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

These NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 Questions and Answers are prepared by our highly skilled subject experts.

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

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

Question 1.
2x + 3y = sinx
Solution:
2x + 3y = sinx
Differentiating w.r.t x,
\(2+3\frac { dy }{ dx } =cosx \)
⇒ \(\frac { dy }{ dx } =\frac { 1 }{ 3 } (cosx-2)\)

Question 2.
2x + 3y = siny
Solution:
2x + 3y = siny
Differentiating w.r.t x,
\(2+3.\frac { dy }{ dx } =cosy\frac { dy }{ dx } \)
⇒ \(\frac { dy }{ dx } =\frac { 2 }{ cosy-3 } \)

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

Question 3.
ax + by² = cosy
Solution:
ax + by² = cosy
Differentiate both sides w.r.t. x,
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 1

Question 4.
xy + y² = tan x + y
Solution:
xy + y² = tan x + y
Differentiate both sides w.r.t. x,
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 2

Question 5.
x² + xy + y² = 100
Solution:
x² + xy + y² = 100
Differentiate both sides w.r.t. x,
 NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 3

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

Question 6.
x³ + x²y + xy² + y³ = 81
Solution:
x³ + x²y + xy² + y³ = 81
Differentiate both sides w.r.t. x,
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 4

Question 7.
sin² y + cos xy = π
Solution:
sin² y + cos xy = π
Differentiate both sides w.r.t. x,
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 5

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

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 6

Question 9.
y = \({ sin }^{ -1 }\left( \frac { 2x }{ { 1+x }^{ 2 } } \right)\)
Solution:
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 6a

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

Question 10.
y = \({ tan }^{ -1 }\left( \frac { { 3x-x }^{ 3 } }{ { 1-3x }^{ 2 } } \right) ,-\frac { 1 }{ \sqrt { 3 } } <x<\frac { 1 }{ \sqrt { 3 } } \)
Solution:
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 7

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 = \({ cos }^{ -1 }\left( \frac { 1-tan^{ 2 }\quad \theta }{ 1+{ tan }^{ 2 }\quad \theta } \right) ={ cos }^{ -1 }(cos2\theta )=2\theta\)
= 2θ = 2 tan-1 x
i.e., y = 2 tan-1x
Differentiating w.r.t x, \(\frac{d y}{d x}=\frac{2}{1+x^{2}}\)

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

Question 12.
y = \({ sin }^{ -1 }\left( \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) ,0<x<1\)
Solution:
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 8

Question 13.
y = \({ cos }^{ -1 }\left( \frac { 2x }{ 1+{ x }^{ 2 } } \right)\), – 1 < x < 1
Solution:
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 9

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

Question 14.
y = \(sin^{ -1 }\left( 2x\sqrt { 1-{ x }^{ 2 } } \right) ,-\frac { 1 }{ \sqrt { 2 } } <x<\frac { 1 }{ \sqrt { 2 } } \)
Solution:
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 8a

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 = \(sec^{ -1 }(sec2\theta )=2\theta ,\quad 2{ cos }^{ -1 }x \)
i.e., y = 2 cos-1 x
Differentiating w.r.t.x,
∴ \(\frac{d y}{d x}=\frac{-2}{\sqrt{1-x^{2}}}\)

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