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

In this chapter, we provide NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.4 for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.4 pdf, free NCERT solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.4 book pdf download.

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

Differentiate the following w.r.t.x:

Ex 5.4 Class 12 Maths Question 1.
$frac { { e }^{ x } }{ sinx }$
Solution:
$y=frac { { e }^{ x } }{ sinx }$
$forquad y=frac { u }{ v } ,$
$frac { dy }{ dx } =frac { { e }^{ x }{ sin }x-{ e }^{ x }cosx }{ { sin }^{ 2 }x }$
$orfrac { dy }{ dx } =frac { { e }^{ x }{ sin }x-{ e }^{ x }cosx }{ { sin }^{ 2 }x } ,wherequad xneq npi ,xin z$

Ex 5.4 Class 12 Maths Question 2.
${ e }^{ { sin }^{ -1 }x }$
Solution:
${ e }^{ { sin }^{ -1 }x }$
$y={ e }^{ { sin }^{ -1 }x }$
x=sint
$therefore y={ e }^{ t },frac { dt }{ dx } =frac { 1 }{ sqrt { 1-{ x }^{ 2 } } } ,frac { dy }{ dt } ={ e }^{ t }$
$therefore frac { dy }{ dx } =frac { dy }{ dt } .frac { dt }{ dx } ={ e }^{ t }.frac { 1 }{ sqrt { { 1- }x^{ 2 } } } =frac { { e }^{ { sin }^{ -1 }x } }{ sqrt { 1-{ x }^{ 2 } } }$

Ex 5.4 Class 12 Maths Question 3.
${ e }^{ { x }^{ 3 } }=y$
Solution:
${ e }^{ { x }^{ 3 } }=y$
$Putquad { x }^{ 3 }=tquad therefore quad y={ e }^{ t },frac { dy }{ dt } ={ e }^{ t },frac { dt }{ dx } ={ 3x }^{ 2 }$
$therefore frac { dy }{ dx } =frac { dy }{ dt } times frac { dt }{ dx } ={ e }^{ t }times { 3x }^{ 2 }={ 3x }^{ 2 }{ e }^{ { x }^{ 3 } }$

Ex 5.4 Class 12 Maths Question 4.
$sinleft( { tan }^{ -1 }{ e }^{ -x } right) =y$
Solution:
$sinleft( { tan }^{ -1 }{ e }^{ -x } right) =y$
$frac { dy }{ dx } =cosleft( { tan }^{ -1 }{ e }^{ -x } right) frac { d }{ dx } left( { tan }^{ -1 }{ e }^{ -x } right)$
$=cosleft( { tan }^{ -1 }{ e }^{ -x } right) frac { 1 }{ 1+{ e }^{ -2x } } frac { d }{ dx } left( { e }^{ -x } right)$
$=-cosleft( { tan }^{ -1 }{ e }^{ -x } right) frac { 1 }{ 1+{ e }^{ -2x } } .left( { e }^{ -x } right)$

Ex 5.4 Class 12 Maths Question 5.
$log(cosquad { e }^{ x })=y$
Solution:
$frac { dy }{ dx } =frac { 1 }{ cosquad { e }^{ x } } left( -sin{ e }^{ x } right) .{ e }^{ x }quad =-tanleft( { e }^{ x } right)$

Ex 5.4 Class 12 Maths Question 6.
${ e }^{ x }+{ e }^{ { x }^{ 2 } }+$$+{ e }^{ { x }^{ 5 } }=y(say)$
Solution:
$letquad u={ e }^{ { x }^{ n } },putquad { x }^{ n }=t,u={ e }^{ t },t={ x }^{ n }$
${ e }^{ x }+{ e }^{ { x }^{ 2 } }+$$+{ e }^{ { x }^{ 5 } }=y(say)$

Ex 5.4 Class 12 Maths Question 7.
$sqrt { { e }^{ sqrt { x } } } ,x>0″>
Solution:
y =

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

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

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