NCERT Solutions For Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3
Here, Below you all know about NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 Question Answer. I know many of you confuse about finding Chapter 5 Continuity and Differentiability Ex 5.3 Of Class 12 NCERT Solutions. So, Read the full post below and get your solutions.
Textbook | NCERT |
Board | CBSE |
Category | NCERT Solutions |
Class | Class 12 |
Subject | Maths |
Chapter | Chapter 5 |
Exercise | Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.3 |
Number of Questions Solved |

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3
NCERT TEXTBOOK EXERCISES
Find 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,
$a+2\quad by\quad \frac { dy }{ dx } =-siny\frac { dy }{ dx }$
=> $or\quad (2b+siny)\frac { dy }{ dx } =-a=>\frac { dy }{ dx } =-\frac { a }{ 2b+siny }$

Ex 5.3 Class 12 Maths Question 4.
xy + y² = tan x + y
Solution:
xy + y² = tanx + y
Differentiating w.r.t. x,

Ex 5.3 Class 12 Maths Question 5.
x² + xy + y² = 100
Solution:
x² + xy + xy = 100
Differentiating w.r.t. x,

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

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
$2\quad sin\quad y\frac { d\quad siny }{ dx } +(-sinxy)\frac { d(xy) }{ dx } =0$

Ex 5.3 Class 12 Maths Question 8.
sin²x + cos²y = 1
Solution:
Given that
sin²x + cos²y = 1
Differentiating both sides, we get

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
$y={ sin }^{ -1 }\left( \frac { 2tan\theta }{ { 1+tan }^{ 2 }\theta } \right) ={ sin }^{ -1 }(sin2\theta )=2\theta$
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

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
$y={ cos }^{ -1 }\left( \frac { 1-tan^{ 2 }\quad \theta }{ 1+{ tan }^{ 2 }\quad \theta } \right) ={ cos }^{ -1 }(cos2\theta )=2\theta$
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

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

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
$y=sin^{ -1 }\left( 2sin\quad \theta \sqrt { 1-{ x }^{ 2 } } \right)$
$y=sin^{ -1 }\left( 2sin\theta \quad cos\theta \right) \quad ={ sin }^{ -1 }(sin2\theta )\quad =2\theta$
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
$y=sec^{ -1 }\left( \frac { 1 }{ { 2cos }^{ 2 }\theta -1 } \right) ={ sec }^{ -1 }\left( \frac { 1 }{ cos2\theta } \right)$
$y=sec^{ -1 }(sec2\theta )=2\theta ,\quad y=2{ cos }^{ -1 }x$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.3 PDF
For NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3, you may click on the link below and get your NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise pdf file.
Finally, You all know about NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3. If you have any questions, then comment below and share this post with others.
Other Chapter of Class 12 Maths Chapter 5 Continuity and Differentiability
- Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1
- Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.2
- Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3
- Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.4
- Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5
- Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.6
- Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.7
- Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.8