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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 Ex 5.4

Get Free NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.4 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Continuity and Differentiability Exercise 5.4 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 5 Class 12 Continuity and Differentiability Ex 5.4 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.4 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

The topics and sub-topics included in the Continuity and Differentiability chapter are the following:

  • Continuity and Differentiability
  • Introduction
  • Algebra of continuous functions
  • Differentiability
  • Derivatives of composite functions
  • Derivatives of implicit functions
  • Derivatives of inverse trigonometric functions
  • Exponential and Logarithmic Functions
  • Logarithmic Differentiation
  • Derivatives of Functions in Parametric Forms
  • Second Order Derivative
  • Mean Value Theorem
  • Summary

There are total eight exercises and one misc exercise(144 Questions fully solved) in the class 12th maths chapter 5 Continuity and Differentiability.

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

Differentiate the following w.r.t.x:

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

Solution:
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y=\frac { { e }^{ x } }{ sinx }

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for\quad y=\frac { u }{ v } ,

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\frac { dy }{ dx } =\frac { { e }^{ x }{ sin }x-{ e }^{ x }cosx }{ { sin }^{ 2 }x }

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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.
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{ e }^{ { sin }^{ -1 }x }

Solution:
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{ e }^{ { sin }^{ -1 }x }

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

x=sint
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\therefore y={ e }^{ t },\frac { dt }{ dx } =\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } ,\frac { dy }{ dt } ={ e }^{ t }

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\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.
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{ e }^{ { x }^{ 3 } }=y

Solution:
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{ e }^{ { x }^{ 3 } }=y

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

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\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.
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sin\left( { tan }^{ -1 }{ e }^{ -x } \right) =y

Solution:
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sin\left( { tan }^{ -1 }{ e }^{ -x } \right) =y

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\frac { dy }{ dx } =cos\left( { tan }^{ -1 }{ e }^{ -x } \right) \frac { d }{ dx } \left( { tan }^{ -1 }{ e }^{ -x } \right)

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

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=-cos\left( { tan }^{ -1 }{ e }^{ -x } \right) \frac { 1 }{ 1+{ e }^{ -2x } } .\left( { e }^{ -x } \right)

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

Solution:
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\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.
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{ e }^{ x }+{ e }^{ { x }^{ 2 } }+
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+{ e }^{ { x }^{ 5 } }=y(say)

Solution:
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let\quad u={ e }^{ { x }^{ n } },put\quad { x }^{ n }=t,u={ e }^{ t },t={ x }^{ n }

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{ e }^{ x }+{ e }^{ { x }^{ 2 } }+
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+{ e }^{ { x }^{ 5 } }=y(say)

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NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 6

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

Solution:
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\sqrt { { e }^{ \sqrt { x } } } ,x>0

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

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NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 7

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
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NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 8

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

Solution:
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y=\frac { cosx }{ logx }

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tiwari academy class 12 maths Chapter 5 Continuity and Differentiability 9

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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NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 10

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

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

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NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.4

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NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.4 in English medium

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NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.4 in Hindi medium

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NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.4 in PDF form

NCERT Class 12 Maths Solutions

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