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

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.4 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.4.

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Maths
Chapter	Chapter 5
Chapter Name	Continuity and Differentiability
Exercise	Ex 5.4
Number of Questions Solved	10
Category	NCERT Solutions

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exc 5.4

Differentiate the following w.r.t.x:

Question 1.
$\frac { { e }^{ x } }{ sinx }$
Solution:
$y=\frac { { e }^{ x } }{ sinx }$
$for\quad y=\frac { u }{ v } ,$
$\frac { dy }{ dx } =\frac { { e }^{ x }{ sin }x-{ e }^{ x }cosx }{ { sin }^{ 2 }x }$
$or\frac { dy }{ dx } =\frac { { e }^{ x }{ sin }x-{ e }^{ x }cosx }{ { sin }^{ 2 }x } ,where\quad x\neq n\pi ,x\in z$

Question 2.
${ e }^{ { sin }^{ -1 }x }$
Solution:
${ e }^{ { sin }^{ -1 }x }$
$y={ e }^{ { sin }^{ -1 }x }$
x=sint
$\therefore y={ e }^{ t },\frac { dt }{ dx } =\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } ,\frac { dy }{ dt } ={ e }^{ t }$
$\therefore \frac { dy }{ dx } =\frac { dy }{ dt } .\frac { dt }{ dx } ={ e }^{ t }.\frac { 1 }{ \sqrt { { 1- }x^{ 2 } } } =\frac { { e }^{ { sin }^{ -1 }x } }{ \sqrt { 1-{ x }^{ 2 } } }$

Question 3.
${ e }^{ { x }^{ 3 } }=y$
Solution:
${ e }^{ { x }^{ 3 } }=y$
$Put\quad { x }^{ 3 }=t\quad \therefore \quad y={ e }^{ t },\frac { dy }{ dt } ={ e }^{ t },\frac { dt }{ dx } ={ 3x }^{ 2 }$
$\therefore \frac { dy }{ dx } =\frac { dy }{ dt } \times \frac { dt }{ dx } ={ e }^{ t }\times { 3x }^{ 2 }={ 3x }^{ 2 }{ e }^{ { x }^{ 3 } }$

Question 4.
$sin\left( { tan }^{ -1 }{ e }^{ -x } \right) =y$
Solution:
$sin\left( { tan }^{ -1 }{ e }^{ -x } \right) =y$
$\frac { dy }{ dx } =cos\left( { tan }^{ -1 }{ e }^{ -x } \right) \frac { d }{ dx } \left( { tan }^{ -1 }{ e }^{ -x } \right)$
$=cos\left( { tan }^{ -1 }{ e }^{ -x } \right) \frac { 1 }{ 1+{ e }^{ -2x } } \frac { d }{ dx } \left( { e }^{ -x } \right)$
$=-cos\left( { tan }^{ -1 }{ e }^{ -x } \right) \frac { 1 }{ 1+{ e }^{ -2x } } .\left( { e }^{ -x } \right)$

Question 5.
$log(cos\quad { e }^{ x })=y$
Solution:
$\frac { dy }{ dx } =\frac { 1 }{ cos\quad { e }^{ x } } \left( -sin{ e }^{ x } \right) .{ e }^{ x }\quad =-tan\left( { e }^{ x } \right)$

Question 6.
${ e }^{ x }+{ e }^{ { x }^{ 2 } }+$ … $+{ e }^{ { x }^{ 5 } }=y(say)$
Solution:
$let\quad u={ e }^{ { x }^{ n } },put\quad { x }^{ n }=t,u={ e }^{ t },t={ x }^{ n }$
${ e }^{ x }+{ e }^{ { x }^{ 2 } }+$ … $+{ e }^{ { x }^{ 5 } }=y(say)$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 6

Question 7.
$\sqrt { { e }^{ \sqrt { x } } } ,x>0$
Solution:
y = $\sqrt { { e }^{ \sqrt { x } } } ,x>0$
$y=\sqrt { { e }^{ \sqrt { x } } } ,let\quad y=\sqrt { s } ,s={ e }^{ t },t=\sqrt { x }$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 7

Question 8.
log(log x),x>1
Solution:
y = log(log x),
put y = log t, t = log x,
differentiating
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 8

Question 9.
$\frac { cosx }{ logx } =y(say),x>0$
Solution:
let $y=\frac { cosx }{ logx }$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 9

Question 10.
cos(log x+e^x),x>0
Solution:
y = cos(log x+e^x),x>0
put y = cos t,t = log x+e^x
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 10

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

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exc 5.4

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