RD Sharma Solutions Class 12 Mathematics Chapter 11 MCQ

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RD Sharma Solutions Class 12 Mathematics Chapter 11 MCQ

RD Sharma Solutions Class 12 Mathematics Chapter 11 MCQ

Satyajeet KumarUpdated on 27 Jan 2022, 02:55 PM IST

RD Sharma books are considered the best source of information for CBSE students. They contain comprehensive material that is helpful for students to get a good insight into the subject. They are widely used all over the country and contain detailed exercises on all concepts. Moreover, many faculties use the RD Sharma book as a medium for setting up question papers. This is why it is beneficial for students.

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RD Sharma Class 12 Solutions Chapter 11 MCQ Higher Order Derivatives - Other Exercise
Higher Order Derivatives Excercise: MCQ
RD Sharma Chapter-wise Solutions

RD Sharma Class 12th Chapter 11 MCQ contains 28 questions that are easy to solve and based on fundamentals. Students can breeze through this exercise with no problem if they are familiar with the basics of this chapter. RD Sharma Solutions It covers topics like finding second-order derivatives and evaluating differential equations. Students can get good knowledge about evaluating trigonometric and logarithmic equations through this exercise.

RD Sharma Class 12 Solutions Chapter 11 MCQ Higher Order Derivatives - Other Exercise

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Higher Order Derivatives Excercise: MCQ

Higher Order Derivatives Exercise Multiple Choice Questions Question 1

Answer:
(b)
Hint:
We must know the derivative of $\cos x$ and $\sin x$ .
Given:
$\begin{aligned} &x=a \cos n t-b \sin n t \\\\ &\end{aligned}$
Explanation:
$\begin{aligned} &x=a \cos n t-b \sin n t \\\\ &\end{aligned}$
$\frac{d x}{d t}=a(-\sin n t) \times n-b(\cos n t) \times n \\\\$
$\frac{d x}{d t}=-n a \sin n t-n b \cos n t$
Differentiate on both sides,
$\begin{aligned} &\frac{d^{2} x}{d t^{2}}=-n a(\cos n t) \times n-n b(-\sin n t) \times n \\\\ \\ \end{aligned}$
$\frac{d^{2} x}{d t^{2}}=-n^{2} a(\cos n t)+n^{2} b(\sin n t)$
$\frac{d^{2} x}{d t^{2}}=n^{2}[-a \cos n t+b \sin n t] \\\\$
$\frac{d^{2} x}{d t^{2}}=-n^{2} x$

Higher Order Derivatives Exercise Multiple Choice Questions Question 2

Answer:
(d)
Hint:
We must know about the derivative of $t$ .
Given:
$x=a t^{2}, y=2 a t$

Explanation:
$x=a t^{2}$
$\Rightarrow \quad \frac{d x}{d t}=2 a t$
$y=2 a t$
$\Rightarrow \frac{d y}{d t}=2 a$
$\begin{aligned} & \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{2 a}{2 a t} \\ \end{aligned}$
$\Rightarrow \frac{d y}{d x}=\frac{1}{t}$
Differentiate on both sides,
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{d y}{d x}\right) \\ & \end{aligned}$
$=\frac{d}{d t}\left(\frac{d y}{d x}\right) \times \frac{d t}{d x}$
$\begin{aligned} &=\frac{d}{d t}\left(\frac{1}{t}\right) \times \frac{1}{2 a t} \\ \end{aligned}$
$=\frac{-1}{t^{2}} \times \frac{1}{2 a t} \\$
$=\frac{-1}{2 a t^{3}}$

Higher Order Derivatives Exercise Multiple Choice Questions Question 3

Answer:
(b)
Hint:
We must know the derivative of $x$ and $y$ .
Given:
$\begin{aligned} &y=a x^{n+1}+b x^{-n} \\ \end{aligned}$
Explanation:
$\begin{aligned} &y=a x^{n+1}+b x^{-n} \\ \end{aligned}$
$\frac{d y}{d x}=(n+1) a x^{n+1-1}+(-n) b x^{-n-1} \\$
$\frac{d y}{d x}=(n+1) a x^{n}+(-n) b x^{-(n+1)}$
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=n(n+1) a x^{n-1}-n b x^{-(n+1)-1}(-(n+1)) \\ \end{aligned}$
$\frac{d^{2} y}{d x^{2}}=n(n+1) a x^{n-1}+n(n+1) b x^{-n-2}$
$\begin{aligned} &x^{2} \frac{d^{2} y}{d x^{2}}=n(n+1)\left[a x^{n-1}+b x^{-(n+2)}\right] x^{2} \\ & \end{aligned}$
$x^{2} \frac{d^{2} y}{d x^{2}}=n(n+1)\left[a x^{n-1+2}+b x^{-n-2+2}\right] \\$
$x^{2} \frac{d^{2} y}{d x^{2}}=n(n+1)\left[a x^{n+1}+b x^{-n}\right]$

Higher Order Derivatives Exercise Multiple Choice Question question 4

Answer:
(b)
Hint:
We must know the derivative of $\cos x$ .
Given:
$= \frac{d^{20}}{dx^{20}}\left ( 2 \cos x \cos 3x \right )$
Explanation:
First let us solve the inner function,
$= \frac{d}{dx}\left ( 2 \cos x \cos 3x \right )$
Using identity $\left ( 2 \cos x \cos 3x \right )$
$\begin{aligned} &=\cos (-2 x)+\cos (4 x) \\ & \end{aligned}$
$=\cos (2 x)+\cos (4 x)$
So, now we have to compute $20^{th}$ derivative of $\cos \left ( 2x \right )+\cos\left ( 4x \right )$
Since, successive derivative of $\cos x$ cycle in $4:-\sin x,-\cos x, \sin x, \cos x\cdot \cdot \cdot$
$4^{th}$ derivative of $\cos x$ is $20=\left ( 5.4 \right )^{th}$ derivative of $\cos x$ is also $\cos x$ .
Keep chain in mind,
$\begin{aligned} & & 2^{20} \cos (2 x)+4^{20} \cos (4 x) \\ \end{aligned}$
$\Rightarrow \quad 2^{20}\left[\cos (2 x)+2^{20} \cos (4 x)\right]$

Higher Order Derivatives Exercise Multiple Choice Question question 5

Answer:
(b)
Hint:
We must know the derivative of $x$ and $y$
Given:
$x=t^{2}, y=t^{3}$
Explanation:
$x=t^{2}, y=t^{3}$
$\begin{aligned} &\frac{d x}{d t}=2 t, \frac{d y}{d t}=3 t^{2} \\ \end{aligned}$
$\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{3 t^{2}}{2 t} \\$
$=\frac{3}{2} t$
Differentiate on both sides,
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{3}{2} t\right) \\ & \end{aligned}$
$=\frac{d}{d t}\left(\frac{3}{2} t\right) \times \frac{d t}{d x} \\$
$=\frac{3}{2} t^{2} \times \frac{1}{2 t} \\$
$=\frac{3}{4} t$

Higher Order Derivatives Exercise Multiple Choice Question question 6

Answer:
(b)
Hint:
We must know the derivative of $x$ .
Given:
$\begin{aligned} &y=a+b x^{2} \\ & \end{aligned}$ ,$a,b$ are arbitrary constant.
Explanation:
$\begin{aligned} &y=a+b x^{2} \\ & \end{aligned}$
$\frac{d y}{d x}=2 b x$
$\frac{1}{x} \cdot \frac{d y}{d x}=2 b$
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=2 b=\frac{1}{x} \cdot \frac{d y}{d x} \\ & \end{aligned}$
$=x^{2} \frac{d^{2} y}{d x^{2}}=\frac{d y}{d x}$

Higher Order Derivatives Excercise Multiple Choice Questions Question 7

Answer:
(c)
Hint:
We must know about the derivative of $\cos x$ and $\sin x$ .
Given:
$\begin{aligned} & f(x)=(\cos x+i \sin x)(\cos 2 x+i \sin 2 x)(\cos 3 x+i \sin 3 x) \ldots .(\cos n x+i \sin n x) \\ \end{aligned}$ and $f\left ( 1 \right )=1$ , then $f^{11}\left ( 1 \right )$ is equal to.
Explanation:
$\begin{aligned} & f(x)=(\cos x+i \sin x)(\cos 2 x+i \sin 2 x)(\cos 3 x+i \sin 3 x) \ldots .(\cos n x+i \sin n x) \\ \end{aligned}$
$\Rightarrow \quad \cos (x+2 x+3 x+\ldots .+n x)+i \sin (x+2 x+3 x+\ldots .+n x)$
Using Fourier series,
$\Rightarrow \quad \left[\cos \frac{n(n+1)}{2} x+i \sin \frac{n(n+1)}{2} x\right]$
Differentiate on both sides,
$\begin{aligned} &f^{\prime}(x)=\frac{n(n+1)}{2}\left[-\sin \frac{n(n+1)}{2} x+i \cos \frac{n(n+1)}{2} x\right] \\ & \end{aligned}$
$f^{\prime \prime}(x)=-\left(\frac{n(n+1)}{2}\right)^{2}\left[\cos \frac{n(n+1)}{2} x+i \sin \frac{n(n+1)}{2} x\right]$
$\begin{aligned} f^{\prime \prime}(x) &=-\left(\frac{n(n+1)}{2}\right)^{2} f(x) \\ \end{aligned}$
$\therefore f^{\prime \prime}(x) =-\left(\frac{n(n+1)}{2}\right)^{2} f(1)$
$=-\left(\frac{n(n+1)}{2}\right)^{2}$

Higher Order Derivatives Excercise Multiple Choice Questions Question 8

Answer:
(a)
Hint:
We must know about the derivative of $\cos x$ and $\sin x$ .
Given:
$\begin{aligned} &y=a \sin m x+b \cos m x \\ & \end{aligned}$
Explanation:
$\begin{aligned} &y=a \sin m x+b \cos m x \\ & \end{aligned}$
$\frac{d y}{d x}=a(\cos m x) m+b(-\sin m x) m \\$
$\frac{d y}{d x}=m a \cos m x-m b \sin m x$
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=m a(-\sin m x) \times m-m b(\cos m x) \times m \\ & \end{aligned}$
$\frac{d^{2} y}{d x^{2}}=-a m^{2}(\sin m x)-m^{2} b(\cos m x)$
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=-m^{2}[a \sin m x+b \cos m x] \\ & \end{aligned}$
$\frac{d^{2} y}{d x^{2}}=-m^{2} y$

Higher Order Derivatives Excercise Multiple Choice Questions Question 9

Answer:
(a)
Hint:
We must know about the derivative of $\sin^{-1} x$ .
Given:
$f(x)=\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}$ , then $\left(1-x^{2}\right) f^{\prime}(x)-x f(x)=?$
Explanation:
$f(x)=\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}$
Differentiate with respect to $x$ ,
$\begin{aligned} &y \sqrt{1-x^{2}}=\sin ^{-1} x \\ & \end{aligned}$
$y \cdot \frac{1}{2 \sqrt{1-x^{2}}}(-2 x)+\sqrt{1-x^{2}} \frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}} \\$
$-x y+\left(1-x^{2}\right) \frac{d y}{d x}=1$
Therefore, $\left(1-x^{2}\right) f^{\prime}(x)-x f(x)=1$

Higher Order Derivatives Exercise Multiple Choice Question Question 10

Answer:
(c)
Hint:
We must know about the derivative of logarithm and $\tan x$ .
Given:
$y=\tan ^{-1}\left\{\frac{\log _{e}\left(\frac{e}{x^{2}}\right)}{\log _{e}\left(e x^{2}\right)}+\tan ^{-1}\left(\frac{3+2 \log _{e} x}{1-6 \log _{e} x}\right)\right\}$
Explanation:
$y=\tan ^{-1}\left\{\frac{\log _{e}\left(\frac{e}{x^{2}}\right)}{\log _{e}\left(e x^{2}\right)}+\tan ^{-1}\left(\frac{3+2 \log _{e} x}{1-6 \log _{e} x}\right)\right\}$
$\begin{aligned} &y=\tan ^{-1}\left(\frac{1-2 \log _{e} x}{1+2 \log _{e} x}\right)+\tan ^{-1}\left(\frac{3+2 \log _{e} x}{1-6 \log _{e} x}\right) \\ & \end{aligned}$
$y=\tan ^{-1}\left(\frac{\left(\frac{1-2 \log _{e} x}{1+2 \log _{e} x}\right)+\left(\frac{3+2 \log _{e} x}{1-6 \log _{e} x}\right)}{1-\left(\frac{1-2 \log _{e} x}{1+2 \log _{e} x}\right)\left(\frac{3+2 \log _{e} x}{1-6 \log _{e} x}\right)}\right)$
$\begin{aligned} &y=\tan ^{-1}\left\{\frac{\left(1-2 \log _{e} x\right)\left(1-6 \log _{e} x\right)+\left(3+2 \log _{e} x\right)\left(1+2 \log _{e} x\right)}{\left(1+2 \log _{e} x\right)\left(1-6 \log _{e} x\right)-\left(1-2 \log _{e} x\right)\left(3+2 \log _{e} x\right)}\right\} \\ & \end{aligned}$
$y=\tan ^{-1}\left\{\frac{1-8 \log _{e} x+12\left(\log _{e} x\right)^{2}+3+8 \log _{e} x+4\left(\log _{e} x\right)^{2}}{1-4 \log _{e} x-12\left(\log _{e} x\right)^{2}-3+4 \log _{e} x+4\left(\log _{e} x\right)^{2}}\right\}$
$\begin{aligned} &y=\tan ^{-1}\left\{\frac{4+16\left(\log _{e} x\right)^{2}}{-2-8\left(\log _{e} x\right)^{2}}\right\} \\ & \end{aligned}$
$y=\tan ^{-1}\left\{\frac{4\left(1+4\left(\log _{e} x\right)^{2}\right)}{-2\left(1+4\left(\log _{e} x\right)^{2}\right)}\right\}$
$\begin{aligned} &y=\tan ^{-1}[-2] \\ & \end{aligned}$
$\frac{d y}{d x}=0, \frac{d^{2} y}{d x^{2}}=0$

Higher Order Derivatives Exercise Multiple Choice Question Question 11

Answer:
(a)
Hint:
We know about the derivative of polynomials.
Given:
$f\left ( x \right )$ be a polynomial, $f\left ( e \right )^{x}$ .
Explanation:
$y=f\left ( e \right )^{x}$
$\begin{gathered} \frac{d}{d x}\left(f\left(e^{x}\right)\right)=f^{\prime}\left(e^{x}\right) \times \frac{d}{d x}\left(e^{x}\right) \\ \end{gathered}$
$=f^{\prime}\left(e^{x}\right) \cdot e^{x}$
Similarly,
$\begin{aligned} & \frac{d^{2}}{d x^{2}}\left(f\left(e^{x}\right)\right)=f "\left(e^{x}\right) \frac{d}{d x}\left(e^{x}\right) e^{x}+\frac{d}{d x}\left(e^{x}\right) f^{\prime}\left(e^{x}\right) \\ \end{aligned}$
$\Rightarrow \quad f^{\prime \prime}\left(e^{x}\right) e^{2 x}+f^{\prime}\left(e^{x}\right)\left(e^{x}\right)$

Higher Order Derivatives Exercise Multiple Choice Question Question 12

Answer: Option (c)
Hint:
We must know about the derivative of $\cos x$ and logarithm.
Given:
$y=a \cos \left(\log _{e} x\right)+b \sin \left(\log _{e} x\right)$
Explanation:
$y=a \cos \left(\log _{e} x\right)+b \sin \left(\log _{e} x\right)$
Differentiate both side with respect to $x$
$\frac{d y}{d x}=\frac{a[-\sin (\log x)]}{x}-\frac{b[\cos (\log x)]}{x} \\$
$\frac{d y}{d x}=\frac{-[a \sin (\log x)+b \cos (\log x)]}{x} \$
$\begin{aligned} \ &x \frac{d y}{d x}=-[a \sin (\log x)+b \cos (\log x)] \end{aligned}$
Again differentiating with respect to $x$ ,
$\begin{aligned} &x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=-\left[\frac{a \cos (\log x)}{x}-\frac{b \sin (\log x)}{x}\right] \\ \end{aligned}$
$\begin{aligned} &x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=-\left[\frac{a \cos (\log x)}{x}-\frac{b \sin (\log x)}{x}\right] \\ \end{aligned}$
$x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=\frac{-y}{x} \\$
$x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}+\frac{y}{x}=0$
$\begin{aligned} \\ &x^{2} \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}+y=0 \end{aligned}$ or
$x^{2}y_{2}+xy_{1}+y=0$
Hence option the value of $x^{2}y_{2}+xy_{1}$ is $(-y)$

Higher Order Derivatives Exercise Multiple Choice Questions Question 13

Answer:
(a)
Hint:
We must know about the derivative of $x$
Given:
$x=2 a t, y=a t^{2} \\$ , where $a$ is constant.
Explanation:
$x=2 a t, y=a t^{2} \\$
$\frac{d x}{d t}=2 a, \frac{d y}{d t}=2 a t \\$
$\begin{aligned} &\frac{d y}{\frac{d t}{d x}}{d t}=\frac{2 a t}{2 a} \end{aligned}$
$=t \\$
$\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{d y}{d x}\right) \\$
$\begin{aligned} & &\frac{d^{2} y}{d x^{2}}=\frac{d}{d t}\left(\frac{d y}{d x}\right) \frac{d t}{d x} \end{aligned}$
$=\frac{d}{d t}(t) \frac{d t}{d x} \\$
$=1\left(\frac{1}{2 a}\right) \\$
$\begin{aligned} &=\frac{1}{2 a} \end{aligned}$

Higher Order Derivatives Exercise Multiple Choice Questions Question 14

Answer:
(a)
Hint:
We must know about the derivative.
Given:
$x=f(t), y=g(t)$
Explanation:
$x=f(t), y=g(t)$
$\\ \frac{d x}{d t}=f^{\prime}(t), \frac{d y}{d t}=g^{\prime}(t) \$
$\ \frac{d y}{dt}\cdot{\frac{d t}{d x}}=\frac{g^{\prime}(t)}{f^{\prime}(t)} \\$
$\begin{aligned} &\frac{d y}{d x}=\frac{g^{\prime}(t)}{f^{\prime}(t)} \end{aligned}$
Differentiate on both sides,
$\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{d y}{d x}\right) \\$
$=\frac{d}{d t}\left(\frac{g^{\prime}(t)}{f^{\prime}(t)}\right) \times \frac{d t}{d x} \\$
$=\frac{f^{\prime}(t) g^{\prime \prime}(t)-g^{\prime}(t) f^{\prime \prime}(t)}{\left(f^{\prime}(t)\right)^{2}} \times \frac{1}{f^{\prime}(t)}$
$=\frac{f^{\prime}(t) g^{\prime \prime}(t)-g^{\prime}(t) f^{\prime \prime}(t)}{\left(f^{\prime}(t)\right)^{3}}$

Higher Order Derivatives Exercise Multiple Choice Questions Question 15

Answer:
(c)
Hint:
We must have known about the derivative of inverse trigonometric functions like $\sin^{-1}x$ .
Given:
$y=\sin \left(m \sin ^{-1} x\right)$
Explanation:
$y=\sin \left(m \sin ^{-1} x\right)$
Differentiating both sides with respect $x$,
$\frac{d y}{d x}=\cos \left(m \sin ^{-1} x\right) \times \frac{m}{\sqrt{1-x^{2}}}$
$\begin{aligned} &\\ &\frac{d y}{d x}=\frac{m \cos \left(m \sin ^{-1} x\right)}{\sqrt{1-x^{2}}} \end{aligned}$
Again differentiate with respect to $x$ ,
$\frac{d^{2} y}{d x^{2}}=m\left[\left(\sqrt{1-x^{2}}\right)\left(-\sin \left(m\left(\sin ^{-1} x\right)\right) \times \frac{m}{\sqrt{1-x^{2}}}\right)-\cos \left(m\left(\sin ^{-1} x\right)\right) \times \frac{1}{2 \sqrt{1-x^{2}}} \times(0-2 x)\right] \\$
$\frac{d^{2} y}{d x^{2}}=m\left[\frac{\left(-m \sin \left(m\left(\sin ^{-1} x\right)\right)\right)+\cos \left(m\left(\sin ^{-1} x\right)\right) \times \frac{x}{\sqrt{1-x^{2}}}}{1-x^{2}}\right]$
$\begin{aligned} & \\ &\frac{d^{2} y}{d x^{2}}=m\left[\frac{\left(-m \sin \left(m\left(\sin ^{-1} x\right)\right)\right)}{1-x^{2}}+\frac{x \cos \left(m\left(\sin ^{-1} x\right)\right)}{\left(1-x^{2}\right) \sqrt{1-x^{2}}}\right] \end{aligned}$
$\frac{d^{2} y}{d x^{2}}=m\left[\frac{\left(-m \sin \left(m\left(\sin ^{-1} x\right)\right)\right)}{1-x^{2}}+\frac{x \frac{d y}{d x}}{\left(1-x^{2}\right)}\right]$
$\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}=m x \frac{d y}{d x}-m^{2} \sin \left(m\left(\sin ^{-1} x\right)\right)$
$\begin{aligned} &\\ &\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}=m x \frac{d y}{d x}-m^{2} y \end{aligned}$

$\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-m x \frac{d y}{d x}+m^{2} y=0$ or
$\begin{aligned} & \\ &\left(1-x^{2}\right) y^{2}-m x y^{1}=-m^{2} y \end{aligned}$

Higher Order Derivatives Exercise Multiple Choice Question question 16

Answer:
(a)
Hint:
We must have known about the derivative of trigonometric function like $\sin^{-1}x$ .
Given:
$y=\left(\sin ^{-1} x\right)^{2}$
Explanation:
$y=\left(\sin ^{-1} x\right)^{2}$
Differentiate with respect to $x$
$\frac{d y}{d x}=2 \sin ^{-1} x \times \frac{1}{\sqrt{1-x^{2}}}$
Again differentiate with respect to $x$
$\frac{d^{2} y}{d x^{2}}=\frac{2 \frac{1}{\sqrt{1-x^{2}}} \times \sqrt{1-x^{2}}}{\left(1-x^{2}\right)}-\sin ^{-1}\left(\frac{x-2 x}{2 \sqrt{1-x^{2}}}\right)$
$\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}=2\left[1+\frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}}\right] \\$
$\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-\frac{2 x \sin ^{-1} x}{\sqrt{1-x^{2}}}=2 \\$
$\begin{gathered} \left(1-x^{2}\right) y_{2}=2+x y_{1} \end{gathered}$

Higher Order Derivatives Exercise Multiple Choice Question question 17

Answer:
(c)
Hint:
We must have known about the derivative of $\sin x,\cos x$ and $\tan x$ .
Given:
$y=e^{\tan x} \$
Explanation:
$y=e^{\tan x} \$
$y_{1}=\frac{d y}{d x}=e^{\tan x} \times \sec ^{2} x \\$
$\begin{aligned} & &\frac{d y}{d x}=e^{\tan x}\left(\sec ^{2} x\right) \end{aligned}$
Again differentiate with respect to $x$ ,
$\frac{d^{2} y}{d x^{2}}=\left[e^{\tan x}\left(\sec ^{2} x\right)\left(\sec ^{2} x\right)+e^{\tan x}\left(2 \sec ^{2} x \tan x\right)\right] \$
$\begin{aligned} &\ &\frac{d^{2} y}{d x^{2}}=e^{\tan x}\left[\sec ^{4} x+2 \sec ^{2} x \tan x\right] \end{aligned}$
$\frac{d^{2} y}{d x^{2}}=e^{\tan x} \cdot \sec ^{2} x\left[\sec ^{2} x+2 \tan x\right] \\$
$\frac{d^{2} y}{d x^{2}}=e^{\tan x} \cdot \sec ^{2} x\left[\frac{1}{\cos ^{2} x}+2 \frac{\sin x}{\cos x}\right]$
$\begin{aligned} & \\ &\frac{d^{2} y}{d x^{2}}=e^{\operatorname{lan} x} \cdot \sec ^{2} x\left[\frac{1+2 \sin x \cos x}{\cos ^{2} x}\right] \end{aligned}$
$\cos ^{2} x \frac{d^{2} y}{d x^{2}}=y_{1}[1+2 \sin x \cos x]$
$\begin{aligned} &\\ &\cos ^{2} x \frac{d^{2} y}{d x^{2}}=y_{1}[1+\sin 2 x] \end{aligned}$ $[\because 2 \sin x \cos x=\sin 2 x]$

Higher Order Derivatives Exercise Multiple Choice Question question 18

Answer:
(b)
Hint:
We must know about the derivative of trigonometric function.
Given:
$y=\frac{2}{\sqrt{a^{2}-b^{2}}} \tan ^{-1}\left(\sqrt{\frac{a-b}{a+b}} \tan \frac{x}{2}\right)$
Explanation:
$y=\frac{2}{\sqrt{a^{2}-b^{2}}} \tan ^{-1}\left(\sqrt{\frac{a-b}{a+b}} \tan \frac{x}{2}\right)$
$\begin{aligned} &\\ &\frac{d y}{d x}=\frac{2}{\sqrt{a^{2}-b^{2}}} \cdot \frac{1}{1+\left(\frac{a-b}{a+b}\right) \tan ^{2}\left(\frac{x}{2}\right)} \times \sqrt{\frac{a-b}{a+b}} \frac{1}{2} \sec ^{2}\left(\frac{x}{2}\right) \end{aligned}$
$\frac{d y}{d x}=\frac{2}{\sqrt{a^{2}-b^{2}}} \cdot \frac{(a+b)}{(a+b)+(a-b) \tan ^{2}\left(\frac{x}{2}\right)} \times \sqrt{\frac{a-b}{a+b}} \frac{1}{2} \sec ^{2}\left(\frac{x}{2}\right)$
$\frac{d y}{d x}=\sqrt{\frac{1}{a^{2}-b^{2}} \times \frac{a-b}{a+b} \times(a+b)^{2}} \times \frac{1+\tan ^{2}\left(\frac{x}{2}\right)}{a\left(1+\tan ^{2}\left(\frac{x}{2}\right)+b\left(1-\tan ^{2}\left(\frac{x}{2}\right)\right)\right)}$
$\frac{d y}{d x}= \frac{1}{a+b\left(\frac{1-\tan ^{2}\left(\frac{x}{2}\right)}{1+\tan ^{2}\left(\frac{x}{2}\right)}\right)}$
$\begin{aligned} \\ \frac{d y}{d x}=& \frac{1}{a+b \cos x} \end{aligned}$
Differentiate again,
$\frac{d^{2} y}{d x^{2}}=-\frac{-b \sin x}{(a+b \cos x)^{2}} \\$
$\begin{aligned} & &\frac{d^{2} y}{d x^{2}}=\frac{b \sin x}{(a+b \cos x)^{2}} \end{aligned}$

Higher Order Derivatives Exercise Multiple Choice Question question 19

Answer:
(a)
Hint:
We must know about the derivative of $x$
Given:
$y=\frac{a x+b}{x^{2}+c} \$
Explanation:
$y=\frac{a x+b}{x^{2}+c} \$
$\begin{aligned} &\ &\left(x^{2}+c\right) y=a x+b \end{aligned}$
Differentiate with respect to $x$
$2 x y+\left(x^{2}+c\right) \frac{d y}{d x}=a$
Again differentiate,
$2 y+2 x y_{1}+2 x y_{1}+\left(x^{2}+c\right) y_{2}=0 \\$
$\begin{aligned} &2 y+4 x y_{1}+\left(x^{2}+c\right) y_{2}=0 \end{aligned}$
Differentiate again with respect to $x$
$2 y_{1}+4 y_{1}+4 x y_{2}+\left(x^{2}+c\right) y_{3}+2 x y_{2}=0 \\$
$\begin{aligned} &6 y_{1}+6 x y_{2}+\left(x^{2}+c\right) y_{3}=0 \end{aligned}$
$6 y_{1}+6 x y_{2}+\left(\frac{-2 y-4 x y_{1}}{y_{2}}\right) y_{3}=0 \\$
$6 y_{1} y_{2}+6 x\left(y_{2}\right)^{2}-2 y-4 x y_{1} y_{3}=0 \\$
$3 y_{1} y_{2}+3 x\left(y_{2}\right)^{2}-y-2 x y_{1} y_{3}=0$
$\begin{aligned} \\ &\left(2 x y_{1}+y\right) y_{3}=\left(y_{1}+x y_{2}\right) 3 y_{2} \end{aligned}$

Higher Order Derivatives Exercise Multiple Choice Question question 20

Answer:
(a)
Hint:
We must know about the derivative of logarithm.
Given:
$y=\log _{e}\left(\frac{x}{a+b x}\right)^{x}$
Explanation:
$y=\log _{e}\left(\frac{x}{a+b x}\right)^{x}$
$\\ y=x \log _{e}\left(\frac{x}{a+b x}\right) \$
$\begin{aligned} &\ &\frac{y}{x}=\log x-\log (a+b x) \end{aligned}$
Differentiate with respect to $x$
$\frac{x \frac{d y}{d x}-y}{x^{2}}=\frac{1}{x}-\frac{b}{a+b x} \$
$\begin{aligned} &\ &x \frac{d y}{d x}-y=x-\frac{b x^{2}}{a+b x} \end{aligned}$
Differentiate with respect to $x$
$x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}-\frac{d y}{d x}=1-\left(\frac{(a+b x) 2 b x-b x^{2}(b)}{(a+b x)^{2}}\right) \\$
$x \frac{d^{2} y}{d x^{2}}=\left(\frac{(a+b x)^{2}-(a+b x) 2 b x-b^{2} x^{2}}{(a+b x)^{2}}\right) \\$
$\begin{aligned} & &x \frac{d^{2} y}{d x^{2}}=\left(\frac{(a+b x)[a+b x-2 b x]+b^{2} x^{2}}{(a+b x)^{2}}\right) \end{aligned}$
$x \frac{d^{2} y}{d x^{2}}=\left(\frac{(a+b x)(a-b x)+b^{2} x^{2}}{(a+b x)^{2}}\right)$
$\begin{aligned} &\\ &x \frac{d^{2} y}{d x^{2}}=\left(\frac{a^{2}-b^{2} x^{2}+b^{2} x^{2}}{(a+b x)^{2}}\right) \end{aligned}$
$x \frac{d^{2} y}{d x^{2}}=\left(\frac{a^{2}}{(a+b x)^{2}}\right)=\left(\frac{a}{a+b x}\right)^{2}$
$\\ x \frac{d^{2} y}{d x^{2}}=\left(\frac{a}{a+b x}\right)^{2} \$ … (i)
And $\begin{aligned} \ &y=x \log \left(\frac{x}{a+b x}\right) \end{aligned}$
Differentiate with respect to $x$
$\frac{d y}{d x}=\frac{x(a+b x-b x)}{(a+b x)^{2}}\left(\frac{a+b x}{x}\right)+\log \left(\frac{x}{a+b x}\right) \\$
$\frac{d y}{d x}=\frac{a}{a+b x}+\frac{y}{x}$
$\begin{aligned} \\ &\frac{d y}{d x}-\frac{y}{x}=\frac{a}{a+b x} \end{aligned}$
From (i)
$x \frac{d^{2} y}{d x^{2}}=\left(\frac{d y}{d x}-\frac{y}{x}\right)^{2}$
$\begin{aligned} & \\ &x^{2} \frac{d^{2} y}{d x^{2}}=\frac{1}{x^{2}}\left(x \frac{d y}{d x}-y\right)^{2} \end{aligned}$
$x^{3} \frac{d^{2} y}{d x^{2}}=\left(x \frac{d y}{d x}-y\right)^{2}$

Higher Order Derivatives Exercise Multiple Choice Question question 21

Answer: (c)
Hint:
We must know about the derivative of trigonometric function.
Given:
$x=f(t) \cos t-f^{\prime}(t) \sin t \\$
$y=f(t) \sin t-f^{\prime}(t) \cos t \\$
Explanation:
$x=f(t) \cos t-f^{\prime}(t) \sin t \\$
$y=f(t) \sin t-f^{\prime}(t) \cos t \\$
$\frac{d x}{d t}=f^{\prime}(t) \cos t-f(t) \sin t-f^{\prime \prime}(t) \sin t-f^{\prime}(t) \cos t \\$
$\begin{aligned} & &\frac{d x}{d t}=-f(t) \sin t-f^{\prime \prime}(t) \sin t \end{aligned}$
$\frac{d x}{d t}=-\sin t\left[f(t)+f^{\prime \prime}(t)\right] \\$
$\frac{d y}{d t}=f^{\prime}(t) \sin t-f(t) \cos t-f^{\prime \prime}(t) \cos t-f^{\prime}(t) \sin t \\$
$\begin{aligned} &\frac{d y}{d t}=-f(t) \cos t-f^{\prime \prime}(t) \cos t \end{aligned}$
$\frac{d y}{d t}=-\cos t\left[f(t)+f^{\prime \prime}(t)\right]$
$\\ \left(\frac{d x}{d t}\right)^{2}=\left[-\sin t\left[f(t)+f^{\prime \prime}(t)\right]\right]^{2} \\$
$\begin{aligned} & &\left(\frac{d y}{d t}\right)^{2}=\left[-\cos t\left[f(t)+f^{\prime \prime}(t)\right]\right]^{2} \end{aligned}$
$\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}=(\sin t)^{2}\left[f(t)+f^{\prime \prime}(t)\right]^{2}+(\cos t)^{2}\left[f(t)+f^{\prime \prime}(t)\right]^{2} \\$
$\begin{aligned} & &\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}=\left[f(t)+f^{\prime \prime}(t)\right]^{2}\left[(\sin t)^{2}+(\cos t)^{2}\right] \end{aligned}$
$=\left[f(t)+f^{\prime \prime}(t)\right]^{2}$ $\left[\because(\sin t)^{2}+(\cos t)^{2}=1\right]$

Higher Order Derivatives Exercise Multiple Choice Question question 22

Answer:
(b)
Hint:
We must have known about the derivative of $x$
Given:
$y^{\frac{1}{n}}+y^{\frac{-1}{n}}=2 x$
Explanation:
$y^{\frac{1}{n}}+y^{\frac{-1}{n}}=2 x$
Differentiate with respect to $x$
${\left[\frac{1}{n} y^{\frac{1}{n}-1}-\frac{1}{n} y^{\frac{-1}{n}-1}\right] \frac{d y}{d x}=2} \\$
$\frac{1}{n} y^{-1}\left[y^{\frac{1}{n}}-y^{\frac{-1}{n}}\right] \frac{d y}{d x}=2 \\$
$\frac{1}{n y}\left[y^{\frac{1}{n}}-y^{\frac{-1}{n}}\right] \frac{d y}{d x}=2$
$\begin{aligned} \\ &{\left[y^{\frac{1}{n}}-y^{\frac{-1}{n}}\right] \frac{d y}{d x}=2 x y} \end{aligned}$
Squaring both sides,
$\left[y^{\frac{2}{n}}-y^{\frac{-2}{n}}-2\right]\left(\frac{d y}{d x}\right)^{2}=4 x^{2} y^{2}$ … (i)
Now
Given $y^{\frac{1}{n}}+y^{\frac{-1}{n}}=2 x$
Squaring both sides,
$y^{\frac{2}{n}}+y^{\frac{-2}{n}}+2=4 x^{2}$
$\begin{aligned} & \\ &y^{\frac{2}{n}}+y^{\frac{-2}{n}}=4 x^{2}-2 \end{aligned}$
Putting this value in (i)
$4 x^{2} y^{2}=\left[4 x^{2}-2-2\right]\left(\frac{d y}{d x}\right)^{2} \\$
$4 x^{2} y^{2}=\left[4 x^{2}-4\right]\left(\frac{d y}{d x}\right)^{2}$
$\begin{aligned} \\ &x^{2} y^{2}=\left[x^{2}-1\right]\left(\frac{d y}{d x}\right)^{2} \end{aligned}$
Differentiate with respect to $x$ ,
$2 n^{2} y\left(\frac{d y}{d x}\right)=2\left(\frac{d y}{d x}\right)\left(\frac{d^{2} y}{d x^{2}}\right)\left[n^{2} y-1\right]+2 x\left(\frac{d y}{d x}\right)^{2}$
$\\ n^{2} y=\left(\frac{d^{2} y}{d x^{2}}\right)\left[n^{2} y-1\right]+x\left(\frac{d y}{d x}\right) \\$
${\left[n^{2} y-1\right] y_{2}+x y_{1}=n^{2} y}$
$\begin{aligned} & \\ &{\left[x^{2}-1\right] y_{2}+x y_{1}=n^{2} y} \end{aligned}$

Higher Order Derivatives Exercise Multiple Choice Question question 23

Answer:
(c)
Hint:
We must know about the rules of finding the derivative.
Given:
$\frac{d}{d x}\left\{x^{n}-a_{1} x^{n-1}+a_{2} x^{n-2}+\ldots .+(-1)^{n} a_{n}\right\} e^{x}=x^{n} e^{x} \\$
Explanation:
$\frac{d}{d x}\left\{x^{n}-a_{1} x^{n-1}+a_{2} x^{n-2}+\ldots .+(-1)^{n} a_{n}\right\} e^{x}=x^{n} e^{x} \\$
$\frac{d}{d x}\left[x^{n}-a_{1} x^{n-1}+a_{2} x^{n-2}+\ldots .+(-1)^{n} a_{n}\right] e^{x} \\$
$\begin{aligned} & &\frac{d}{d x}\left[x^{n}-n x^{n-1}+n(n-1) x^{n-2}+\ldots . .+(-1)^{n} a_{n}\right] e^{x} \end{aligned}$
Comparing the coefficients of above two equations,
$a_{1}=n, a_{2}=n(n-1)$
Similarly,
$a_{r}=n(n-1)(n-2)(n-3) \ldots(n-r+1) \\$
$\begin{aligned} & &a_{r}=\frac{n !}{(n-r) !} \end{aligned}$

Higher Order Derivatives Exercise Multiple Choice Question question 24

Answer:
(a)
Hint:
We must know about the derivative of logarithm.
Given:
$y=x^{n-1} \log x$
Explanation:
$y=x^{n-1} \log x$
$\\ \frac{d y}{d x}=y_{1}=(n-1) x^{n-2} \log x+\frac{x^{n-1}}{x} \\$
$\begin{aligned} & &y_{1}=\frac{(n-1) x^{n-1} \log x+x^{n-1}}{x} \end{aligned}$
$x y_{1}=(n-1) y+x^{n-1} \\$ $\left[\because y=x^{n-1} \log x\right]$
$x y_{2}+y_{1}=(n-1) y_{1}+(n-1) x^{n-2}$
$\begin{aligned} & \\ &x y_{2}+y_{1}=(n-1) y_{1}+\frac{(n-1) x^{n-1}}{x} \end{aligned}$
$x^{2} y_{2}+x y_{1}=x(n-1) y_{1}+(n-1) x^{n-1}$
$\\ x^{2} y_{2}+x y_{1}=x(n-1) y_{1}+(n-1)\left[x y_{1}-(n-1) y\right] \\$
$x^{2} y_{2}+x y_{1}=x(n-1) y_{1}+(n-1) x y_{1}-(n-1)^{2} y \\$
$\begin{aligned} & &x^{2} y_{2}+x y_{1}=2 x(n-1) y_{1}-(n-1)^{2} y \end{aligned}$
$x^{2} y_{2}+x y_{1}-2 x(n-1) y_{1}=-(n-1)^{2} y \\$
$x^{2} y_{2}+x y_{1}(1-2 n+2)=-(n-1)^{2} y \\$
$\begin{aligned} & &x^{2} y_{2}+(3-2 n) x y_{1}=-(n-1)^{2} y \end{aligned}$

Higher Order Derivatives Exercise Multiple Choice Questions Question 25

Answer:
(c)
Hint:
We must know about the derivative of logarithm.
Given:
$x y-\log _{e} y=1 \\$ satisfy the equation $x\left(y y_{2}+y_{1}^{2}\right)-y_{2}+\lambda y y_{1}=0$
Explanation:
$x y-\log _{e} y=1 \\$
$\begin{aligned} & &x y_{1}+y-\frac{y_{1}}{y}=0 \end{aligned}$
$\frac{x y y_{1}+y^{2}-y_{1}}{y}=0$
$\begin{aligned} &\\ &x y y_{1}+y^{2}-y_{1}=0 \end{aligned}$
$y y_{1}+x y_{1} y_{1}+x y y_{2}+2 y y_{1}-y_{2}=0 \\$
$\begin{aligned} & &x\left(y_{1}^{2}+y y_{2}\right)-y_{2}+3 y y_{1}=0 \end{aligned}$
Compare with given equation,
$x\left(y y_{2}+y_{1}^{2}\right)-y_{2}+\lambda y y_{1}=0$
$\begin{aligned} & \\ &\lambda=3 \end{aligned}$

Higher Order Derivatives Exercise Multiple Choice Questions Question 26

Answer:
(a)
Hint:
We must know about the derivative values of every function.
Given:
$y^{2}=a x^{2}+b x+c$
Explanation:
$y^{2}=a x^{2}+b x+c$
$\\ 2 y \frac{d y}{d x}=2 a x+b$
$\begin{aligned} & \\ &2\left(\frac{d y}{d x}\right)^{2}+2 y \frac{d^{2} y}{d x^{2}}=2 a \end{aligned}$
$y \frac{d^{2} y}{d x^{2}}=a-\left(\frac{d y}{d x}\right)^{2} \\$
$\begin{aligned} & &y \frac{d^{2} y}{d x^{2}}=a-\left(\frac{2 a x+b}{2 y}\right)^{2} \end{aligned}$
$y \frac{d^{2} y}{d x^{2}}=\frac{4 a y^{2}-(2 a x+b)^{2}}{4 y^{2}} \$
$\begin{aligned} &\ &4 y^{3} \frac{d^{2} y}{d x^{2}}=4 a\left(a x^{2}+b x+c\right)-\left(4 a^{2} x^{2}+4 a x b+b^{2}\right) \end{aligned}$
$4 y^{3} \frac{d^{2} y}{d x^{2}}=4 a c-b^{2} \$
$\begin{aligned} &\ &y^{3} \frac{d^{2} y}{d x^{2}}=\frac{4 a c-b^{2}}{4} \end{aligned}$
= Constant

Higher Order Derivatives Excercise: 1.3

Higher Order Derivatives Exercise Multiple Choice Questions Question 27

Answer:
(a)
Hint:
We must know about the derivative rules of exponential functions.
Given:
$y=A e^{5 x}+B e^{-5 x} \\$
Explanation:
$y=A e^{5 x}+B e^{-5 x} \\$
$\frac{d y}{d x}=A e^{5 x} \cdot 5+B e^{-5 x} \cdot(-5) \\$
$\begin{aligned} & &\frac{d y}{d x}=5 A e^{5 x}-5 B e^{-5 x} \end{aligned}$
Again differentiate with respect to $x$
$\frac{d^{2} y}{d x^{2}}=5 A e^{5 x}(5)-5 B e^{-5 x}(-5) \\$
$\begin{aligned} & &\frac{d^{2} y}{d x^{2}}=25 A e^{5 x}+25 B e^{-5 x} \end{aligned}$
$\frac{d^{2} y}{d x^{2}}=25\left[A e^{5 x}+B e^{-5 x}\right] \$
$\begin{aligned} &\ &\frac{d^{2} y}{d x^{2}}=25 y \end{aligned}$

Higher Order Derivatives Exercise Multiple Choice Question Question 28

Answer:
(d)
Hint:
We must know about the derivative rules of logarithm.
Given:
$y=\log _{e}\left(\frac{x^{2}}{e^{2}}\right)$
Explanation:
$y=\log _{e}\left(\frac{x^{2}}{e^{2}}\right)$
Differentiate with respect to $x$
$\frac{d y}{d x}=\frac{1}{\frac{x^{2}}{e^{2}}} \cdot \frac{1}{e^{2}} 2 x$
$\begin{aligned} &\\ &\frac{d y}{d x}=\frac{2}{x} \end{aligned}$
Again differentiate,
$\frac{d^{2} y}{d x^{2}}=2\left(\frac{-1}{x^{2}}\right)$
$\begin{aligned} & \\ &\frac{d^{2} y}{d x^{2}}=\frac{-2}{x^{2}} \end{aligned}$

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