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

Here, Below you all know about NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.4 Question Answer. I know many of you confuse about finding Chapter 5 Continuity and Differentiability Ex 5.4 Of Class 12 NCERT Solutions. So, Read the full post below and get your solutions.

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

NCERT TEXTBOOK EXERCISES

Ex 5.4 Class 12 Maths Question 1.
$y=\frac { { e }^{ x } }{ sinx }$

Solution:

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

Ex 5.4 Class 12 Maths Question 2.
${ e }^{ { sin }^{ -1 }x }$

Solution:

${ e }^{ { sin }^{ -1 }x }$
$y={ e }^{ { sin }^{ -1 }x }$
x=sint

Ex 5.4 Class 12 Maths Question 3.
${ e }^{ { x }^{ 3 } }=y$

Solution:

${ e }^{ { x }^{ 3 } }=y$

Ex 5.4 Class 12 Maths Question 4.
$sin\left( { tan }^{ -1 }{ e }^{ -x } \right) =y$

Solution:

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

Ex 5.4 Class 12 Maths Question 5.
$log(cos\quad { e }^{ x })=y$

Solution:

$\frac { dy }{ dx } =\frac { 1 }{ cos\quad { e }^{ x } } \left( -sin{ e }^{ x } \right) .{ e }^{ x }\quad =-tan\left( { e }^{ x } \right)$

Ex 5.4 Class 12 Maths Question 6.
ex+ex2+â€¦+ex5=y(say)

Solution:

$let\quad u={ e }^{ { x }^{ n } },put\quad { x }^{ n }=t,u={ e }^{ t },t={ x }^{ n }$
ex+ex2+â€¦+ex5=y(say)

Ex 5.4 Class 12 Maths Question 7.
$\sqrt { { e }^{ \sqrt { x } } } ,x>0$

Solution:

y = $\sqrt { { e }^{ \sqrt { x } } } ,x>0$
$y=\sqrt { { e }^{ \sqrt { x } } } ,let\quad y=\sqrt { s } ,s={ e }^{ t },t=\sqrt { x }$

Ex 5.4 Class 12 Maths Question 8.
log(log x),x>1

Solution:

y = log(log x),
put y = log t, t = log x,
differentiating

Ex 5.4 Class 12 Maths Question 9.
$\frac { cosx }{ logx } =y(say),x>0$

Solution:

Let $y=\frac { cosx }{ logx }$

Ex 5.4 Class 12 Maths Question 10.
cos(log x+ex),x>0

Solution:

y = cos(log x+ex),x>0
put y = cos t,t = log x+ex

## NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.4 PDF

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