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

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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+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 10

 

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