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

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

Board CBSE
Textbook NCERT
Class Class 12
Subject Maths
Chapter Chapter 5
Chapter Name Continuity and Differentiability
Exercise Ex 5.3
Number of Questions Solved 15
Category NCERT Solutions

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exc  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 Ex 5.3 Q3.1

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 Ex 5.3 Q4.1

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 Ex 5.3 Q5.1

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
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 Q6.1

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 Ex 5.3 Q7.1

Ex 5.3 Class 12 Maths Question 8.
sin²x + cos²y = 1
Solution:
Given that
sin²x + cos²y = 1
Differentiating both sides, we get
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 Q8.1

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 Ex 5.3 Q10.1

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 Ex 5.3 Q12.1

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 Ex 5.3 Q13.1

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 } } }

 

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