RD Sharma Class 12 Exercise 11.1 Higher Order Derivatives Solutions Maths

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

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RD Sharma Class 12 Solutions Chapter 11 Higher Order Derivatives - Other Exercise
Higher Order Derivatives Excercise:11.1
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Higher Order Derivatives Excercise:11.1

Higher Order Derivatives exercise 11.1 question 1(i)

Answer:

6 x + 2 s e c^{2} x t a n x

Hint:
You must know about derivative of tan x
Given:

x^{3} + t a n x

Solution:

L e t y = x^{3} + t a n x

\begin{aligned} \frac{d y}{d x} = 3 x^{2} + \sec^{2} x [\frac{d (\tan x)}{d x} = \sec^{2} x and \frac{d x^{3}}{d x} = 3 x^{2}] \\ \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)

Answer:

\frac{- [\sin (\log x) + \cos (\log x)]}{x^{2}}

Hint:
You must know about derivative of sin x & log x
Given:

s i n (l o g x)

Solution:

L e t y = s i n (l o g x)

\begin{aligned} \frac{d y}{d x} = \cos (\log x) \times \frac{d}{d x} \log x (\frac{d \sin x}{d x} = \cos x) \\ \frac{d y}{d x} = \cos (\log x) \times \frac{1}{x} (\frac{d \log x}{d x} = \frac{1}{x}) \\ \frac{d y}{d x} = \frac{\cos (\log x)}{x} \end{aligned}

Use quotient rule

A s \frac{u}{v} = \frac{u^{'} v - v^{'} u}{v^{2}}

W h e r e v = x a n d u = c o s (l o g x)

\begin{aligned} \frac{d^{2} y}{d x^{2}} = \frac{x \frac{d}{d x} (\cos (\log x)) - \cos (\log x)) \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}} (\frac{d}{d x} x = 1) \\ \frac{d^{2} y}{d x^{2}} = \frac{- \sin (\log x) \cdot \frac{1}{x} x - \cos (\log x)}{x^{2}} (\frac{d \log x}{d x} = \frac{1}{x}) \\ \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)

Answer:

- c o s e c^{2} x

Hint:
You must know about derivative of sin x & log x
Given:

l o g (s i n x)

Solution:
$L e t y = l o g (s i n x)$
$\begin{array}{ll} \frac{d y}{d x} = \frac{1}{\sin x} \frac{d \sin x}{d x} & (\frac{d \log x}{d x} = \frac{1}{x}) \\ \frac{d y}{d x} = \frac{1}{\sin x} \cos x & (\frac{d \sin x}{d x} = \cos x) \\ \frac{d y}{d x} = \cot x & (\frac{\cos x}{\sin x} = \cot x) \\ \frac{d^{2} y}{d x^{2}} = \cot x \\ \frac{d^{2} y}{d x^{2}} = - c o s e c^{2} x & (\frac{d}{d x} \cot x = - c o s e c^{2} x) \end{array}$

Higher Order Derivatives exercise 11.1 question 1(iv)

Answer:

- 24 e^{x} s i n 5 x + 10 e^{x} c o s 5 x

Hint:
You must know about derivative of sin x & e^x
Given:

e^{x} s i n 5 x

Solution:

L e t y = e^{x} s i n 5 x

Use multiplicative rule

A s U V = U V^{1} + U^{1} V

Where U=e^x & 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 (\frac{d}{d x} e^{x} = e^{x}) (\frac{d \sin 5 x}{d x} = \cos 5 x) \\ \frac{d y}{d x} = e^{x} \cdot \cos 5 x \cdot 5 + e^{x} \cdot \sin 5 x (\frac{d}{d x} 5 x = 5) \\ \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

A s U V = U V^{1} + U^{1} V

Where U=e^x & V=sin 5x
U=e^x & V=cos 5x

\begin{aligned} \frac{d^{2} y}{d x^{2}} = 5 [e^{x} \frac{d}{d x} \cos 5 x + \frac{d}{d x} e^{x} \cdot \cos 5 x] + [e^{x} \frac{d}{d x} \sin 5 x + \frac{d}{d x} e^{x} \cdot \sin 5 x] \\ \frac{d^{2} y}{d x^{2}} = 5 [e^{x} \cdot (- \sin 5 x) \frac{d}{d x} 5 x + e^{x} \cdot \cos 5 x] + [e^{x} \cos 5 x \frac{d}{d x} 5 x + e^{x} \cdot \sin 5 x] \\ (\frac{d \sin x}{d x} = \cos x, \frac{d \cos x}{d x} = - \sin x, \frac{d 5 x}{d x} = 5) \end{aligned}

\begin{aligned} \frac{d^{2} y}{d x^{2}} = 5 [e^{x} \cdot (- \sin 5 x) 5 + e^{x} \cdot \cos 5 x] + [e^{x} \cos 5 x .5 + e^{x} \cdot \sin 5 x] \\ \frac{d^{2} y}{d x^{2}} = [25 e^{x} \sin 5 x + 5 e^{x} \cdot \cos 5 x] + [5 e^{x} \cos 5 x + e^{x} \cdot \sin 5 x] \\ \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)

Answer:

27 e^{6 x} c o s 3 x - 36 e^{6 x} s i n 3 x

Hint:
You must know about derivative of cos3x & e^6x
Given:

e^{6 x} c o s 3 x

Solution:
$L e t y = e^{6 x} c o s 3 x$
Use multiplicative rule

A s U V = U V^{1} + U^{1} V

Where U=e^6x 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 (\frac{d \cos x}{d x} = - \sin x, \frac{d}{d x} e^{6 x} = e^{6 x} .6) \\ \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}

A s U V = U V^{1} + U^{1} V

Where U=e^6x and V=sin3x

\begin{aligned} \frac{d^{2} y}{d x^{2}} = - 3 [e^{6 x} \cos 3 x (3) + e^{6 x} \cdot \sin 3 x .6] + 6 [e^{6 x} \cos 3 x \cdot 6 + e^{6 x} \cdot (- \sin 3 x)] \\ (\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}) \\ \frac{d^{2} y}{d x^{2}} = - 3 [3 e^{6 x} \cos 3 x + 6 e^{6 x} \cdot \sin 3 x] + 6 [e^{6 x} \cos 3 x \cdot 6 - 3 e^{6 x} \cdot \sin 3 x] \\ \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)

Answer:

5 x + 6 x l o g x

HInt:
You must know about derivative of log x & x³
Given:

x^{3} l o g x

Solution:

L e t y = x^{3} l o g 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 (\frac{d}{d x} \log x = \frac{1}{x}, \frac{d}{d x} x^{3} = 3 x^{2}) \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 [x^{2} \frac{d}{d x} \log x + \frac{d}{d x} x^{2} \cdot \log x] \\ \frac{d^{2} y}{d x^{2}} = 2 x + 3 [x + 2 x \cdot \log x] (\frac{d}{d x} x^{2} = 2 x, \frac{d}{d x} \log x = \frac{1}{x}) \\ \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)

Answer:

\frac{- 2 x}{(1 + x^{2})^{2}}

Hint:
You must know about derivative of tan^-1x
Given:
tan^-1x
Solution:
$L e t y = t a n^{- 1} x$
$\begin{aligned} \frac{d y}{d x} = \frac{1}{1 + x^{2}} (\frac{d \tan^{- 1} x}{d x} = \frac{1}{1 + x^{2}}) \end{aligned}$
Use Quotient rule

\begin{aligned} As \frac{u}{v} = \frac{u^{1} v - v^{1} u}{v^{2}} \\ Where v = 1 + x^{2} & u = 1 \\ \frac{d^{2} y}{d x^{2}} = \frac{(1 + x^{2}) \frac{d}{d x} 1 - 1 \cdot \frac{d}{d x} (1 + x^{2})}{{(1 + x^{2})}^{2}} \\ \frac{d^{2} y}{d x^{2}} = \frac{0 (1 + x^{2}) - 1.2 x}{{(1 + x^{2})}^{2}} (\frac{d}{d x} 1 = 0, \frac{d}{d x} (1 + x^{2}) = 2 x) \\ \frac{d^{2} y}{d x^{2}} = \frac{0 - 2 x}{{(1 + x^{2})}^{2}} \\ \frac{d^{2} y}{d x^{2}} = \frac{- 2 x}{{(1 + x^{2})}^{2}} \end{aligned}

Higher Order Derivatives exercise 11.1 question 1(viii) maths

Answer:

- x c o s x - 2 s i n x

Hint:
You must know about derivative of x cos x
Given:
x cos x
Solution:
$L e t x c o s x$
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 (\frac{d}{d x} x = 1, \frac{d}{d x} \cos x = - \sin x) \\ \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}} = - (x \frac{d}{d x} \sin x + \frac{d}{d x} x \cdot \sin x) + \frac{d \cos x}{d x} \\ \frac{d^{2} y}{d x^{2}} = - (x \cos x + \sin x) - \sin x (\frac{d}{d x} \cos x = - \sin x) \\ \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)

Answer:

\frac{- (1 + l o g x)}{x^{2} (l o g x)^{2}}

Hint:
You must know about derivative of log(log x)
Given:

l o g (l o g x)

Solution:

L e t y = l o g (l o g x)

\begin{aligned} \frac{d y}{d x} = \frac{d}{d x} \log (\log x) (\frac{d \log x}{d x} = \frac{1}{x}) \\ \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} As \frac{u}{v} = \frac{u^{1} v - v^{1} u}{v^{2}} \\ Where u = 1 & v = x \log x \\ Use multiplicative rule \\ As U V = U V^{1} + U^{1} V \\ 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}} (\frac{d}{d x} \log x = \frac{1}{x}, \frac{d}{d x} 1 = 0, \frac{d}{d x} x = 1) \\ \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

Answer:

2 e^{- x} s i n x

Hint:
You have to show

\frac{d^{2} y}{d x^{2}} = 2 e^{- x} s i n x

Given:
If y=e^-xcos x, show that

\frac{d^{2} y}{d x^{2}} = 2 e^{- x} s i n x

Solution:
Let y=e^-xcos x
Use multiplicative rule

\begin{aligned} As U V = U V^{1} + U^{1} V \\ 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} & (\frac{d \cos x}{d x} = - \sin x, \frac{d e^{- x}}{d x} = - 1 e^{- x}) \end{aligned}

Again differentiating w.r.t.x we get
Use multiplicative rule

\begin{aligned} As U V = U V^{1} + U^{1} V \\ Where U = \cos x & V = e^{- x} \\ \frac{d^{2} y}{d x^{2}} = - [\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] \\ \frac{d^{2} y}{d x^{2}} = - [\sin x \cdot (- e^{- x}) + e^{- x} \cos x + \cos x (- e^{- x}) + e^{- x} (- \sin x)] \end{aligned}

\begin{aligned} \frac{d^{2} y}{d x^{2}} = - [- e^{- x} \sin x + e^{- x} \cos x - e^{- x} \cos x - e^{- x} \sin x] \\ (\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}) \\ \frac{d^{2} y}{d x^{2}} = - [- 2 \sin x e^{- x}] \\ \frac{d^{2} y}{d x^{2}} = 2 e^{- x} \sin x \end{aligned}

Hence proved

Higher Order Derivatives exercise 11.1 question 3

Answer:

\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 + t a n x (\frac{d t a n x}{d x} = s e c^{2} x, \frac{d}{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) (\frac{d}{d x} \sec^{2} x = 2 \sec x (\sec x \tan x) \\ \frac{d^{2} y}{d x^{2}} = 2 \sec^{2} x \tan x (\sec^{2} x = \frac{1}{\cos^{2} x}, \tan x = \frac{\sin x}{\cos x}) \\ \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} (\frac{2 \sin x}{\cos x}) - 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

Answer:

\frac{6}{x}

HInt:
You must know about derivative of x³ & log x
Given:
If y=x³log x, Prove that

\frac{d^{4} y}{d x^{4}} = \frac{6}{x}

Solution:
Let y=x³log x
Use multiplicative rule

A s U V = U V^{1} + U^{1} V

\begin{aligned} 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 (\frac{d \log x}{d x} = \frac{1}{x}, \frac{d}{d x} x^{3} = 3 x^{2}) \\ \frac{d y}{d x} = x^{2} + 3 x^{2} \log x \end{aligned}

Use multiplicative rule

A s U V = U V^{1} + U^{1} V

\begin{aligned} Where U = x^{2} & V = \log x \end{aligned}

\begin{aligned} \frac{d^{2} y}{d x^{2}} = 3 (x^{2} \frac{d}{d x} \log x + \frac{d}{d x} x^{2} \log x) + \frac{d}{d x} x^{2} \\ \frac{d^{2} y}{d x^{2}} = 3 (x^{2} \frac{1}{x} + 2 x \log x) + 2 x (\frac{d \log x}{d x} = \frac{1}{x}, \frac{d}{d x} x^{2} = 2 x) \\ \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

A s U V = U V^{1} + U^{1} V

\begin{aligned} Where U = x & V = \log x \end{aligned}

\begin{aligned} \frac{d^{3} y}{d x^{3}} = \frac{d}{d x} 5 x + 6 (x \frac{d}{d x} \log x + \frac{d}{d x} x \log x) \\ \frac{d^{3} y}{d x^{3}} = 5 + 6 (x \frac{1}{x} + \log x) (\frac{d \log x}{d x} = \frac{1}{x}, \frac{d}{d x} 5 x = 5) \\ \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 \\ \frac{d^{4} y}{d x^{4}} = 0 + \frac{6}{x} (\frac{d \log x}{d x} = \frac{1}{x}, \frac{d}{d x} 11 = 0) \\ \frac{d^{4} y}{d x^{4}} = \frac{6}{x} \end{aligned}

Hence proved

Higher Order Derivatives exercise 11.1 question 5

Answer:

2 c o s x c o s e c^{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}{d x^{3}} = 2 c o s x c o s e c^{3} x

Solution:

L e t y = l o g (s i n x) (\frac{d l o g x}{d x} = \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}} = - c o s e c^{2} x

(\frac{d c o t x}{d x} = - c o s e c^{2} x)

Again differentiating w.r.t. x we get

\begin{aligned} \frac{d^{3} y}{d x^{3}} = - 2 cosec x \cdot (- \cos e c x \cdot \cot x) (\frac{d {cosec}^{2} x}{d x} = - c o s e c x \cot x) \\ \frac{d^{3} y}{d x^{3}} = - 2 \cos x {cosec}^{3} x \end{aligned}

Higher Order Derivatives exercise 11.1 question 6

Answer:

\frac{d^{2} y}{d x^{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}{d x^{2}} + y = 0

Solution:
Let y=2sin x+3cos x

\begin{aligned} \frac{d y}{d x} = 2 \cos x - 3 \sin x (\frac{d \cos x}{d x} = - \sin x, \frac{d \sin x}{d x} = \cos x) \\ \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

Answer:

\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}, s h o w t h a t

\frac{d^{2} y}{d x^{2}} = \frac{2 \log x - 3}{x^{3}}

Solution:

L e t 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}} (\frac{d \log x}{d x} = \frac{1}{x}) \\ \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}} \\ 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)}{{(x^{2})}^{2}} \\ \frac{d^{2} y}{d x^{2}} = \frac{x^{2} (\frac{- 1}{x}) - (1 - \log x) 2 x}{x^{4}} (\frac{d - \log x}{d x} = \frac{- 1}{x}, \frac{d}{d x} x^{2} = 2 x) \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

Answer:

\frac{- b^{4}}{a^{2} y^{3}}

Hint:
You must know about derivative of

t a n θ a n d s e c θ

Given:

I f x = a s e c θ, y = b t a n θ

Prove that

\frac{d^{2} y}{d x^{2}} = \frac{- b^{4}}{a^{2} y^{3}}

Solution:

L e t x = a s e c θ y = b t a n θ

\begin{aligned} \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} \\ \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{d y}{d x}) \\ x = a \sec θ, y = b \tan θ \end{aligned}

\begin{aligned} \frac{d x}{d θ} = a \sec θ \tan θ (\frac{d \tan θ}{d x} = \sec^{2} x, \frac{d \sec θ}{d x} = \sec θ \tan θ) \\ \frac{d y}{d θ} = b \sec^{2} θ \\ \frac{d y}{d x} = \frac{b}{a} \frac{\sec θ}{\tan θ} = \frac{b}{a \sin θ} \\ \frac{d}{d θ} (\frac{d y}{d x}) = - \frac{b}{a} cosec θ \cot θ \end{aligned}

\begin{aligned} = \frac{- \frac{b}{a} cosec θ \cot θ}{a \sec θ \tan θ} \\ \frac{d^{2} y}{d x^{2}} = \frac{- b}{a^{2} \tan^{3} θ} y = b \tan θ \\ \frac{d^{2} y}{d x^{2}} = \frac{- b}{a^{2} {(\frac{y}{b})}^{3} θ} y = \frac{- b^{4}}{a^{2} y^{3}} \end{aligned}

Higher Order Derivatives exercise 11.1 question 9

Answer:

\frac{s e c^{3} θ}{a θ}

Hint:
You must know about derivative of

c o s θ a n d s i n θ

Given:

\begin{aligned} If x = a (\cos θ + θ \sin θ), y = a (\sin θ - θ \cos θ) Prove that \\ \frac{d^{2} x}{d θ^{2}} = a (\cos θ - θ \sin θ), \frac{d^{2} y}{d θ^{2}} = a (\sin θ - θ \cos θ), \frac{d^{2} y}{d x^{2}} = \frac{\sec^{3} θ}{a θ} \end{aligned}

Solution:

\begin{aligned} Let x = a (\cos θ + θ \sin θ), y = a (\sin θ - θ \cos θ) \end{aligned}

Use multiplicative rule

\begin{aligned} As U V = U V^{1} + U^{1} V \\ Where U = θ & V = \sin θ \\ \begin{matrix} \frac{d x}{d θ} = a [- \sin θ + θ \cos θ + \sin θ] [\frac{d \sin θ}{d x} = \cos θ, \frac{d \cos θ}{d x} = - \sin θ] \\ \frac{d x}{d θ} = a θ \cos θ \end{matrix} \end{aligned}

again
Use multiplicative rule

\begin{aligned} As U V = U V^{1} + U^{1} V \\ Where U = θ & V = \cos θ \\ \frac{d y}{d θ} = a [\cos θ + θ \sin θ - \cos θ] \\ \frac{d y}{d θ} = a θ \sin θ \end{aligned}

Again Use multiplicative rule

\begin{aligned} As U V = U V^{1} + U^{1} V \\ Where U = θ & V = \cos θ \\ \frac{d^{2} x}{d θ^{2}} = a [- θ \sin θ + \cos θ] \\ \frac{d^{2} x}{d θ^{2}} = a [\cos θ - θ \sin θ] \end{aligned}

Again Use multiplicative rule

\begin{aligned} As U V = U V^{1} + U^{1} V \\ Where U = θ & V = \sin θ \\ \frac{d^{2} y}{d θ^{2}} = a {θ \cos θ + \sin θ} \\ \frac{d^{2} x}{d θ^{2}} = a (\cos θ - θ \sin θ) \end{aligned}

\begin{aligned} \frac{d^{2} y}{d θ^{2}} = a (θ \cos θ + \sin θ) \\ \frac{d y}{d x} = \frac{a θ \sin θ}{a θ \cos θ} = \tan θ \\ \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d θ} (\frac{d y}{d x}) \frac{d θ}{d x} \end{aligned}

\begin{aligned} \frac{d}{d θ} (\tan θ) \times \frac{1}{a θ \cos θ} \\ \frac{\sec^{2} θ}{a θ \cos θ} \\ \frac{d^{2} y}{d x^{2}} = \frac{\sec^{3} θ}{a θ} \end{aligned}

Higher Order Derivatives exercise 11.1 question 10

Answer:

2 e^{x} c o s (x + \frac{π}{2})

Hint:
You must know about derivative of e^x cos x
Given:
If y = e^x cos x, prove that

\frac{d^{2} y}{d x^{2}} = 2 e^{x} c o s (x + \frac{π}{2})

Solution:
Let y = e^x cos x
Use multiplicative rule

\begin{aligned} As U V = U V^{1} + U^{1} V \\ Where U = e^{x} & V = \cos x \\ \frac{d y}{d x} = e^{x} (- \sin x) + e^{x} \cos x (\frac{d}{d x} \cos x = \sin x) \end{aligned}

Again Use multiplicative rule
Differentiating again

\begin{aligned} \frac{d^{2} y}{d x^{2}} = - [e^{x} \cos x + \sin x e^{x}] + [- e^{x} \sin x + e^{x} \cos x] \\ \frac{d^{2} y}{d x^{2}} = - 2 \sin x e^{x} \\ \frac{d^{2} y}{d x^{2}} = 2 e^{x} \cos (x + \frac{π}{2}) \end{aligned}

Higher Order Derivatives exercise 11.1 question 11

Answer:

\frac{- b}{a^{2} y^{3}}

Hint:
You must know about derivative of second order
Given:

\begin{aligned} If x = a \cos θ, y = b \sin θ \\ Show that \frac{d^{2} y}{d x^{2}} = \frac{- b}{a^{2} y^{3}} \end{aligned}

Solution:

\begin{aligned} Let x = a \cos θ, y = b \sin θ \end{aligned}

\begin{aligned} \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} \\ \frac{d^{2} y}{d x^{2}} = \frac{\frac{d}{d θ} (\frac{d y}{d x})}{\frac{d x}{d θ}} \\ x = a \cos θ, y = b \sin θ \\ \frac{d x}{d θ} = - a \sin θ \frac{d y}{d θ} = b \cos θ \\ \frac{d y}{d x} = \frac{- b}{a} \cot θ \end{aligned}

\begin{aligned} \frac{\frac{d}{d θ} (\frac{d y}{d x})}{\frac{d x}{d θ}} = \frac{\frac{b}{a} \cos e c^{2} θ}{- a \sin θ} (\frac{d}{d θ} \cot θ = \cos e c^{2} θ) \\ \frac{- b \times b^{3}}{a^{2} \sin^{3} θ \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

Answer:

\frac{32}{27 a}

Hint:
You must know about derivative of

\frac{c o s^{3} θ}{s i n^{3} θ}

Given:

\begin{aligned} If x = a (1 - \cos^{3} θ), y = a \sin^{3} θ \\ Prove that \frac{d^{2} y}{d x^{2}} = \frac{32}{27 a} a t θ = \frac{π}{6} \end{aligned}

Solution:

\begin{aligned} Let x = a (1 - \cos^{3} θ), y = a \sin^{3} θ \end{aligned}

\begin{aligned} \frac{d y}{d θ} = 3 a \sin^{2} θ \cos θ (\frac{d}{d θ} \sin^{3} θ = 3 a \sin^{2} θ \cos θ) \\ \frac{d y}{d θ} = 3 \cos^{2} θ \sin θ (\frac{d}{d θ} \cos^{3} θ = 3 \cos^{2} θ \sin θ) \\ \frac{d y}{d x} = \tan θ (\frac{d}{d θ} \tan θ = \sec^{2} θ) \end{aligned}

\begin{aligned} \frac{\frac{d}{d θ} (\frac{d y}{d x})}{\frac{d x}{d θ}} = \frac{\sec^{2} θ}{3 a \cos^{2} θ \sin θ} \\ \frac{d^{2} y}{d x^{2}} = \frac{\sec^{4} θ}{3 a \sin θ} (θ = \frac{π}{6}) \\ \frac{d^{2} y}{d x^{2}} = \frac{\sec^{4} (\frac{π}{6})}{3 a \sin (\frac{π}{6})} = \frac{32}{27 a} \end{aligned}

Hence proved

Higher Order Derivatives exercise 11.1 question 13

Answer:

\frac{- a}{y^{2}}

HInt:
You must know about derivative of second order
Given:

\begin{aligned} If x = a (θ + \sin θ), y = a (1 + \cos θ) \\ Prove that \frac{d^{2} y}{d x^{2}} = \frac{- a}{y^{2}} \end{aligned}

Solution:

\begin{aligned} Let x = a (θ + \sin θ), y = a (1 + \cos θ) \end{aligned}

\begin{aligned} \frac{d x}{d θ} = a [1 + \cos θ] (\frac{d}{d θ} θ = 1, \frac{d}{d θ} \sin θ = \cos θ) \\ \frac{d y}{d θ} = a (- \sin θ) \\ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{- \sin θ}{(1 + \cos θ)} \end{aligned}

\begin{aligned} \sin 2 θ = 2 \sin θ \cos θ \\ \cos 2 θ = 2 \cos^{2} θ - 1 \\ \frac{d y}{d x} = \frac{- 2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \cos^{2} \frac{θ}{2}} \end{aligned}

\begin{aligned} \frac{d y}{d x} = - \tan \frac{θ}{2} \\ \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d θ} (\frac{d y}{d x}) \frac{d θ}{d x} \\ \frac{d}{d θ} (- \tan \frac{θ}{2}) \cdot \frac{1}{a (1 + \cos θ)} \end{aligned}

\begin{aligned} \frac{d^{2} y}{d x^{2}} = \frac{- 1}{2} \sec^{2} \frac{θ}{2} \cdot \frac{1}{a (1 + \cos θ)} \\ = \frac{- \sec^{2} \frac{θ}{2}}{2 a (1 + \cos θ)} \\ \frac{- a}{y^{2}} = \frac{- a}{a (1 + \cos θ)^{2}} = \frac{- a}{a^{2} (1 + \cos θ) (1 + \cos θ)} \end{aligned}

\begin{aligned} \frac{- 1}{a (1 + 2 \cos^{2} \frac{θ}{2} - 1) (1 - \cos θ)} \\ = \frac{- 1}{2 a \cos^{2} \frac{θ}{2} (1 + \cos θ)} = \frac{- \sec^{2} \frac{θ}{2}}{2 a (1 + \cos θ)} \end{aligned}

\begin{aligned} R H L : - \frac{- a}{y^{2}} = \frac{- a}{a (1 + \cos θ)^{2}} = \frac{- 1}{a (1 + 2 \cos^{2} \frac{θ}{2} - 1) (1 - \cos θ)} \\ \frac{d^{2} y}{d x^{2}} = \frac{- \sec^{2} \frac{θ}{2}}{2 a (1 + \cos θ)} = \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

Answer:

\frac{- c o s e c x \frac{4 θ}{2}}{4 a}

Hint:
You must know about derivative of

c o s θ a n d s i n θ

Given:

\begin{aligned} If x = a (θ - \sin θ), y = a (1 + \cos θ) \\ Find \frac{d^{2} y}{d x^{2}} \end{aligned}

Solution:

\begin{aligned} Let x = a (θ - \sin θ), y = a (1 + \cos θ) \end{aligned}

\begin{aligned} \frac{d x}{d θ} = a [1 - \cos θ] . . . . . . (1) \\ \frac{d y}{d θ} = a (- \sin θ) . . . . . . (2) \\ Now \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{\sin θ}{(1 - \cos θ)} \end{aligned}

Using identity

\begin{aligned} \sin 2 θ = 2 \sin \frac{θ}{2} \cos \frac{θ}{2} \\ 2 \sin^{2} \frac{θ}{2} = 1 - \cos θ \\ \cos 2 θ = 1 - 2 \sin^{2} θ \\ 2 \sin^{2} θ = 1 - \cos 2 θ \end{aligned}

\begin{aligned} \frac{d y}{d x} = \frac{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \sin^{2} \frac{θ}{2}} \\ \frac{d y}{d x} = \cot \frac{θ}{2} \end{aligned}

Now diff on both sides

\begin{aligned} \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d θ} (\frac{d y}{d x}) \frac{d θ}{d x} \\ \frac{d^{2} y}{d x^{2}} = - {cosec}^{2} θ \cdot \frac{1}{2} ∙ \frac{1}{a (1 - \cos θ)} \\ \frac{d^{2} y}{d x^{2}} = {cosec}^{2} \frac{θ}{2} ∙ \frac{1}{2} ∙ \frac{1}{a 2 \sin^{2} \frac{θ}{2}} \end{aligned}

\begin{aligned} = \frac{- \cos e c^{2} \frac{θ}{2} \times \cos e c^{2} \frac{θ}{2}}{4 a} \\ \frac{d^{2} y}{d x^{2}} = \frac{\cos e c^{4} \frac{θ}{2}}{4 a} \end{aligned}

Hence proved

Higher Order Derivatives exercise 11.1 question 15

Answer:

\frac{- 1}{a}

Hint:

\begin{aligned} You must know about derivative of \cos θ & \sin θ \end{aligned}

Given:

\begin{aligned} If x = a (1 - \cos θ), y = a (θ + \sin θ) \\ Prove that \frac{d^{2} y}{d x^{2}} = \frac{- 1}{a} a n d θ = \frac{π}{4} \end{aligned}

Solution:
$\begin{aligned} Let x = a (1 - \cos θ), y = a (θ + \sin θ) \end{aligned}$

\begin{aligned} \frac{d x}{d θ} = a \sin θ \\ \frac{d y}{d θ} = a (1 + \cos θ) \\ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{1 + \cos θ}{\sin θ} \end{aligned}

Use Quotient rule

\begin{aligned} As \frac{u}{v} = \frac{u^{1} v - v^{1} u}{v^{2}} \\ Where u = 1 + \cos θ & v = \sin θ \\ \frac{d}{d θ} (\frac{d y}{d x}) = \frac{\sin θ (- \sin θ) - (1 + \cos θ) \cos θ}{\sin^{2} θ} \end{aligned}

\begin{aligned} \frac{d}{d θ} (\frac{d y}{d x}) = \frac{- 1 + \cos θ}{(1 - \cos θ) (1 + \cos θ)} \\ \frac{d^{2} y}{d x^{2}} = \frac{\frac{- 1}{1 - \cos θ}}{a \sin^{2} θ} \\ \frac{d^{2} y}{d x^{2}} = \frac{- 1}{a \sin θ (1 - \cos θ)} (θ = \frac{π}{2}) \end{aligned}

\begin{aligned} \frac{d^{2} y}{d x^{2}} & = \frac{- 1}{a \sin \frac{π}{2} (1 - \cos \frac{π}{2})} \\ \frac{d^{2} y}{d x^{2}} & = \frac{- 1}{a} \end{aligned}

Hence proved

Higher Order Derivatives exercise 11.1 question 16 maths

Answer:

\frac{- 1}{a}

Hint:

\begin{aligned} You must know about derivative of \cos θ & \sin θ \end{aligned}

Given:

\begin{aligned} If x = a (1 + \cos θ), y = a (θ + \sin θ) \\ Prove that \frac{d^{2} y}{d x^{2}} = \frac{- 1}{a} a n d θ = \frac{π}{4} \end{aligned}

Solution:

\begin{aligned} Let x = a (1 + \cos θ), y = a (θ + \sin θ) \end{aligned}

\begin{aligned} \frac{d x}{d θ} = - a \sin θ \\ \frac{d y}{d θ} = a (1 + \cos θ) \\ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{1 + \cos θ}{- \sin θ} \end{aligned}

Use Quotient rule

\begin{aligned} As \frac{u}{v} = \frac{u^{1} v - v^{1} u}{v^{2}} \\ Where u = 1 + \cos θ & v = \sin θ \\ \frac{d}{d θ} (\frac{d y}{d x}) = \frac{- \sin θ (- \sin θ) + (1 + \cos θ) \cos θ}{\sin^{2} θ} \end{aligned}

\begin{aligned} \frac{d^{2} y}{d x^{2}} = \frac{1 + \cos θ}{\sin^{2} θ} \cdot \frac{- 1}{a \sin θ} \\ \frac{d^{2} y}{d x^{2}} = \frac{- (1 + \cos θ)}{a \sin^{2} θ} \\ \frac{d^{2} y}{d x^{2}} = \frac{- (1 + \cos \frac{π}{2})}{a \sin^{2} \frac{π}{2}} (θ = \frac{π}{2}) \\ \frac{d^{2} y}{d x^{2}} = \frac{- 1}{a} \end{aligned}

Hence proved

Higher Order Derivatives exercise 11.1 question 17

Answer:

\begin{aligned} 3 \sin^{2} θ [5 \cos^{2} θ - 1] \end{aligned}

Hint:

\begin{aligned} You must know about derivative of \cos θ & \sin θ \end{aligned}

Given:

\begin{aligned} If x = \cos θ, y = \sin^{3} θ \\ Prove that y \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2} = 3 \sin^{2} θ [5 \cos^{2} θ - 1] \end{aligned}

Solution:

\begin{aligned} Let x = \cos θ, y = \sin^{3} θ \end{aligned}

\begin{aligned} \frac{d x}{d θ} = - \sin θ \\ \frac{d y}{d θ} = 3 \sin^{2} θ \cos θ \\ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{3 \sin^{2} θ \cos θ}{- \sin θ} = 3 \sin θ \cos θ \end{aligned}

\begin{aligned} \frac{d^{2} y}{d x^{2}} = d (\frac{\frac{d y}{d x}}{d x}) = \frac{d (\frac{d y}{d θ}) d θ}{\frac{d x}{d θ}} \\ = \frac{- \cos^{2} θ \cdot 3 + 3 \sin^{2} θ}{- \sin θ} \end{aligned}

\begin{aligned} y \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2} = \frac{\sin^{3} θ (3 \cos^{2} θ - 3 \sin^{2} θ)}{\sin θ} \\ y \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2} = 3 \sin^{2} θ [5 \cos^{2} θ - 1] \end{aligned}

Hence proved

Higher Order Derivatives exercise 11.1 question 18

Answer:
0
Hint:
You must know about derivative of

c o s θ a n d t a n θ

Given:

I f y = s i n (s i n x)

\begin{aligned} P r o v e t h a t \frac{d^{2} y}{d x^{2}} + \tan x \frac{d y}{d x} + y \cos^{2} x = 0 \end{aligned}

Solution:

L e t y = s i n (s i n 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

Answer:
0
HInt:
You must know about derivative of sin pt
Given:

I f x = s i n t, y = s i n p t

Prove that (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + p^{2} y = 0

Solution:

L e t x = s i n t, y = s i n p t

\begin{aligned} \frac{d x}{d t} = \cos t and \frac{d y}{d t} = p \cos p t \\ 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} {(\frac{d y}{d x})}^{2} (1 - \sin^{2} t) = p^{2} (1 - \sin^{2} p t) \\ {(\frac{d y}{d x})}^{2} (1 - x^{2}) = p^{2} (1 - y^{2}) \end{aligned}

Differentiating with x

\begin{aligned} (1 - x^{2}) 2 \frac{d y}{d x} \times \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2} (- 2 x) \\ = - p^{2} \times 2 y \frac{d y}{d x} \\ 2 \frac{d y}{d x} [(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x}] = p^{2} y 2 \frac{d y}{d x} \\ (1 - x^{2}) \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

Answer:
0
Hint:
You must know about derivative of sin^-1 x
Given:

I f y = (s i n^{- 1} x)^{2} P r o v e t h a t (1 - x^{2}) y_{2} - x y_{1} - 2 = 0

Solution:

L e t y = (s i n^{- 1} x)^{2}

\begin{aligned} \frac{d y}{d x} = 2 \sin^{- 1} x (\frac{1}{\sqrt{1 - x^{2}}}) (\frac{d}{d x} \sin^{- 1} x = (\frac{1}{\sqrt{1 - x^{2}}})) \\ \frac{d^{2} y}{d x^{2}} = \frac{2 \sqrt{1 - x^{2}} (\frac{1}{\sqrt{1 - x^{2}}}) - \sin^{- 1} x \times \frac{1}{2} (\frac{- 2 x}{\sqrt{1 - x^{2}}})}{1 - x^{2}} \end{aligned}

\begin{aligned} \frac{d^{2} y}{d x^{2}} = \frac{2 (1 + \frac{x \sin^{- 1} x}{\sqrt{1 - x^{2}}})}{1 - x^{2}} \\ (1 - x^{2}) y_{2} = \frac{2 + 2 x \sin^{- 1} x}{\sqrt{1 - x^{2}}} \\ (1 - x^{2}) y_{2} - y_{1} x - 2 = 0 \end{aligned}

Hence proved

Higher Order Derivatives exercise 11.1 question 21

Answer:
Proved
Hint:
You must know about the derivative of exponential function and tangent inverse

x

Given:
$y = e^{t a n^{- 1} x}, P r o v e (1 + x^{2}) y_{2} + (2 x - 1) y_{1} = 0$
Solution:

L e t y = e^{t a n^{- 1} x}

\begin{aligned} \frac{d y}{d x} = e^{\tan^{- 1} x} \times \frac{1}{(1 + x^{2})} \\ (x^{2} + 1) \frac{d y}{d x} = e^{\tan^{- 1} x} \\ (x^{2} + 1) \frac{d^{2} y}{d x^{2}} + 2 x \frac{d y}{d x} = e^{\tan^{- 1} x} \times \frac{1}{(1 + x^{2})} \end{aligned}

\begin{aligned} (x^{2} + 1) \frac{d^{2} y}{d x^{2}} + 2 x \frac{d y}{d x} = \frac{d y}{d x} \\ (x^{2} + 1) \frac{d^{2} y}{d x^{2}} + (2 x - 1) \frac{d y}{d x} = 0 \\ or \\ (x^{2} + 1) y_{2} + (2 x - 1) y_{1} = 0 \end{aligned}

Higher Order Derivatives exercise 11.1 question 22

Answer:
Proved
Hint:
You must know about the derivative of logarithm function and cos x and sin x
Given:

y = 3 c o s (l o g x) + 4 s i n (l o g x)

Solution:

L e t y = 3 c o s (l o g x) + 4 s i n (l o g x)

\begin{aligned} 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} 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} Again differentiating both sides w.r.t x, \\ x \frac{d}{d x} (\frac{d y}{d x}) + \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} 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} ∴ x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + y = 0 \\ or \\ x^{2} y_{2} + x y_{1} + y = 0 \end{aligned}

Higher Order Derivatives exercise 11.1 question 23

Answer:
Proved
Hint:
You must know about the derivative of exponential function
Given:

y = e^{2 x} (a x + b), s h o w t h a t y_{2} - 4 y_{1} + 4 y = 0

Solution:

y = e^{2 x} (a x + b) . . . . . . (1)

\begin{aligned} 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} Again differentiating both sides w.r.t x, 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} \dots . (3) \end{aligned}

I n o r d e r t o p r o v e t h e e x p r e s s i o n t r y t o g e t t h e r e q u i r e d f o r m S u b t r a c t i n g 4 \times e q u a t i o n (2) f r o m e q u a t i o n (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} 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

Answer:
Proved
Hint:
You must know about the derivative of sin function and logarithm function
Given:

x = \sin (\frac{1}{a} \log y) show that (1 - x^{2}) y_{2} - x y_{1} - a^{2} y = 0

Solution:

\begin{aligned} x = \sin (\frac{1}{a} \log y) \\ \log y = a \sin^{- 1} x \\ y = e^{a \sin^{- 1} x} . . . . . (1) \end{aligned}

\begin{aligned} Let t = a \sin^{- 1} x \\ \frac{d t}{d x} = \frac{a}{\sqrt{1 - x^{2}}} [\frac{d}{d x} \sin^{- 1} x = \frac{1}{\sqrt{1 - x^{2}}}] \\ 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} Again differentiating both sides \\ \frac{d^{2} y}{d x^{2}} = a e^{a \sin^{- 1} x} \frac{d}{d x} (\frac{1}{\sqrt{1 - x^{2}}}) + \frac{a}{\sqrt{1 - x^{2}}} \frac{d}{d x} e^{a \sin^{- 1} x} \end{aligned}

\begin{aligned} Using chain rule and equation \\ \frac{d^{2} y}{d x^{2}} = \frac{- a e^{a \sin^{- 1} x}}{2 (1 - x^{2}) \sqrt{1 - x^{2}}} (- 2 x) + \frac{a^{2} e^{a \sin^{- 1} x}}{(1 - x^{2})} \\ \frac{d^{2} y}{d x^{2}} = \frac{x a e^{a \sin^{- 1} x}}{(1 - x^{2}) \sqrt{1 - x^{2}}} + \frac{a^{2} e^{a \sin^{- 1} x}}{(1 - x^{2})} \\ (1 - x^{2}) \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} Using equation (1) and (2) \\ (1 - x^{2}) \frac{d^{2} y}{d x^{2}} = a^{2} y + x \frac{d y}{d x} \\ (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - a^{2} y - x \frac{d y}{d x} = 0 \\ (1 - x^{2}) y_{2} - x y_{1} - a^{2} y = 0 \end{aligned}

Higher Order Derivatives exercise 11.1 question 25

Answer:
Proved
HInt:
You must know the derivative of logarithm and tangent inverse

x

Given:

\log y = \tan^{- 1} x, show (1 + x^{2}) y_{2} + (2 x - 1) y_{1} = 0

Solution:

\begin{aligned} \log y = \tan^{- 1} x \\ 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}

D i f f e r e n t i a t e a g a i n (1 + x^{2}) \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} (2 x) = \frac{d y}{d x} (1 + x^{2}) \frac{d^{2} y}{d x^{2}} + (2 x - 1) \frac{d y}{d x} = 0 o r (1 + x^{2}) y_{2} + (2 x - 1) y_{1} = 0

Higher Order Derivatives exercise 11.1 question 26

Answer:
Proved
Hint:
You must know the derivative of logarithm and tangent inverse

x

Given:

y = \tan^{- 1} x, show (1 + x^{2}) \frac{d^{2} y}{d x^{2}} + (2 x) \frac{d y}{d x} = \frac{d y}{d x}

Solution:

y = \tan^{- 1} x

D i f f e r e n t i a t e t h e e q u a t i o n w . r . t x \frac{d y}{d x} = \frac{1}{1 + x^{2}} (1 + x^{2}) \frac{d y}{d x} = 1 D i f f e r e n t i a t e a g a i n (1 + x^{2}) \frac{d^{2} y}{d x^{2}} + (2 x) \frac{d y}{d x} = 0 o r (1 + x^{2}) y_{2} + (2 x) y_{1} = 0

Higher Order Derivatives exercise 11.1 question 27

Answer:
Proved
Hint:
You must know the derivative of logarithm and tangent inverse

x

Given:

y = {\log (x + \sqrt{x^{2} + 1})}^{2}, show (1 + x^{2}) \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} = 2

Solution:

y = {\log (x + \sqrt{x^{2} + 1})}^{2}

\begin{aligned} Differentiate the equation w.r.t x \\ \frac{d y}{d x} = 2 \log (x + \sqrt{x^{2} + 1}) \cdot \frac{1}{x + \sqrt{x^{2} + 1}} \times (1 + \frac{2 x}{2 \sqrt{x^{2} + 1}}) \\ = \frac{2 \log (x + \sqrt{x^{2} + 1})}{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 (x + \sqrt{x^{2} + 1}) \end{aligned}

\begin{aligned} 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 (1 + \frac{2 x}{2 \sqrt{x^{2} + 1}}) \\ \frac{(x^{2} + 1) \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}} \\ (1 + x^{2}) \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

Answer:
Proved
Hint:
You must know the derivative of logarithm and tangent inverse

x

Given:

y = {(\tan^{- 1} x)}^{2}, show {(1 + x^{2})}^{2} y_{2} + 2 x (1 + x^{2}) y_{1} = 2

Solution:

\begin{aligned} y = {(\tan^{- 1} x)}^{2} \\ Differentiate the equation w.r.t x \\ \frac{d y}{d x} = 2 \tan^{- 1} x \cdot \frac{1}{1 + x^{2}} \\ (1 + x^{2}) \frac{d y}{d x} = 2 \tan^{- 1} x \end{aligned}

\begin{aligned} Differentiate again \\ (1 + x^{2}) \frac{d^{2} y}{d x^{2}} + (2 x) \frac{d y}{d x} = \frac{2}{1 + x^{2}} \\ {(1 + x^{2})}^{2} \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} (1 + x^{2}) (2 x) = 2 \\ o r \\ {(1 + x^{2})}^{2} y_{2} + 2 x (1 + x^{2}) y_{1} = 2 \end{aligned}

Higher Order Derivatives exercise 11.1 question 29

Answer:
Proved
Hint:
You must know the derivative of cot x function
Given:

y = \cot x, show \frac{d^{2} y}{d x^{2}} + (2 y) \frac{d y}{d x} = 0

Solution:

L e t y = \cot x

\begin{aligned} Differentiate the equation w.r.t x \\ \frac{d y}{d x} = \frac{d (\cot x)}{d x} \\ \frac{d y}{d x} = - {cosec}^{2} x \end{aligned}

\begin{aligned} Differentiate again \\ \frac{d^{2} y}{d x^{2}} = - [2 c o s e c x (- c o s e c x \cot x)] \\ \frac{d^{2} y}{d x^{2}} = - 2 {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

Answer:

\frac{- 2}{x^{2}}

Hint:
You must know the derivative of logarithm function
Given:

y = \log (\frac{x^{2}}{e^{2}}), find \frac{d^{2} y}{d x^{2}}

Solution:
$L e t y = \log (\frac{x^{2}}{e^{2}})$
$D i f f e r e n t i a t e t h e e q u a t i o n w . r . t x \frac{d y}{d x} = \frac{1}{\frac{x^{2}}{e^{2}}} \cdot \frac{1}{e^{2}} \cdot 2 x = \frac{2}{x} D i f f e r e n t i a t e a g a i n \frac{d^{2} y}{d x^{2}} = - 2 [\frac{1}{x^{2}}] = \frac{- 2}{x^{2}} H e n c e, \frac{d^{2} y}{d x^{2}} = \frac{- 2}{x^{2}}$

Higher Order Derivatives exercise 11.1 question 31

Answer:
Proved
Hint:
You must know the derivative of exponential function
Given:

y = a e^{2 x} + b e^{- x}, 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} 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} 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 = RHS \end{aligned}

Higher Order Derivatives exercise 11.1 question 31

Answer:
Proved
Hint:
You must know the derivative of exponential function
Given:

y = a e^{2 x} + b e^{- x}, 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} 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} 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 = RHS \end{aligned}

Higher Order Derivatives exercise 11.1 question 32 maths

Answer:
Proved
Hint:
You must know the derivative of exponential sin and cos functions
Given:

y = e^{x} (\sin x + \cos x), 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} 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} 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} \\ ∴ \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

Answer:

- c o t y . c o s e c^{2} y

Hint:
You must know the derivative of cos inverse function
Given:

y = \cos^{- 1} x, find \frac{d^{2} y}{d x^{2}} in terms of y alone

Solution:

y = \cos^{- 1} x

\begin{aligned} Differentiate the equation w.r.t x \\ \frac{d y}{d x} = \frac{d (\cos^{- 1} x)}{d x} \\ = \frac{- 1}{\sqrt{1 - x^{2}}} = - {(1 - x^{2})}^{\frac{- 1}{2}} \\ \frac{d^{2} y}{d x^{2}} = \frac{d [- {(1 - x^{2})}^{\frac{- 1}{2}}]}{d x} \end{aligned}

\begin{aligned} = - (\frac{- 1}{2}) {(1 - x^{2})}^{\frac{- 3}{2}} \times (- 2 x) \\ = \frac{1}{2 \sqrt{{(1 - x^{2})}^{3}}} \times (- 2 x) \\ \frac{d^{2} y}{d x^{2}} = \frac{- x}{\sqrt{{(1 - x^{2})}^{3}}} \end{aligned}

\begin{aligned} y = \cos^{- 1} x \\ x = \cos y \\ Put in above equation \\ \frac{d^{2} y}{d x^{2}} = \frac{- \cos y}{\sqrt{{(1 - \cos^{2} y)}^{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} \\ ∴ \frac{d^{2} y}{d x^{2}} = - \cot y \cdot {cosec}^{2} y \end{aligned}

Higher Order Derivatives exercise 11.1 question 34

Answer:
Proved
Hint:
You must know the derivative of exponential and cos inverse function
Given:

y = e^{a \cos^{- 1} x}, prove (1 - x^{2}) \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} 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 (\frac{- 1}{\sqrt{1 - x^{2}}}) \\ \frac{d y}{d x} = \frac{a y}{\sqrt{1 - x^{2}}} \end{aligned}

\begin{aligned} Squaring both sides, \\ {(\frac{d y}{d x})}^{2} = \frac{a^{2} y^{2}}{(1 - x^{2})} \\ (1 - x^{2}) {(\frac{d y}{d x})}^{2} = a^{2} y^{2} \end{aligned}

\begin{aligned} Again differentiate, \\ {(\frac{d y}{d x})}^{2} (- 2 x) + (1 - x^{2}) \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} + (1 - x^{2}) \frac{d^{2} y}{d x^{2}} = a^{2} y \\ (1 - x^{2}) \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

Answer:
Proved
HInt:
You must know the derivative of exponential function
Given:

y = 500 e^{7 x} + 600 e^{- 7 x}, 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) [(500) e^{7 x} + (600) e^{- 7 x}] \\ \frac{d^{2} y}{d x^{2}} = 49 y \end{aligned}

Higher Order Derivatives exercise 11.1 question 36 maths

Answer:

\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, find \frac{d^{2} y}{d x^{2}} at t = \frac{π}{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} 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} 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} \\ {\frac{d^{2} y}{d x^{2}}]}_{t = \frac{π}{2}} = \frac{- 3}{2} \end{aligned}

Higher Order Derivatives exercise 11.1 question 37

Answer:

\frac{- 7}{64 z^{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, 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 \\ 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}

A g a i n d i f f e r e n t i a t i n g w . r . t z \frac{d^{2} y}{d x^{2}} \times \frac{d x}{d z} = - \frac{7}{8 z^{2}} o r \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

Answer:
Proved
Hint:
You must know the derivative of logarithm and cos function
Given:

y = \log (1 + \cos x), 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} \\ 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}} + (\frac{- 1}{1 + \cos x}) (\frac{- \sin x}{1 + \cos x}) = 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

Answer:
Proved
Hint:
You must know the derivative of sin and logarithm function
Given:

y = \sin (\log x), 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} \\ Again differentiating \\ \frac{d^{2} y}{d x^{2}} = \frac{x [- \sin (\log x) \times \frac{1}{x}] - \cos (\log x)}{x^{2}} \\ = \frac{- \cos (\log x) - \sin (\log x)}{x^{2}} \end{aligned}

\begin{aligned} Now, \\ 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 = RHS \end{aligned}

Higher Order Derivatives exercise 11.1 question 40 maths

Answer:
Proved
Hint:
You must know the derivative of exponential function
Given:

y = 3 e^{2 x} + 2 e^{3 x}, 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 (y - 3 e^{2 x})}{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} Again differentiating \\ \frac{d^{2} y}{d x^{2}} = \frac{3 d y}{d x} - 6 e^{2 x} \dots \dots \dots . (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} Put in (1), \\ \frac{d^{2} y}{d x^{2}} = \frac{3 d y}{d x} - 6 (\frac{\frac{d y}{d x} - 3 y}{- 3}) \\ \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

Answer:
Proved
Hint:
You must know the derivative of cot inverse

x

Given:

y = {(\cot^{- 1} x)}^{2}, prove y_{2} {(x^{2} + 1)}^{2} + 2 x (x^{2} + 1) y_{1} = 2

Solution:

y = {(\cot^{- 1} x)}^{2}

\begin{aligned} \frac{d y}{d x} = 2 \cot^{- 1} x (\frac{- 1}{1 + x^{2}}) \\ (1 + x^{2}) \frac{d y}{d x} = - 2 \cot^{- 1} x \\ Again differentiating \\ {(1 + x^{2})}^{2} \frac{d^{2} y}{d x^{2}} + 2 x \frac{d y}{d x} = - 2 (\frac{- 1}{1 + x^{2}}) \end{aligned}

\begin{aligned} {(1 + x^{2})}^{2} \frac{d^{2} y}{d x^{2}} + 2 x \frac{d y}{d x} = (\frac{2}{1 + x^{2}}) \\ {(1 + x^{2})}^{2} \frac{d^{2} y}{d x^{2}} + 2 x (1 + x^{2}) \frac{d y}{d x} = 2 \\ y_{2} {(x^{2} + 1)}^{2} + 2 x (x^{2} + 1) y_{1} = 2 \end{aligned}

Higher Order Derivatives exercise 11.1 question 42

Answer:
Proved
Hint:
You must know the derivative of cosec^-1x
Given:

y = ({cosec}^{- 1} x), x > 1, prove x (x^{2} - 1) \frac{d^{2} y}{d x^{2}} + (2 x^{2} - 1) \frac{d y}{d x} = 0

Solution:

\begin{aligned} y = (\cos e c^{- 1} x) \\ \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} 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 (x^{2} - 1) \frac{d^{2} y}{d x^{2}} + (2 x^{2} - 1) \frac{d y}{d x} = 0 \end{aligned}

Higher Order Derivatives exercise 11.1 question 43

Answer:

2 \sqrt{2} = \frac{d^{2} y}{d x^{2}} 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 find \frac{d^{2} y}{d x^{2}} & \frac{d^{2} y}{d x^{2}} at t = \frac{π}{4} \end{aligned}

Solution:

\begin{aligned} y = \sin t \frac{d y}{d t} = \cos t \\ \frac{d^{2} y}{d t^{2}} = - \sin t \\ {\frac{d^{2} y}{d t^{2}}]}_{t = \frac{π}{4}} = - \sin \frac{π}{4} = \frac{- 1}{\sqrt{2}} \end{aligned}

\begin{aligned} 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 + cosec t \\ Now, \frac{d y}{d x} = \frac{d y}{d t} \times \frac{d t}{d x} = \frac{\cos t}{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 (\frac{d y}{d x})}{d x} \end{aligned}

\begin{aligned} = \frac{\frac{d}{d t} (\frac{d y}{d x})}{\frac{d x}{d t}} = \frac{\sec^{2} t}{cosec t - \sin t} \\ = \frac{\sec^{2} t \cdot \sin t}{\cos^{2} t} = \sec^{2} t \tan t \\ {\frac{d^{2} y}{d x^{2}}]}_{- \frac{π}{4}} = 2 \sqrt{2} \times 1 \\ = 2 \sqrt{2} \end{aligned}

Higher Order Derivatives exercise 11.1 question 44 maths

Answer:

\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 (\cos t + \log \tan \frac{t}{2}) 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 (\cos t + \log \tan \frac{t}{2}) \\ \frac{d y}{d t} = a (- \sin t + \cot \frac{t}{2} \times \sec^{2} \frac{t}{2} \times \frac{t}{2}) \end{aligned}

\begin{aligned} = a (- \sin t + \frac{1}{2 \sin (\frac{t}{2}) \times \cos (\frac{t}{2})}) \\ \frac{d y}{d t} = a (- \sin t + \frac{1}{\sin t}) \\ = a (\frac{- \sin^{2} t + 1}{\sin t}) \Rightarrow \frac{a \cos^{2} t}{\sin t} \end{aligned}

\begin{aligned} ∴ \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 \\ Again \frac{d^{2} y}{d x^{2}} = - c o s e c^{2} t \frac{d t}{d x} = c o s e 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

Answer:

\frac{8 \sqrt{2}}{π 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) find \frac{d^{2} y}{d x^{2}} at t = \frac{π}{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} ∴ \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 \\ 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} \\ {\frac{d^{2} y}{d x^{2}}]}_{t = \frac{π}{4}} = \frac{\sec^{2} \frac{π}{4}}{a \frac{π}{4}} = \frac{2 \sqrt{2} \times 4}{a π} \\ = \frac{8 \sqrt{2}}{π a} \end{aligned}

Higher Order Derivatives exercise 11.1 question 46

Answer:

\frac{8 \sqrt{3}}{a}

Hint:
You must know the derivative of cos, sin, tan and logarithm function
Given:

\begin{aligned} x = a (\cos t + \log \tan \frac{t}{2}) \\ y = a \sin t find \frac{d^{2} y}{d x^{2}} at t = \frac{π}{3} \end{aligned}

Solution:

\begin{aligned} x = a (\cos t + \log \tan \frac{t}{2}) \\ \frac{d x}{d t} = a [- \sin t + \frac{1}{\tan \frac{t}{2}} \sec^{2} \frac{t}{2} \cdot \frac{1}{2}] \\ \frac{d x}{d t} = a [- \sin t + \frac{\cos \frac{t}{2}}{2 \sin \frac{t}{2}} \frac{1}{\cos^{2} \frac{t}{2}}] \end{aligned}

\begin{aligned} \frac{d x}{d t} = a [- \sin t + \frac{1}{2 \sin \frac{t}{2} \cos \frac{t}{2}}] \\ \frac{d x}{d t} = a [- \sin t + \frac{1}{\sin t}] \\ \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 \\ ∴ \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 \\ Again \frac{d^{2} y}{d x^{2}} = \sec^{2} t \times \frac{\sin t}{a \cos^{2} t} \end{aligned}

\begin{aligned} {\frac{d^{2} y}{d x^{2}}]}_{t = \frac{π}{3}} = \frac{\sec^{2} \frac{π}{3} \cdot \sin \frac{π}{3}}{a \cos^{2} \frac{π}{4}} = \frac{(2)^{2} \cdot (\frac{\sqrt{3}}{2})}{a {(\frac{1}{2})}^{2}} \\ {\frac{d^{2} y}{d x^{2}}]}_{t = \frac{π}{3}} = \frac{8 \sqrt{3}}{a} \end{aligned}

Higher Order Derivatives exercise 11.1 question 47

Answer:

\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) 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} 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

Answer:

\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 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 + (6 \cos^{2} t - 3) \\ y = 3 \sin t - 2 \sin^{3} t \\ \frac{d y}{d t} = 3 \cos t - 6 \sin^{2} t \cos t \\ = \cos t (3 - 6 \sin^{2} t) \end{aligned}

\begin{aligned} ∴ \frac{d y}{d x} = \frac{d y}{d t} \times \frac{d x}{d t} = \frac{\cos t (3 - 6 \sin^{2} t)}{\sin t + (6 \cos^{2} t - 3)} \\ = \frac{\cos t (1 - 2 \sin^{2} t)}{3 (2 \cos^{2} t - 1)} \\ = \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}} = - c o s e c^{2} t \end{aligned}

Higher Order Derivatives exercise 11.1 question 49

Answer:

\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 Prove \frac{d^{2} y}{d x^{2}} = \frac{- (x^{2} + y^{2})}{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} \\ Use quotient rule \\ \frac{U}{V} = \frac{U V^{'} - U^{'} V}{V^{2}} \end{aligned}

\begin{aligned} \frac{d^{2} y}{d x^{2}} = \frac{\frac{d (\frac{d y}{d x})}{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{- (x^{2} + y^{2})}{y^{3}} \end{aligned}

Higher Order Derivatives exercise 11.1 question 50

Answer:

A = \frac{2}{3} a n d B = \frac{- 1}{3}

Hint:
You must know the derivative of second order
Given:

Find A & B so that y = A \sin 3 x + B \cos 3 x 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} 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) [\begin{matrix} \frac{d \cos 3 x}{d x} = - 3 \sin 3 x \\ \frac{d \sin 3 x}{d x} = 3 \cos 3 x \end{matrix}] \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 \dots \dots . . (1) \\ - 6 B + 12 A = 0 \dots \dots . . (2) \\ 6 A = - 12 B \\ A = - 2 B \dots \dots (3) \end{aligned}

\begin{aligned} Put (3) in (2) \\ - 6 B + (- 2 B) 12 = 10 \\ - 6 B - 24 B = 10 \\ - 30 B = 10 \\ B = \frac{- 1}{3} \\ A = - 2 (\frac{- 1}{3}) = \frac{2}{3} \\ A = \frac{2}{3} & B = \frac{- 1}{3} \end{aligned}

Higher Order Derivatives exercise 11.1 question 51

Answer:

\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:

If y = A e^{k t} \cos (p t + c) prove that \frac{d^{2} y}{d t^{2}} + 2 k \frac{d y}{d t} + n^{2} y = 0 where n^{2} = p^{2} + k^{2}

Solution:

\begin{aligned} Let y = A e^{- k t} \cos (p t + c) \dots . . (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)) \dots . . (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}} = (k^{2} - p^{2}) A e^{- k t} \cos (p t + c) + 2 k (A p e^{- k t} \sin (p t + c)) \dots . . (3) \end{aligned}

\begin{aligned} Using (1) & (2) & n^{2} = k^{2} + p^{2} \\ \frac{d^{2} y}{d t^{2}} = (k^{2} - p^{2}) y + 2 k (- k y - \frac{d y}{d t}) \\ \frac{d^{2} y}{d t^{2}} = (k^{2} - p^{2} - 2 k^{2}) y + 2 k (\frac{d y}{d t}) \\ \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

Answer:

x^{2} \frac{d^{2} y}{d x^{2}} + x (1 - 2 n) \frac{d y}{d x} + (1 + n^{2}) y = 0

Hint:
You must know the derivative of

c o s (l o g x) a n d s i n (l o g x)

Given:

If y = x^{n} {acos (\log x) + b \sin (\log x)} prove that x^{2} \frac{d^{2} y}{d x^{2}} + x (1 - 2 n) \frac{d y}{d x} + (1 + n^{2}) y = 0

Solution:

\begin{aligned} Let y = x^{n} {a \cos (\log x) + b \sin (\log x)} \\ Use multiplicative rule \\ UV = U^{'} V + {UV}^{'} \\ 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} {acos (\log x) + b \sin (\log x)} + x^{n} {\frac{a \sin (\log x)}{x} + \frac{b \cos (\log x)}{x}} \\ \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} {\frac{a \sin (\log x)}{x} + \frac{b \cos (\log x)}{x}} \\ \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} {\frac{a \sin (\log x)}{x} + \frac{b \cos (\log x)}{x}} \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} (x \frac{d y}{d x} - n y) - \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) - (1 + n^{2}) y = 0 \\ x^{2} \frac{d^{2} y}{d x^{2}} + x (1 - 2 n) \frac{d y}{d x} + (1 + n^{2}) 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.

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Frequently Asked Questions (FAQs)

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.

3. How will the students be able to use the RD Sharma books when they do not have the access to the internet?

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.

4. What is the price quoted for the RD Sharma books at the Career 360 website?

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

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