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.

• Continuity and Differentiability Class 12 Ex 5.1
• Continuity and Differentiability Class 12 Ex 5.2
• Continuity and Differentiability Class 12 Ex 5.3
• Continuity and Differentiability Class 12 Ex 5.5
• Continuity and Differentiability Class 12 Ex 5.6
• Continuity and Differentiability Class 12 Ex 5.7
• Continuity and Differentiability Class 12 Ex 5.8

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:

Ex 5.4 Class 12 Maths Question 1.
\(\frac { { e }^{ x } }{ sinx } \)
Solution:
\(y=\frac { { e }^{ x } }{ sinx } \)
\(for\quad y=\frac { u }{ v } ,\)
\(\frac { dy }{ dx } =\frac { { e }^{ x }{ sin }x-{ e }^{ x }cosx }{ { sin }^{ 2 }x } \)
\(or\frac { dy }{ dx } =\frac { { e }^{ x }{ sin }x-{ e }^{ x }cosx }{ { sin }^{ 2 }x } ,where\quad x\neq n\pi ,x\in z \)

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:
\({ e }^{ { x }^{ 3 } }=y \)
\(Put\quad { x }^{ 3 }=t\quad \therefore \quad y={ e }^{ t },\frac { dy }{ dt } ={ e }^{ t },\frac { dt }{ dx } ={ 3x }^{ 2 } \)
\(\therefore \frac { dy }{ dx } =\frac { dy }{ dt } \times \frac { dt }{ dx } ={ e }^{ t }\times { 3x }^{ 2 }={ 3x }^{ 2 }{ e }^{ { x }^{ 3 } }\)

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.
\({ e }^{ x }+{ e }^{ { x }^{ 2 } }+\)…\(+{ e }^{ { x }^{ 5 } }=y(say)\)
Solution:
\(let\quad u={ e }^{ { x }^{ n } },put\quad { x }^{ n }=t,u={ e }^{ t },t={ x }^{ n }\)
\({ e }^{ x }+{ e }^{ { x }^{ 2 } }+\)…\(+{ e }^{ { x }^{ 5 } }=y(say)\)

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+e^x),x>0
Solution:
y = cos(log x+e^x),x>0
put y = cos t,t = log x+e^x

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