NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability (Ex 5.4) Exercise 5.4

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NCERT Solutions for Class 12 Maths Chapter 5 Other Exercises

Chapter 5 – Continuity and Differentiability Exercises
Exercise 5.1	10 Short Questions and 24 Long Questions
Exercise 5.2	2 Short Questions 8 Long Questions
Exercise 5.3	9 short Questions and 6 long Questions
Exercise 5.5	4 Short Questions and 14 Long Questions
Exercise 5.6	1 Short Question and 1 Long Question
Exercise 5.7	10 Short Questions and 7 Long Questions
Exercise 5.8	Questions with Solutions

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Q.1

$Differentiate the following w . r . t . x : \frac{e^{x}}{sinx}$

Ans

$\begin{array}{l} Let y = \frac{e^{x}}{sinx} \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} \frac{e^{x}}{sinx} \\ = \frac{sinx \frac{d}{dx} e^{x} - e^{x} \frac{d}{dx} sinx}{{(sinx)}^{2}} [By Quotient Rule] \\ = \frac{sinx . e^{x} - e^{x} cosx}{{(sinx)}^{2}} \\ = \frac{e^{x} (sinx - cosx)}{\sin^{2} x}, x \neq nπ, n \in Z \end{array}$

Q.2

$D i f f e r e n t i a t e t h e f o l l o w i n g w . r . t . x : e^{s i n^{- 1} x}$

Ans

$\begin{array}{l} Let y = e^{\sin^{- 1} x} \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} e^{\sin^{- 1} x} \\ = e^{\sin^{- 1} x} \frac{d}{dx} \sin^{- 1} x [By Chain Rule] \\ = e^{\sin^{- 1} x} \times \frac{1}{\sqrt{1 - x^{2}}} \\ = \frac{e^{\sin^{- 1} x}}{\sqrt{1 - x^{2}}}, x \in (- 1, 1) \end{array}$

Q.3

$D i f f e r e n t i a t e t h e f o l l o w i n g w . r . t . x : e^{x^{3}}$

Ans

$\begin{array}{l} Let y = e^{x^{3}} \\ Differentiating both sides w .r .t . x, we get \\ \frac{d y}{d x} = \frac{d}{d x} e^{x^{3}} \\ = e^{x^{3}} \frac{d}{d x} x^{3} [By Chain Rule] \\ = e^{x^{3}} \times 3 x^{2} \\ = 3 x^{2} {.e}^{x^{3}} \end{array}$

Q.4

$Differentiate the following w . r . t . x : sin {(tan}^{- 1} e^{- x})$

Ans

$\begin{array}{l} Let y = \sin (\tan^{- 1} e^{- x}) \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} \sin (\tan^{- 1} e^{- x}) \\ = \cos (\tan^{- 1} e^{- x}) \frac{d}{dx} \tan^{- 1} e^{- x} [By Chain Rule] \\ = \cos (\tan^{- 1} e^{- x}) \times \frac{1}{1 + {(e^{- x})}^{2}} \times \frac{d}{dx} e^{- x} \\ = \frac{\cos (\tan^{- 1} e^{- x})}{1 + {(e^{- x})}^{2}} \times e^{- x} \frac{d}{dx} (- x) \\ = \frac{e^{- x} \cos (\tan^{- 1} e^{- x})}{1 + {(e^{- x})}^{2}} \times (- 1) \\ = - \frac{e^{- x} \cos (\tan^{- 1} e^{- x})}{1 + e^{- 2 x}} \end{array}$

Q.5

$Differentiate the following w . r . t . x : log {(cos e}^{x})$

Ans

$\begin{array}{l} Let y = \log ({cose}^{x}) \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} \log ({cose}^{x}) \\ = \frac{1}{{cose}^{x}} \times \frac{d}{dx} {cose}^{x} [By Chain Rule] \\ = \frac{1}{{cose}^{x}} \times - {sine}^{x} \times \frac{d}{dx} e^{x} \\ = \frac{- {sine}^{x}}{{cose}^{x}} \times e^{x} \\ = \frac{- e^{x} {sine}^{x}}{{cose}^{x}} \\ = - e^{x} {tane}^{x}, e^{x} \neq (2 n + 1) \frac{π}{2}, n \in N \end{array}$

Q.6

$\begin{array}{l} Differentiate the following w . r . t . x : \\ e^{x} + e^{x^{2}} + . .. + e^{x^{5}} \end{array}$

Ans

$\begin{array}{l} Let y = e^{x} + e^{x^{2}} + . .. + e^{x^{5}} \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} (e^{x} + e^{x^{2}} + . .. + e^{x^{5}}) \\ = \frac{d}{dx} e^{x} + \frac{d}{dx} e^{x^{2}} + \frac{d}{dx} e^{x^{3}} + \frac{d}{dx} e^{x^{4}} + \frac{d}{dx} e^{x^{5}} \\ = e^{x} + e^{x^{2}} \frac{d}{dx} (x^{2}) + e^{x^{3}} \frac{d}{dx} (x^{3}) + e^{x^{4}} \frac{d}{dx} (x^{4}) + e^{x^{5}} \frac{d}{dx} (x^{5}) \\ [Using chain rule] \\ = e^{x} + 2 {xe}^{x^{2}} + 3 x^{2} e^{x^{3}} + 4 x^{3} e^{x^{4}} + 5 x^{4} e^{x^{5}} \end{array}$

Q.7

$\begin{array}{l} Differentiate the following w . r . t . x : \\ \sqrt{e^{\sqrt{x}}}, x > 0 \end{array}$

Ans

$\begin{array}{l} Let y = \sqrt{e^{\sqrt{x}}}, x > 0 \\ then, \\ y^{2} = e^{\sqrt{x}} \\ Differentiating both sides w . r . t . x, we get \\ \frac{d}{dx} y^{2} = \frac{d}{dx} (e^{\sqrt{x}}) \\ 2 y \frac{dy}{dx} = e^{\sqrt{x}} \frac{d}{dx} (\sqrt{x}) [Using chain rule] \\ \frac{dy}{dx} = \frac{e^{\sqrt{x}}}{2 y} . \frac{1}{2 \sqrt{x}} \\ = \frac{e^{\sqrt{x}}}{4 \sqrt{x} \sqrt{e^{\sqrt{x}}}} \\ ∴ \frac{dy}{dx} = \frac{e^{\sqrt{x}}}{4 \sqrt{{xe}^{\sqrt{x}}}}, x > 0 \end{array}$

Q.8

$\begin{array}{l} Differentiate the following w . r . t . x : \\ \log (logx), x > 1 \end{array}$

Ans

$\begin{array}{l} Let y = \log (logx), x > 1 \\ Differentiating g both sides w . r . t . x, we get \\ \frac{d}{dx} y = \frac{d}{dx} {\log (logx)} \\ \frac{dy}{dx} = \frac{1}{logx} \frac{d}{dx} (logx) [Using chain rule] \\ \frac{dy}{dx} = \frac{1}{logx} . \frac{1}{x} \\ \frac{dy}{dx} = \frac{1}{xlogx}, x > 1 \end{array}$

Q.9

$\begin{array}{l} Differentiate the following w . r . t . x : \\ \frac{cosx}{logx}, x > 0 \end{array}$

Ans

$\begin{array}{l} Let y = \frac{cosx}{logx}, x > 0 \\ Differentiating both sides w . r . t . x, we get \\ \frac{d}{dx} y = \frac{d}{dx} (\frac{cosx}{logx}) \\ \frac{dy}{dx} = \frac{logx \frac{d}{dx} cosx - cosx \frac{d}{dx} logx}{{(logx)}^{2}} [By Quotient Rule] \\ \frac{dy}{dx} = \frac{logx \times - sinx - cosx \times \frac{1}{x}}{{(logx)}^{2}} \\ \frac{dy}{dx} = \frac{- (xlogx . sinx + cosx)}{x {(logx)}^{2}}, x > 0 \end{array}$

Q.10

$\begin{array}{l} D i f f e r e n t i a t e t h e f o l l o w i n g w . r . t . x : \\ c o s (l o g x + e^{x}), x > 0 \end{array}$

Ans

$\begin{array}{l} L e t y = \cos (\log x + e^{x}), x > 0 \\ D i f f e r e n t i a t i n g both sides w .r .t . x, we get \\ \frac{d}{d x} y = \frac{d}{d x} \cos (\log x + e^{x}) \\ \frac{d y}{d x} = - \sin (\log x + e^{x}) \frac{d}{d x} (\log x + e^{x}) [Using c h a i n rule] \\ \frac{d y}{d x} = - \sin (\log x + e^{x}) (\frac{1}{x} + e^{x}) \\ \frac{d y}{d x} = - (\frac{1}{x} + e^{x}) . \sin (\log x + e^{x}), x > 0 \end{array}$

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NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability (Ex 5.4) Exercise 5.4