NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3

Get Free NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Continuity and Differentiability Exercise 5.3 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 5.3 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 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 5.3

Find $\\ \frac { dy }{ dx }$ in the following

Ex 5.3 Class 12 Maths Question 1.
2x + 3y = sinx
Solution:
2x + 3y = sinx
Differentiating w.r.t x,
$2+3\frac { dy }{ dx } =cosx$
=> $\frac { dy }{ dx } =\frac { 1 }{ 3 } (cosx-2)$

Ex 5.3 Class 12 Maths Question 2.
2x + 3y = siny
Solution:
2x + 3y = siny
Differentiating w.r.t x,
$2+3.\frac { dy }{ dx } =cosy\frac { dy }{ dx }$
=> $\frac { dy }{ dx } =\frac { 2 }{ cosy-3 }$

Ex 5.3 Class 12 Maths Question 3.
ax + by² = cosy
Solution:
ax + by² = cosy
Differentiate w.r.t. x,
$a+2\quad by\quad \frac { dy }{ dx } =-siny\frac { dy }{ dx }$
=> $or\quad (2b+siny)\frac { dy }{ dx } =-a=>\frac { dy }{ dx } =-\frac { a }{ 2b+siny }$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 3

Ex 5.3 Class 12 Maths Question 4.
xy + y² = tan x + y
Solution:
xy + y² = tanx + y
Differentiating w.r.t. x,
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 4

Ex 5.3 Class 12 Maths Question 5.
x² + xy + y² = 100
Solution:
x² + xy + xy = 100
Differentiating w.r.t. x,
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 5

Ex 5.3 Class 12 Maths Question 6.
x³ + x²y + xy² + y³ = 81
Solution:
Given that
x³ + x²y + xy² + y³ = 81
Differentiating both sides we get
byjus class 12 maths Chapter 5 Continuity and Differentiability 6

Ex 5.3 Class 12 Maths Question 7.
sin² y + cos xy = π
Solution:
Given that
sin² y + cos xy = π
Differentiating both sides we get
$2\quad sin\quad y\frac { d\quad siny }{ dx } +(-sinxy)\frac { d(xy) }{ dx } =0$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 7

Ex 5.3 Class 12 Maths Question 8.
sin²x + cos²y = 1
Solution:
Given that
sin²x + cos²y = 1
Differentiating both sides, we get
byjus class 12 maths Chapter 5 Continuity and Differentiability 8

Ex 5.3 Class 12 Maths Question 9.
$y={ sin }^{ -1 }\left( \frac { 2x }{ { 1+x }^{ 2 } } \right)$
Solution:
$y={ sin }^{ -1 }\left( \frac { 2x }{ { 1+x }^{ 2 } } \right)$
put x = tanθ
$y={ sin }^{ -1 }\left( \frac { 2tan\theta }{ { 1+tan }^{ 2 }\theta } \right) ={ sin }^{ -1 }(sin2\theta )=2\theta$
$y={ 2sin }^{ -1 }x\quad \therefore \frac { dy }{ dx } =\frac { 2 }{ 1+{ x }^{ 2 } }$

Ex 5.3 Class 12 Maths Question 10.
$y={ tan }^{ -1 }\left( \frac { { 3x-x }^{ 3 } }{ { 1-3x }^{ 2 } } \right) ,-\frac { 1 }{ \sqrt { 3 } } <x<\frac { 1 }{ \sqrt { 3 } }$
Solution:
$y={ tan }^{ -1 }\left( \frac { { 3x-x }^{ 3 } }{ { 1-3x }^{ 2 } } \right)$
put x = tanθ
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 10

Ex 5.3 Class 12 Maths Question 11.
$y={ cos }^{ -1 }\left( \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) ,0<x<1$
Solution:
$y={ cos }^{ -1 }\left( \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) ,0<x<1$
put x = tanθ
$y={ cos }^{ -1 }\left( \frac { 1-tan^{ 2 }\quad \theta }{ 1+{ tan }^{ 2 }\quad \theta } \right) ={ cos }^{ -1 }(cos2\theta )=2\theta$
$y={ 2tan }^{ -1 }x\quad \therefore \frac { dy }{ dx } =\frac { 2 }{ 1+{ x }^{ 2 } }$

Ex 5.3 Class 12 Maths Question 12.
$y={ sin }^{ -1 }\left( \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) ,0<x<1$
Solution:
$y={ sin }^{ -1 }\left( \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) ,0<x<1$
put x = tanθ
we get
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 12

Ex 5.3 Class 12 Maths Question 13.
$y={ cos }^{ -1 }\left( \frac { 2x }{ 1+{ x }^{ 2 } } \right) ,-1<x<1$
Solution:
$y={ cos }^{ -1 }\left( \frac { 2x }{ 1+{ x }^{ 2 } } \right) ,-1<x<1$
put x = tanθ
we get
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 13

Ex 5.3 Class 12 Maths Question 14.
$y=sin^{ -1 }\left( 2x\sqrt { 1-{ x }^{ 2 } } \right) ,-\frac { 1 }{ \sqrt { 2 } } <x<\frac { 1 }{ \sqrt { 2 } }$
Solution:
$y=sin^{ -1 }\left( 2x\sqrt { 1-{ x }^{ 2 } } \right) ,-\frac { 1 }{ \sqrt { 2 } } <x<\frac { 1 }{ \sqrt { 2 } }$
put x = tanθ
we get
$y=sin^{ -1 }\left( 2sin\quad \theta \sqrt { 1-{ x }^{ 2 } } \right)$
$y=sin^{ -1 }\left( 2sin\theta \quad cos\theta \right) \quad ={ sin }^{ -1 }(sin2\theta )\quad =2\theta$
$y=2sin^{ -1 }x\quad \therefore \frac { dy }{ dx } =\frac { 2 }{ \sqrt { { 1-x }^{ 2 } } }$

Ex 5.3 Class 12 Maths Question 15.
$y=sin^{ -1 }\left( \frac { 1 }{ { 2x }^{ 2 }-1 } \right) ,0<x<\frac { 1 }{ \sqrt { 2 } }$
Solution:
$y=sin^{ -1 }\left( \frac { 1 }{ { 2x }^{ 2 }-1 } \right) ,0<x<\frac { 1 }{ \sqrt { 2 } }$
put x = tanθ
we get
$y=sec^{ -1 }\left( \frac { 1 }{ { 2cos }^{ 2 }\theta -1 } \right) ={ sec }^{ -1 }\left( \frac { 1 }{ cos2\theta } \right)$
$y=sec^{ -1 }(sec2\theta )=2\theta ,\quad y=2{ cos }^{ -1 }x$
$\therefore \frac { dy }{ dx } =\frac { -2 }{ \sqrt { { 1-x }^{ 2 } } }$