\frac{d y}{d x}=e^{x}(\cos x-\sin x)+e^{x}(\sin x+\cos x)\\ &\frac{d y}{d x}=e^{x}(\cos x-\sin x)+y \end{aligned}
\begin{aligned} &\text { Differentiate again }\\ &\frac{d^{2} y}{d x^{2}}=e^{x}(-\sin x-\cos x)+e^{x}(\cos x-\sin x)+\frac{d y}{d x}\\ &\frac{d^{2} y}{d x^{2}}=-y+\frac{d y}{d x}-y+\frac{d y}{d x}\\ &\therefore \frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0 \end{aligned}

Higher Order Derivatives exercise 11.1 question 33

$-cot\: y.cosec^{2}y$
Hint:
You must know the derivative of cos inverse function
Given:
$y=\cos ^{-1} x, \text { find } \frac{d^{2} y}{d x^{2}} \text { in terms of } y \text { alone }$
Solution:
$y=\cos ^{-1} x$
\begin{aligned} &\text { Differentiate the equation w.r.t } x\\ &\frac{d y}{d x}=\frac{d\left(\cos ^{-1} x\right)}{d x}\\ &=\frac{-1}{\sqrt{1-x^{2}}}=-\left(1-x^{2}\right)^{\frac{-1}{2}}\\ &\frac{d^{2} y}{d x^{2}}=\frac{d\left[-\left(1-x^{2}\right)^{\frac{-1}{2}}\right]}{d x} \end{aligned}
\begin{aligned} &=-\left(\frac{-1}{2}\right)\left(1-x^{2}\right)^{\frac{-3}{2}} \times(-2 x) \\ &=\frac{1}{2 \sqrt{\left(1-x^{2}\right)^{3}}} \times(-2 x) \\ &\frac{d^{2} y}{d x^{2}}=\frac{-x}{\sqrt{\left(1-x^{2}\right)^{3}}} \end{aligned}
\begin{aligned} &y=\cos ^{-1} x\\ &x=\cos y\\ &\text { Put in above equation }\\ &\frac{d^{2} y}{d x^{2}}=\frac{-\cos y}{\sqrt{\left(1-\cos ^{2} y\right)^{3}}} \end{aligned}
\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{-\cos y}{\sqrt{\sin ^{3} y}} \\ &=\frac{-\cos y}{\sin ^{3} y} \\ &=\frac{-\cos y}{\sin y} \times \frac{1}{\sin ^{2} y} \\ &\therefore \frac{d^{2} y}{d x^{2}}=-\cot y \cdot \operatorname{cosec}^{2} y \end{aligned}

Higher Order Derivatives exercise 11.1 question 34

Proved
Hint:
You must know the derivative of exponential and cos inverse function
Given:
$y=e^{a \cos ^{-1} x} , \text { prove }\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-a^{2} y=0$
Solution:
$y=e^{a \cos ^{-1} x}$
\begin{aligned} &\text { Taking logarithm both sides }\\ &\log y=a \cos ^{-1} x \log c\\ &\log y=a \cos ^{-1} x\\ &\frac{1}{y} \frac{d y}{d x}=a \times\left(\frac{-1}{\sqrt{1-x^{2}}}\right)\\ &\frac{d y}{d x}=\frac{a y}{\sqrt{1-x^{2}}} \end{aligned}
\begin{aligned} &\text { Squaring both sides, }\\ &\left(\frac{d y}{d x}\right)^{2}=\frac{a^{2} y^{2}}{\left(1-x^{2}\right)}\\ &\left(1-x^{2}\right)\left(\frac{d y}{d x}\right)^{2}=a^{2} y^{2} \end{aligned}
\begin{aligned} &\text { Again differentiate, }\\ &\left(\frac{d y}{d x}\right)^{2}(-2 x)+\left(1-x^{2}\right) \times 2 \frac{d y}{d x} \frac{d^{2} y}{d x^{2}}=a^{2} \cdot 2 y \cdot \frac{d y}{d x}\\ &-x \frac{d y}{d x}+\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}=a^{2} y\\ &\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-a^{2} y=0 \end{aligned}

Higher Order Derivatives exercise 11.1 question 35

Proved
HInt:
You must know the derivative of exponential function
Given:
$y=500 e^{7 x}+600 e^{-7 x}, \text { show } \frac{d^{2} y}{d x^{2}}=49 y$
Solution:
$y=500 e^{7 x}+600 e^{-7 x}$
\begin{aligned} &\frac{d y}{d x}=(7)(500) e^{7 x}+(-7)(600) e^{-7 x} \\ &\frac{d^{2} y}{d x^{2}}=(7)(7)(500) e^{7 x}+(-7)(-7)(600) e^{-7 x} \\ &\frac{d^{2} y}{d x^{2}}=(49)(500) e^{7 x}+(49)(600) e^{-7 x} \\ &\frac{d^{2} y}{d x^{2}}=(49)\left[(500) e^{7 x}+(600) e^{-7 x}\right] \\ &\frac{d^{2} y}{d x^{2}}=49 y \end{aligned}

Higher Order Derivatives exercise 11.1 question 36 maths

$\frac{-3}{2}$
Hint:
You must know the derivative of cos and sin function
\begin{aligned} &x=2 \cos t-\cos 2 t \\ \end{aligned}
Given:
\begin{aligned} &y=2 \sin t-\sin 2 t , \text { find } \frac{d^{2} y}{d x^{2}} \text { at } t=\frac{\pi}{2} \end{aligned}
Solution:
\begin{aligned} &x=2 \cos t-\cos 2 t \\ &y=2 \sin t-\sin 2 t \\ &\frac{d x}{d t}=-2 \sin t+2 \sin 2 t \\ &\frac{d y}{d t}=2 \cos t-2 \cos 2 t \end{aligned}
\begin{aligned} &\text { Now, }\\ &\frac{d y}{d x}=\frac{2 \cos t-2 \cos 2 t}{-2 \sin t+2 \sin 2 t} \Rightarrow \frac{\cos t-\cos 2 t}{\sin 2 t-\sin t}\\ &=\frac{2 \sin \frac{3 t}{2} \sin \frac{t}{2}}{2 \cos \frac{3 t}{2} \sin \frac{t}{2}} \Rightarrow \tan \frac{3 t}{2} \end{aligned}
\begin{aligned} &\text { Therefore, }\\ &\frac{d^{2} y}{d x^{2}}=\sec ^{2} \frac{3 t}{2} \times \frac{3}{2} \times \frac{d t}{d x}\\ &=3 \sec ^{2} \frac{3 t}{2} \cdot \frac{1}{2 \sin 2 t-2 \sin t}\\ &\left.\frac{d^{2} y}{d x^{2}}\right]_{t=\frac{\pi}{2}}=\frac{-3}{2} \end{aligned}

Higher Order Derivatives exercise 11.1 question 37

$\frac{-7}{64z^{3}}$
Hint:
You must know the derivative of $x$ and $y$
Given:
\begin{aligned} &x=4 z^{2}+5 \\ &y=6 z^{2}+7 z+3 , \text { find } \frac{d^{2} y}{d x^{2}} \end{aligned}
Solution:
\begin{aligned} &x=4 z^{2}+5\\ &y=6 z^{2}+7 z+3\\ &\frac{d x}{d z}=8 z+0=8 z\\ &\frac{d y}{d z}=12 z+7\\ &\text { Now, }\\ &\frac{d y}{d x}=\frac{12 z+7}{8 z}\\ &\frac{d y}{d x}=\frac{3}{2}+\frac{7}{8 z} \end{aligned}
$Again\; di\! f\!\! f\! erentiating\; w.r.t\; z \\ \frac{d^{2} y}{d x^{2}} \times \frac{d x}{d z}=-\frac{7}{8 z^{2}}\\ \\ or\\ \\ \frac{d^{2} y}{d x^{2}}=-\frac{7}{8 z^{2}} \times \frac{d z}{d x}\\ \\ =-\frac{7}{8 z^{2}} \times \frac{1}{8 z}\\ \frac{d^{2} y}{d x^{2}}=\frac{-7}{64 z^{3}}$

Higher Order Derivatives exercise 11.1 question 38

Proved
Hint:
You must know the derivative of logarithm and cos function
Given:
$y=\log (1+\cos x) , \text { prove } \frac{d^{3} y}{d x^{3}}+\frac{d^{2} y}{d x^{2}} \times \frac{d y}{d x}=0$
Solution:
$y=\log (1+\cos x)$
\begin{aligned} &\frac{d y}{d x}=\frac{-\sin x}{1+\cos x}\\ &\frac{d^{2} y}{d x^{2}}=\frac{-\cos x-\cos ^{2} x-\sin ^{2} x}{(1+\cos x)^{2}}\\ &=\frac{-(\cos x+1)}{(1+\cos x)}\\ &=\frac{-1}{1+\cos x}\\ &\text { Again differentiating } \end{aligned}
\begin{aligned} &\frac{d^{3} y}{d x^{3}}=\frac{-\sin x}{(1+\cos x)^{2}} \\ &\frac{d^{3} y}{d x^{3}}+\frac{\sin x}{(1+\cos x)^{2}}=0 \\ &\frac{d^{3} y}{d x^{3}}+\left(\frac{-1}{1+\cos x}\right)\left(\frac{-\sin x}{1+\cos x}\right)=0 \\ &\frac{d^{3} y}{d x^{3}}+\frac{d^{2} y}{d x^{2}} \times \frac{d y}{d x}=0 \end{aligned}

Higher Order Derivatives exercise 11.1 question 39

Proved
Hint:
You must know the derivative of sin and logarithm function
Given:
$y=\sin (\log x) , \text { prove } x^{2} \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}+y=0$
Solution:
$y=\sin (\log x)$
\begin{aligned} &\frac{d y}{d x}=\cos (\log x) \times \frac{1}{x}\\ &=\frac{\cos (\log x)}{x}\\ &\text { Again differentiating }\\ &\frac{d^{2} y}{d x^{2}}=\frac{x\left[-\sin (\log x) \times \frac{1}{x}\right]-\cos (\log x)}{x^{2}}\\ &=\frac{-\cos (\log x)-\sin (\log x)}{x^{2}} \end{aligned}
\begin{aligned} &\text { Now, }\\ &\text { LHS }=x^{2} \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}+y\\ &=\frac{x^{2}\{-\cos (\log x)-\sin (\log x)\}}{x^{2}}+\frac{x \cos (\log x)}{x}+\sin (\log x)\\ &=0=\mathrm{RHS} \end{aligned}

Higher Order Derivatives exercise 11.1 question 40 maths

Proved
Hint:
You must know the derivative of exponential function
Given:
$y=3 e^{2 x}+2 e^{3 x}, \text { prove } \frac{d^{2} y}{d x^{2}}-5 \frac{d y}{d x}+6 y=0$
Solution:
$y=3 e^{2 x}+2 e^{3 x}$
\begin{aligned} &\frac{d y}{d x}=(2)(3) e^{2 x}+(3)(2) e^{3 x} \\ &=6 e^{2 x}+6 e^{3 x} \\ &\frac{d y}{d x}=6 e^{2 x}+\frac{6\left(y-3 e^{2 x}\right)}{2} \\ &\frac{d y}{d x}=6 e^{2 x}+3 y-9 e^{2 x} \\ &=-3 e^{2 x}+3 y \end{aligned}
\begin{aligned} &\text { Again differentiating }\\ &\frac{d^{2} y}{d x^{2}}=\frac{3 d y}{d x}-6 e^{2 x} \quad \ldots \ldots \ldots .(1)\\ &\frac{d y}{d x}-3 y=-3 e^{2 x}\\ &\frac{\frac{d y}{d x}-3 y}{-3}=e^{2 x} \end{aligned}
\begin{aligned} &\text { Put in (1), }\\ &\frac{d^{2} y}{d x^{2}}=\frac{3 d y}{d x}-6\left(\frac{\frac{d y}{d x}-3 y}{-3}\right)\\ &\frac{d^{2} y}{d x^{2}}=\frac{3 d y}{d x}+2 \frac{d y}{d x}-6 y\\ &\frac{d^{2} y}{d x^{2}}-5 \frac{d y}{d x}+6 y=0 \end{aligned}

Higher Order Derivatives exercise 11.1 question 41

Proved
Hint:
You must know the derivative of cot inverse $x$
Given:
$y=\left(\cot ^{-1} x\right)^{2}, \text { prove } y_{2}\left(x^{2}+1\right)^{2}+2 x\left(x^{2}+1\right) y_{1}=2$
Solution:
$y=\left(\cot ^{-1} x\right)^{2}$
\begin{aligned} &\frac{d y}{d x}=2 \cot ^{-1} x\left(\frac{-1}{1+x^{2}}\right)\\ &\left(1+x^{2}\right) \frac{d y}{d x}=-2 \cot ^{-1} x\\ &\text { Again differentiating }\\ &\left(1+x^{2}\right)^{2} \frac{d^{2} y}{d x^{2}}+2 x \frac{d y}{d x}=-2\left(\frac{-1}{1+x^{2}}\right) \end{aligned}
\begin{aligned} &\left(1+x^{2}\right)^{2} \frac{d^{2} y}{d x^{2}}+2 x \frac{d y}{d x}=\left(\frac{2}{1+x^{2}}\right) \\ &\left(1+x^{2}\right)^{2} \frac{d^{2} y}{d x^{2}}+2 x\left(1+x^{2}\right) \frac{d y}{d x}=2 \\ &y_{2}\left(x^{2}+1\right)^{2}+2 x\left(x^{2}+1\right) y_{1}=2 \end{aligned}

Higher Order Derivatives exercise 11.1 question 42

Proved
Hint:
You must know the derivative of cosec-1x
Given:
$y=\left(\operatorname{cosec}^{-1} x\right), x>1 , \text { prove } x\left(x^{2}-1\right) \frac{d^{2} y}{d x^{2}}+\left(2 x^{2}-1\right) \frac{d y}{d x}=0$
Solution:
\begin{aligned} &y=\left(\cos e c^{-1} x\right) \\ &\frac{d y}{d x}=\frac{-1}{x \sqrt{x^{2}-1}} \\ &x \sqrt{x^{2}-1} \frac{d y}{d x}=-1 \end{aligned}
\begin{aligned} &\text { Again differentiating }\\ &x \sqrt{x^{2}-1} \frac{d^{2} y}{d x^{2}}+\sqrt{x^{2}-1} \frac{d y}{d x}+x \cdot \frac{2 x}{2 \sqrt{x^{2}-1}} \frac{d y}{d x}=0\\ &x\left(x^{2}-1\right) \frac{d^{2} y}{d x^{2}}+\left(2 x^{2}-1\right) \frac{d y}{d x}=0 \end{aligned}

Higher Order Derivatives exercise 11.1 question 43

$2 \sqrt{2}=\frac{d^{2} y}{d x^{2}} \text { and } \frac{-1}{\sqrt{x}}=\frac{d^{2} y}{d x^{2}}$
Hint:
You must know the derivative of cos, tan, sin and logarithm function
Given:
\begin{aligned} &x=\cos t+\log \tan \frac{t}{2} \\ &y=\sin t \quad \text { find } \frac{d^{2} y}{d x^{2}} \& \frac{d^{2} y}{d x^{2}} \quad \text { at } t=\frac{\pi}{4} \end{aligned}
Solution:
\begin{aligned} &y=\sin t \quad \frac{d y}{d t}=\cos t \\ &\frac{d^{2} y}{d t^{2}}=-\sin t \\ &\left.\frac{d^{2} y}{d t^{2}}\right]_{t=\frac{\pi}{4}}=-\sin \frac{\pi}{4}=\frac{-1}{\sqrt{2}} \end{aligned}
\begin{aligned} &\text { Again differentiating }\\ &x=\cos t+\log \tan \frac{t}{2}\\ &\frac{d x}{d t}=-\sin t+\frac{1}{\tan \frac{t}{2}} \cdot \sec ^{2} \frac{t}{2} \cdot \frac{1}{2}\\ &=-\sin t+\frac{\cos \frac{t}{2}}{2 \times \sin \frac{t}{2}} \cdot \sec ^{2} \frac{t}{2} \cdot \frac{1}{2}\\ &=-\sin t+\frac{1}{\sin 2 \times \frac{t}{2}} \end{aligned}
\begin{aligned} &=-\sin t+\operatorname{cosec\: t} \\ &\text { Now }, \frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=\frac{\cos t}{\operatorname{cosec\: t}-\sin t} \\ &=\frac{\cos t}{1-\sin ^{2} t} \sin t=\frac{\sin t \cos t}{\cos ^{2} t} \\ &\frac{d^{2} y}{d x^{2}}=\frac{d\left(\frac{d y}{d x}\right)}{d x} \end{aligned}
\begin{aligned} &=\frac{\frac{d}{d t}\left(\frac{d y}{d x}\right)}{\frac{d x}{d t}}=\frac{\sec ^{2} t}{\operatorname{cosec\: t}-\sin t} \\ &=\frac{\sec ^{2} t \cdot \sin t}{\cos ^{2} t}=\sec ^{2} t \tan t \\ &\left.\frac{d^{2} y}{d x^{2}}\right]_{-\frac{\pi}{4}}=2 \sqrt{2} \times 1 \\ &=2 \sqrt{2} \end{aligned}

Higher Order Derivatives exercise 11.1 question 44 maths

$\frac{1}{a \sin ^{2} t \cos t}$
Hint:
You must know the derivative of sin, cos, tan and logarithm function
Given:
\begin{aligned} &x=a \sin t \\ &y=a\left(\cos t+\log \tan \frac{t}{2}\right) \text { find } \frac{d^{2} y}{d x^{2}} \end{aligned}
Solution:
\begin{aligned} &x=a \sin t \\ &\frac{d x}{d t}=a \cos t \\ &y=a\left(\cos t+\log \tan \frac{t}{2}\right) \\ &\frac{d y}{d t}=a\left(-\sin t+\cot \frac{t}{2} \times \sec ^{2} \frac{t}{2} \times \frac{t}{2}\right) \end{aligned}
\begin{aligned} &=a\left(-\sin t+\frac{1}{2 \sin \left(\frac{t}{2}\right) \times \cos \left(\frac{t}{2}\right)}\right) \\ &\frac{d y}{d t}=a\left(-\sin t+\frac{1}{\sin t}\right) \\ &=a\left(\frac{-\sin ^{2} t+1}{\sin t}\right) \Rightarrow \frac{a \cos ^{2} t}{\sin t} \end{aligned}
\begin{aligned} &\therefore \frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=\frac{\frac{a \cos ^{2} t}{\sin t}}{a \cos t} \\ &=\frac{\cos t}{\sin t}=\cot t \\ &\text { Again } \frac{d^{2} y}{d x^{2}}=-cosec^{2} t \frac{d t}{d x}=cose c^{2} t \times \frac{1}{a \cos t} \\ &=\frac{1}{a \sin ^{2} t \cos t} \end{aligned}

Higher Order Derivatives exercise 11.1 question 45

$\frac{8\sqrt{2}}{\pi a}$
Hint:
You must know the derivative of cos and sin function
Given:
\begin{aligned} &x=a(\cos t+t \sin t) \\ &y=a(\sin t-t \cos t) \text { find } \frac{d^{2} y}{d x^{2}} \text { at } t=\frac{\pi}{4} \end{aligned}
Solution:
\begin{aligned} &x=a(\cos t+t \sin t) \\ &\frac{d x}{d t}=a(-\sin t+t \cos t+\sin t) \\ &=a t \cos t \\ &y=a(\sin t-t \cos t) \\ &\frac{d y}{d t}=a(\cos t+t \sin t-\cos t) \\ &=a t \sin t \end{aligned}
\begin{aligned} &\therefore \frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=\frac{a t \sin t}{a t \cos t}=\tan t \\ &\text { Again } \frac{d^{2} y}{d x^{2}}=\sec ^{2} t \frac{d t}{d x}=\sec ^{2} t \times \frac{1}{a t \cos t} \end{aligned}
\begin{aligned} &=\frac{\sec ^{2} t}{a t} \\ &\left.\frac{d^{2} y}{d x^{2}}\right]_{t=\frac{\pi}{4}}=\frac{\sec ^{2} \frac{\pi}{4}}{a \frac{\pi}{4}}=\frac{2 \sqrt{2} \times 4}{a \pi} \\ &=\frac{8 \sqrt{2}}{\pi a} \end{aligned}

Higher Order Derivatives exercise 11.1 question 46

$\frac{8\sqrt{3}}{a}$
Hint:
You must know the derivative of cos, sin, tan and logarithm function
Given:
\begin{aligned} &x=a\left(\cos t+\log \tan \frac{t}{2}\right) \\ &y=a \sin t \quad \text { find } \frac{d^{2} y}{d x^{2}} \text { at } t=\frac{\pi}{3} \end{aligned}
Solution:
\begin{aligned} &x=a\left(\cos t+\log \tan \frac{t}{2}\right) \\ &\frac{d x}{d t}=a\left[-\sin t+\frac{1}{\tan \frac{t}{2}} \sec ^{2} \frac{t}{2} \cdot \frac{1}{2}\right] \\ &\frac{d x}{d t}=a\left[-\sin t+\frac{\cos \frac{t}{2}}{2 \sin \frac{t}{2}} \frac{1}{\cos ^{2} \frac{t}{2}}\right] \end{aligned}
\begin{aligned} &\frac{d x}{d t}=a\left[-\sin t+\frac{1}{2 \sin \frac{t}{2} \cos \frac{t}{2}}\right] \\ &\frac{d x}{d t}=a\left[-\sin t+\frac{1}{\sin t}\right] \\ &\frac{d x}{d t}=\frac{a \cos ^{2} t}{\sin t} \\ &y=a \sin t \end{aligned}
\begin{aligned} &\frac{d y}{d t}=a \cos t \\ &\therefore \frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=\frac{a \cos t}{a \cos ^{2} t} \times \sin t=\tan t \\ &\text { Again } \frac{d^{2} y}{d x^{2}}=\sec ^{2} t \times \frac{\sin t}{a \cos ^{2} t} \end{aligned}
\begin{aligned} &\left.\frac{d^{2} y}{d x^{2}}\right]_{t=\frac{\pi}{3}}=\frac{\sec ^{2} \frac{\pi}{3} \cdot \sin \frac{\pi}{3}}{a \cos ^{2} \frac{\pi}{4}}=\frac{(2)^{2} \cdot\left(\frac{\sqrt{3}}{2}\right)}{a\left(\frac{1}{2}\right)^{2}} \\ &\left.\frac{d^{2} y}{d x^{2}}\right]_{t=\frac{\pi}{3}}=\frac{8 \sqrt{3}}{a} \end{aligned}

Higher Order Derivatives exercise 11.1 question 47

$\frac{(\sin 2 t+2 a t \cos 2 t)}{(\cos 2 t-2 a t \sin 2 t)}$
Hint:
You must know the derivative of cos and sin function
Given:
\begin{aligned} &x=a(\cos 2 t+2 t \sin 2 t) \\ &y=a(\sin 2 t-2 t \cos 2 t) \quad \text { find } \frac{d^{2} y}{d x^{2}} \end{aligned}
Solution:
\begin{aligned} &x=a(\cos 2 t+2 t \sin 2 t) \\ &\frac{d x}{d t}=a(-2 \sin 2 t+2 \sin 2 t+4 t \cos 2 t) \\ &=4 a t \cos 2 t \\ &\frac{d^{2} x}{d t^{2}}=4 a \cos 2 t-8 a t \sin 2 t \end{aligned}
\begin{aligned} &y=a(\sin 2 t-2 t \cos 2 t) \\ &\frac{d y}{d t}=a(2 \cos 2 t-2 \cos 2 t+4 t \sin 2 t) \\ &=4 a t \sin 2 t \\ &\frac{d^{2} y}{d t^{2}}=4 a \sin 2 t+8 a t \cos 2 t \end{aligned}
\begin{aligned} &\text { So, } \frac{d^{2} y}{d x^{2}}=\frac{4 \sin 2 t+8 a t \cos 2 t}{4 \cos 2 t-8 a t \sin 2 t} \\ &=\frac{4(\sin 2 t+2 a t \cos 2 t)}{4(\cos 2 t-2 a t \sin 2 t)} \\ &=\frac{(\sin 2 t+2 a t \cos 2 t)}{(\cos 2 t-2 a t \sin 2 t)} \end{aligned}

Higher Order Derivatives exercise 11.1 question 48 maths

$\frac{-1}{3 \sin ^{3} t \cos 2 t}$
Hint:
You must know the derivative of cos t and sin t function
Given:
\begin{aligned} &x=3 \cos t-2 \cos ^{3} t \\ &y=3 \sin t-2 \sin ^{3} t \quad \text { find } \frac{d^{2} y}{d x^{2}} \end{aligned}
Solution:
\begin{aligned} &x=3 \cos t-2 \cos ^{3} t \\ &\frac{d x}{d t}=-3 \sin t+6 \cos ^{2} t \sin t \\ &=\sin t+\left(6 \cos ^{2} t-3\right) \\ &y=3 \sin t-2 \sin ^{3} t \\ &\frac{d y}{d t}=3 \cos t-6 \sin ^{2} t \cos t \\ &=\cos t\left(3-6 \sin ^{2} t\right) \end{aligned}
\begin{aligned} &\therefore \frac{d y}{d x}=\frac{d y}{d t} \times \frac{d x}{d t}=\frac{\cos t\left(3-6 \sin ^{2} t\right)}{\sin t+\left(6 \cos ^{2} t-3\right)} \\ &=\frac{\cos t\left(1-2 \sin ^{2} t\right)}{3\left(2 \cos ^{2} t-1\right)} \\ &=\frac{\cot t(\cos 2 t)}{\cos 2 t}=\cot t \\ &\frac{d y}{d x}=\cot t \\ &\frac{d^{2} y}{d x^{2}}=-cos e c^{2} t \end{aligned}

Higher Order Derivatives exercise 11.1 question 49

$\frac{-(x^{2}+y^{2})}{y^{3}}$
Hint:
You must know the derivative of cos t and sin t function
Given:
\begin{aligned} &x=a \sin t-b \cos t \\ &y=a \cos t-b \sin t \quad \text { Prove } \frac{d^{2} y}{d x^{2}}=\frac{-\left(x^{2}+y^{2}\right)}{y^{3}} \end{aligned}
Solution:
\begin{aligned} &\frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=\frac{a \sin t+b \cos t}{a \cos t+b \sin t}\\ &\text { Use quotient rule }\\ &\frac{U}{V}=\frac{U V^{\prime}-U^{\prime} V}{V^{2}} \end{aligned}
\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{\frac{d\left(\frac{d y}{d x}\right)}{d t}}{\frac{d x}{d t}}\\ &=\frac{(-a \cos t-b \sin t)(a \cos t+b \sin t)-(a \sin t+b \cos t)(-a \sin t+b \cos t)}{(a \cos t+b \sin t)^{2}(a \cos t+b \sin t)}\\ &=\frac{(-a \cos t+b \sin t)^{2}-(b \cos t-a \sin t)^{2}}{y^{3}}\\ &=\frac{-y^{2}-x^{2}}{y^{3}}=\frac{-\left(x^{2}+y^{2}\right)}{y^{3}} \end{aligned}

Higher Order Derivatives exercise 11.1 question 50

$A=\frac{2}{3}\: and\: B=\frac{-1}{3}$
Hint:
You must know the derivative of second order
Given:
$\text { Find } A \& B \text { so that } y=A \sin 3 x+B \cos 3 x \text { satisfies the equation } \frac{d^{2} y}{d x^{2}}+4 \frac{d y}{d x}+3 y=10 \cos 3 x$
Solution:
\begin{aligned} &\text { Let } y=A \sin 3 x+B \cos 3 x \\ &\frac{d y}{d x}=\frac{d(A \sin 3 x+B \cos 3 x)}{d x} \\ &=A \cos 3 x-3+b(-\sin 3 x .3) \quad\left[\begin{array}{l} \frac{d \cos 3 x}{d x}=-3 \sin 3 x \\ \frac{d \sin 3 x}{d x}=3 \cos 3 x \end{array}\right] \end{aligned}
\begin{aligned} &\frac{d y}{d x}=3 A \cos 3 x-3 B \sin 3 x \\ &\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}(3 A \cos 3 x-3 B \sin 3 x) \\ &=3 A(-\sin 3 x \cdot 3)-3 B(\cos 3 x .3) \\ &=-9 A \sin 3 x-9 B \cos 3 x \\ &=-9(A \sin 3 x+3 \cos 3 x) \\ &=-9 y \end{aligned}
\begin{aligned} &\frac{d^{2} y}{d x^{2}}+4 \frac{d y}{d x}+3 y=10 \cos 3 x\\ &-9 y+4(3 A \cos 3 x-3 B \sin 3 x)+3 y=10 \cos 3 x\\ &-6 y+12 A \cos 3 x-12 B \sin 3 x=10 \cos 3 x\\ &-6(A \sin 3 x+B \cos 3 x)+12 A \cos 3 x-12 B \sin 3 x=10 \cos 3 x\\ &\sin 3 x(-6 A-12 B)+\cos 3 x(-6 B+12 A)=10 \cos 3 x\\ &-6 A-12 B=0 \quad \ldots \ldots . .(1)\\ &-6 B+12 A=0 \quad \ldots \ldots . .(2)\\ &6 A=-12 B\\ &A=-2 B \quad \ldots \ldots(3) \end{aligned}
\begin{aligned} &\text { Put (3) in (2) }\\ &-6 B+(-2 B) 12=10\\ &-6 B-24 B=10\\ &-30 B=10\\ &B=\frac{-1}{3}\\ &A=-2\left(\frac{-1}{3}\right)=\frac{2}{3}\\ &A=\frac{2}{3}\; \&\; B=\frac{-1}{3} \end{aligned}

Higher Order Derivatives exercise 11.1 question 51

$\frac{d^{2} y}{d t^{2}}+2 k \frac{d y}{d t}+n^{2} y=0$
Hint:
You must know the derivative of
$A e^{-k t} \cos (p t+c)$
Given:
$\text { If } y=A e^{k t} \cos (p t+c) \text { prove that } \frac{d^{2} y}{d t^{2}}+2 k \frac{d y}{d t}+n^{2} y=0 \text { where } n^{2}=p^{2}+k^{2}$
Solution:
\begin{aligned} &\text { Let } y=A e^{-k t} \cos (p t+c) \quad \ldots . .(1) \\ &\frac{d y}{d x}=A e^{-k t}(-k) \cos (p t+c)+A e^{-k t}(-\sin (p t+c) p) \\ &\frac{d y}{d t}=-A k e^{-k t} \cos (p t+c)-A p e^{-k t}(\sin (p t+c)) \quad \ldots . .(2) \end{aligned}
\begin{aligned} &\frac{d^{2} y}{d x^{2}}=A k^{2} e^{-k t} \cos (p t+c)+A p k e^{-k t} \sin (p t+c)-A p k e^{-k t} \sin (p t+c)-A p^{2} e^{-k t} \cos (p t+c) \\ &\frac{d^{2} y}{d t^{2}}=\left(k^{2}-p^{2}\right) A e^{-k t} \cos (p t+c)+2 k\left(A p e^{-k t} \sin (p t+c)\right) \quad \ldots . .(3) \end{aligned}
\begin{aligned} &\text { Using }(1)\: \&\: (2)\: \&\: n^{2}=k^{2}+p^{2} \\ &\frac{d^{2} y}{d t^{2}}=\left(k^{2}-p^{2}\right) y+2 k\left(-k y-\frac{d y}{d t}\right) \\ &\frac{d^{2} y}{d t^{2}}=\left(k^{2}-p^{2}-2 k^{2}\right) y+2 k\left(\frac{d y}{d t}\right) \\ &\frac{d^{2} y}{d t^{2}}=-n^{2} y-2 k \frac{d y}{d t} \\ &\frac{d^{2} y}{d t^{2}}+2 k \frac{d y}{d t}+n^{2} y=0 \end{aligned}

Higher Order Derivatives exercise 11.1 question 52 maths

$x^{2} \frac{d^{2} y}{d x^{2}}+x(1-2 n) \frac{d y}{d x}+\left(1+n^{2}\right) y=0$
Hint:
You must know the derivative of
$cos(log\: x)\: and\: sin(log\: x)$
Given:
$\text { If } y=x^{n}\{\operatorname{acos}(\log x)+b \sin (\log x)\} \text { prove that } x^{2} \frac{d^{2} y}{d x^{2}}+x(1-2 n) \frac{d y}{d x}+\left(1+n^{2}\right) y=0$
Solution:
\begin{aligned} &\text { Let } y=x^{n}\{a \cos (\log x)+b \sin (\log x)\}\\ &\text { Use multiplicative rule }\\ &\mathrm{UV}=\mathrm{U}^{\prime} \mathrm{V}+\mathrm{UV}^{\prime}\\ &U=x^{n}\\ &V=a \cos (\log x)+b \sin (\log x) \end{aligned}
\begin{aligned} &\frac{d y}{d x}=n x^{n-1}\{\operatorname{acos}(\log x)+b \sin (\log x)\}+x^{n}\left\{\frac{a \sin (\log x)}{x}+\frac{b \cos (\log x)}{x}\right\} \\ &\frac{d y}{d x}=n \frac{y}{x}+\frac{x^{n}}{x}(-a \sin (\log x)+b \cos (\log x)) \\ &x \frac{d y}{d x}=n y+x^{n}(-a \sin (\log x)+b \cos (\log x)) \end{aligned}
\begin{aligned} &\frac{d y}{d x}+x \frac{d^{2} y}{d x^{2}}=n \frac{d y}{d x}+n x^{n-1}(-a \sin (\log x)+b \cos (\log x))+x^{n}\left\{\frac{a \sin (\log x)}{x}+\frac{b \cos (\log x)}{x}\right\} \\ &\frac{d y}{d x}(1-x)+x \frac{d^{2} y}{d x^{2}}=n x^{n-1}(-a \sin (\log x)+b \cos (\log x))+x^{n}\left\{\frac{a \sin (\log x)}{x}+\frac{b \cos (\log x)}{x}\right\} \end{aligned}
\begin{aligned} &\frac{d y}{d x}(1-n)+x \frac{d^{2} y}{d x^{2}}=n x^{n-1}(-a \sin (\log x)+b \cos (\log x))-\frac{1}{x} y \\ &\frac{d y}{d x}(1-n)+x \frac{d^{2} y}{d x^{2}}=\frac{n}{x}\left(x \frac{d y}{d x}-n y\right)-\frac{y}{x} \end{aligned}
\begin{aligned} &\frac{d y}{d x}+x(1-n) \frac{d y}{d x}=n x \frac{d y}{d x}-n^{2} y-y \\ &\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}(x-n x-n x)-\left(1+n^{2}\right) y=0 \\ &x^{2} \frac{d^{2} y}{d x^{2}}+x(1-2 n) \frac{d y}{d x}+\left(1+n^{2}\right) y=0 \end{aligned}

The RD Sharma Class 12 Solutions Higher Order Derivatives Ex 11.1 is available for the students who find difficulty in this chapter. The 11th chapter of class 12 mathematics consists of only 1 exercise, ex 11.1. Therefore, it is the smallest and one of the easiest chapters that is present in the syllabus.

There are 61 questions in total including the sub headings and the topics included in this chapter are meaning and documentations of higher order derivatives, discovering second order derivatives, establishing relations and various other order derivatives.

The additional questions mock test questions present in the Class 12 RD Sharma Chapter 11 Exercise 11.1 book helps the students to assess themselves before facing the real examinations. This makes them fix a target score easily when they prepare for the next exam.

Most of the teachers also refer to the RD Sharma Class 12th Exercise 11.1 Chapter 11 Higher Order Derivatives solution book to find out various tricks that can be applied. The questions for the class tests and the homework sums are also taken from the RD Sharma Class 12 Chapter 11 Exercise 11.1 book.

Many previous batch students have admitted that the contribution of the RD Sharma Class 12 Chapter 11 Exercise 11.1 book was immense in making them score high marks.

The benefits of referring to the RD Sharma solution books are:

• The students gain the access to the best solution book provided by the experts at the Career 360 website.

• They can utilize all the resources for free of cost without paying even a single rupee.

• Many additional sets of questions are present.

• Various tips are given for many sums.

• The sums that can be performed in other methods are also solved.

Therefore, the RD Sharma Class 12 Solutions Chapter 11 ex 11.1 can be downloaded in the form of a PDF at the Career 360 website. The students can easily attain their highest targets by practising with the Class 12 RD Sharma Chapter 11 Exercise 11.1 Solution book.

## RD Sharma Chapter-wise Solutions

1. How can the class 12 students clarify their doubts in the Higher Order Derivatives chapter?

The students can use the RD Sharma Class 12th Exercise 11.1 reference book to clear their nagging doubts regarding this chapter.

2. Where can the RD Sharma solution books be easily found?

The entire collection of the RD Sharna Solution books along with the RD Sharma Class 12 Chapter 11 Exercise 11.1 material are available at the Career 360 website. Everyone can access these reference books.

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As the RD Sharma books provide the access for everyone to download it, these books can be used when there is no access to the internet too.

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The Class 12 RD Sharma Chapter 11 Exercise 11.1 is available for free of cost at the top educational website, Career 360.

5. How many sums are solved in exercise 11.1 at the Class 12 RD Sharma Chapter 11 Exercise 11.1 reference book?

There are 61 questions given in the textbook for the exercise 11.1. All the answers for these questions are available at the Class 12 RD Sharma Chapter 11 Exercise 11.1 reference book.

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##### Geotechnical engineer

The role of geotechnical engineer starts with reviewing the projects needed to define the required material properties. The work responsibilities are followed by a site investigation of rock, soil, fault distribution and bedrock properties on and below an area of interest. The investigation is aimed to improve the ground engineering design and determine their engineering properties that include how they will interact with, on or in a proposed construction.

The role of geotechnical engineer in mining includes designing and determining the type of foundations, earthworks, and or pavement subgrades required for the intended man-made structures to be made. Geotechnical engineering jobs are involved in earthen and concrete dam construction projects, working under a range of normal and extreme loading conditions.

3 Jobs Available
##### Cartographer

How fascinating it is to represent the whole world on just a piece of paper or a sphere. With the help of maps, we are able to represent the real world on a much smaller scale. Individuals who opt for a career as a cartographer are those who make maps. But, cartography is not just limited to maps, it is about a mixture of art, science, and technology. As a cartographer, not only you will create maps but use various geodetic surveys and remote sensing systems to measure, analyse, and create different maps for political, cultural or educational purposes.

3 Jobs Available
##### Finance Executive

A career as a Finance Executive requires one to be responsible for monitoring an organisation's income, investments and expenses to create and evaluate financial reports. His or her role involves performing audits, invoices, and budget preparations. He or she manages accounting activities, bank reconciliations, and payable and receivable accounts.

3 Jobs Available
##### Investment Banker

An Investment Banking career involves the invention and generation of capital for other organizations, governments, and other entities. Individuals who opt for a career as Investment Bankers are the head of a team dedicated to raising capital by issuing bonds. Investment bankers are termed as the experts who have their fingers on the pulse of the current financial and investing climate. Students can pursue various Investment Banker courses, such as Banking and Insurance, and Economics to opt for an Investment Banking career path.

3 Jobs Available
##### Bank Branch Manager

Bank Branch Managers work in a specific section of banking related to the invention and generation of capital for other organisations, governments, and other entities. Bank Branch Managers work for the organisations and underwrite new debts and equity securities for all type of companies, aid in the sale of securities, as well as help to facilitate mergers and acquisitions, reorganisations, and broker trades for both institutions and private investors.

3 Jobs Available
##### Treasurer

Treasury analyst career path is often regarded as certified treasury specialist in some business situations, is a finance expert who specifically manages a company or organisation's long-term and short-term financial targets. Treasurer synonym could be a financial officer, which is one of the reputed positions in the corporate world. In a large company, the corporate treasury jobs hold power over the financial decision-making of the total investment and development strategy of the organisation.

3 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### Underwriter

An underwriter is a person who assesses and evaluates the risk of insurance in his or her field like mortgage, loan, health policy, investment, and so on and so forth. The underwriter career path does involve risks as analysing the risks means finding out if there is a way for the insurance underwriter jobs to recover the money from its clients. If the risk turns out to be too much for the company then in the future it is an underwriter who will be held accountable for it. Therefore, one must carry out his or her job with a lot of attention and diligence.

3 Jobs Available
##### Bank Probationary Officer (PO)

A career as Bank Probationary Officer (PO) is seen as a promising career opportunity and a white-collar career. Each year aspirants take the Bank PO exam. This career provides plenty of career development and opportunities for a successful banking future. If you have more questions about a career as  Bank Probationary Officer (PO), what is probationary officer or how to become a Bank Probationary Officer (PO) then you can read the article and clear all your doubts.

3 Jobs Available
##### Operations Manager

Individuals in the operations manager jobs are responsible for ensuring the efficiency of each department to acquire its optimal goal. They plan the use of resources and distribution of materials. The operations manager's job description includes managing budgets, negotiating contracts, and performing administrative tasks.

3 Jobs Available
##### Transportation Planner

A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.

3 Jobs Available
##### Conservation Architect

A Conservation Architect is a professional responsible for conserving and restoring buildings or monuments having a historic value. He or she applies techniques to document and stabilise the object’s state without any further damage. A Conservation Architect restores the monuments and heritage buildings to bring them back to their original state.

2 Jobs Available
##### Safety Manager

A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.

2 Jobs Available

A Team Leader is a professional responsible for guiding, monitoring and leading the entire group. He or she is responsible for motivating team members by providing a pleasant work environment to them and inspiring positive communication. A Team Leader contributes to the achievement of the organisation’s goals. He or she improves the confidence, product knowledge and communication skills of the team members and empowers them.

2 Jobs Available
##### Structural Engineer

A Structural Engineer designs buildings, bridges, and other related structures. He or she analyzes the structures and makes sure the structures are strong enough to be used by the people. A career as a Structural Engineer requires working in the construction process. It comes under the civil engineering discipline. A Structure Engineer creates structural models with the help of computer-aided design software.

2 Jobs Available
##### Architect

Individuals in the architecture career are the building designers who plan the whole construction keeping the safety and requirements of the people. Individuals in architect career in India provides professional services for new constructions, alterations, renovations and several other activities. Individuals in architectural careers in India visit site locations to visualize their projects and prepare scaled drawings to submit to a client or employer as a design. Individuals in architecture careers also estimate build costs, materials needed, and the projected time frame to complete a build.

2 Jobs Available
##### Landscape Architect

Having a landscape architecture career, you are involved in site analysis, site inventory, land planning, planting design, grading, stormwater management, suitable design, and construction specification. Frederick Law Olmsted, the designer of Central Park in New York introduced the title “landscape architect”. The Australian Institute of Landscape Architects (AILA) proclaims that "Landscape Architects research, plan, design and advise on the stewardship, conservation and sustainability of development of the environment and spaces, both within and beyond the built environment". Therefore, individuals who opt for a career as a landscape architect are those who are educated and experienced in landscape architecture. Students need to pursue various landscape architecture degrees, such as M.Des, M.Plan to become landscape architects. If you have more questions regarding a career as a landscape architect or how to become a landscape architect then you can read the article to get your doubts cleared.

2 Jobs Available
##### Plumber

An expert in plumbing is aware of building regulations and safety standards and works to make sure these standards are upheld. Testing pipes for leakage using air pressure and other gauges, and also the ability to construct new pipe systems by cutting, fitting, measuring and threading pipes are some of the other more involved aspects of plumbing. Individuals in the plumber career path are self-employed or work for a small business employing less than ten people, though some might find working for larger entities or the government more desirable.

2 Jobs Available
##### Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

6 Jobs Available
##### Pathologist

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

5 Jobs Available
##### Veterinary Doctor

A veterinary doctor is a medical professional with a degree in veterinary science. The veterinary science qualification is the minimum requirement to become a veterinary doctor. There are numerous veterinary science courses offered by various institutes. He or she is employed at zoos to ensure they are provided with good health facilities and medical care to improve their life expectancy.

5 Jobs Available
##### Gynaecologist

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth.

4 Jobs Available
##### Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available
##### Surgical Technologist

When it comes to an operation theatre, there are several tasks that are to be carried out before as well as after the operation or surgery has taken place. Such tasks are not possible without surgical tech and surgical tech tools. A single surgeon cannot do it all alone. It’s like for a footballer he needs his team’s support to score a goal the same goes for a surgeon. It is here, when a surgical technologist comes into the picture. It is the job of a surgical technologist to prepare the operation theatre with all the required equipment before the surgery. Not only that, once an operation is done it is the job of the surgical technologist to clean all the equipment. One has to fulfil the minimum requirements of surgical tech qualifications.

3 Jobs Available
##### Critical Care Specialist

A career as Critical Care Specialist is responsible for providing the best possible prompt medical care to patients. He or she writes progress notes of patients in records. A Critical Care Specialist also liaises with admitting consultants and proceeds with the follow-up treatments.

2 Jobs Available
##### Ophthalmologist

Individuals in the ophthalmologist career in India are trained medically to care for all eye problems and conditions. Some optometric physicians further specialize in a particular area of the eye and are known as sub-specialists who are responsible for taking care of each and every aspect of a patient's eye problem. An ophthalmologist's job description includes performing a variety of tasks such as diagnosing the problem, prescribing medicines, performing eye surgery, recommending eyeglasses, or looking after post-surgery treatment.

2 Jobs Available
##### Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs.

4 Jobs Available
##### Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
##### Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages. Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
##### Talent Agent

The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.

If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

3 Jobs Available

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
##### Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
##### Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available
##### Talent Director

Individuals who opt for a career as a talent director are professionals who work in the entertainment industry. He or she is responsible for finding out the right talent through auditions for films, theatre productions, or shows. A talented director possesses strong knowledge of computer software used in filmmaking, CGI and animation. A talent acquisition director keeps himself or herself updated on various technical aspects such as lighting, camera angles and shots.

2 Jobs Available
##### Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook.

5 Jobs Available
##### Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
##### Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
##### Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. Ever since internet cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

3 Jobs Available
##### Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
##### Fashion Journalist

Fashion journalism involves performing research and writing about the most recent fashion trends. Journalists obtain this knowledge by collaborating with stylists, conducting interviews with fashion designers, and attending fashion shows, photoshoots, and conferences. A fashion Journalist  job is to write copy for trade and advertisement journals, fashion magazines, newspapers, and online fashion forums about style and fashion.

2 Jobs Available
##### Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications.

2 Jobs Available
##### Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

2 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
##### Production Manager

Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers.

3 Jobs Available

A Team Leader is a professional responsible for guiding, monitoring and leading the entire group. He or she is responsible for motivating team members by providing a pleasant work environment to them and inspiring positive communication. A Team Leader contributes to the achievement of the organisation’s goals. He or she improves the confidence, product knowledge and communication skills of the team members and empowers them.

2 Jobs Available
##### Quality Systems Manager

A Quality Systems Manager is a professional responsible for developing strategies, processes, policies, standards and systems concerning the company as well as operations of its supply chain. It includes auditing to ensure compliance. It could also be carried out by a third party.

2 Jobs Available
##### Merchandiser

A career as a merchandiser requires one to promote specific products and services of one or different brands, to increase the in-house sales of the store. Merchandising job focuses on enticing the customers to enter the store and hence increasing their chances of buying a product. Although the buyer is the one who selects the lines, it all depends on the merchandiser on how much money a buyer will spend, how many lines will be purchased, and what will be the quantity of those lines. In a career as merchandiser, one is required to closely work with the display staff in order to decide in what way a product would be displayed so that sales can be maximised. In small brands or local retail stores, a merchandiser is responsible for both merchandising and buying.

2 Jobs Available
##### Procurement Manager

The procurement Manager is also known as  Purchasing Manager. The role of the Procurement Manager is to source products and services for a company. A Procurement Manager is involved in developing a purchasing strategy, including the company's budget and the supplies as well as the vendors who can provide goods and services to the company. His or her ultimate goal is to bring the right products or services at the right time with cost-effectiveness.

2 Jobs Available
##### Production Planner

Individuals who opt for a career as a production planner are professionals who are responsible for ensuring goods manufactured by the employing company are cost-effective and meets quality specifications including ensuring the availability of ready to distribute stock in a timely fashion manner.

2 Jobs Available
##### ITSM Manager

ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks.

3 Jobs Available
##### Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

3 Jobs Available
##### Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### IT Consultant

An IT Consultant is a professional who is also known as a technology consultant. He or she is required to provide consultation to industrial and commercial clients to resolve business and IT problems and acquire optimum growth. An IT consultant can find work by signing up with an IT consultancy firm, or he or she can work on their own as independent contractors and select the projects he or she wants to work on.

2 Jobs Available
##### Data Architect

A Data Architect role involves formulating the organisational data strategy. It involves data quality, flow of data within the organisation and security of data. The vision of Data Architect provides support to convert business requirements into technical requirements.

2 Jobs Available
##### Security Engineer

The Security Engineer is responsible for overseeing and controlling the various aspects of a company's computer security. Individuals in the security engineer jobs include developing and implementing policies and procedures to protect the data and information of the organisation. In this article, we will discuss the security engineer job description, security engineer skills, security engineer course, and security engineer career path.

2 Jobs Available
##### UX Architect

A UX Architect is someone who influences the design processes and its outcomes. He or she possesses a solid understanding of user research, information architecture, interaction design and content strategy.

2 Jobs Available