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

In this chapter, we provide NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 pdf, free NCERT solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 book pdf download.

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

Find in the following

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

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

Ex 5.3 Class 12 Maths Question 3.
ax + by² = cosy
Solution:
ax + by² = cosy
Differentiate w.r.t. x,
$a+2quad byquad frac { dy }{ dx } =-sinyfrac { dy }{ dx }$
=>$orquad (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
$2quad sinquad yfrac { dquad siny }{ dx } +(-sinxy)frac { d(xy) }{ dx } =0$

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
$y={ sin }^{ -1 }left( frac { 2tantheta }{ { 1+tan }^{ 2 }theta } right) ={ sin }^{ -1 }(sin2theta )=2theta$
$y={ 2sin }^{ -1 }xquad therefore frac { dy }{ dx } =frac { 2 }{ 1+{ x }^{ 2 } }$

Ex 5.3 Class 12 Maths Question 10.
$y={ tan }^{ -1 }left( frac { { 3x-x }^{ 3 } }{ { 1-3x }^{ 2 } } right) ,-frac { 1 }{ sqrt { 3 } }
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
Solution:
$y={ cos }^{ -1 }left( frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } right) ,0
put x = tanθ
$y={ cos }^{ -1 }left( frac { 1-tan^{ 2 }quad theta }{ 1+{ tan }^{ 2 }quad theta } right) ={ cos }^{ -1 }(cos2theta )=2theta$
$y={ 2tan }^{ -1 }xquad therefore frac { dy }{ dx } =frac { 2 }{ 1+{ x }^{ 2 } }$

Ex 5.3 Class 12 Maths Question 12.
$y={ sin }^{ -1 }left( frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } right) ,0
Solution:
$y={ sin }^{ -1 }left( frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } right) ,0
put x = tanθ
we get

Ex 5.3 Class 12 Maths Question 13.
$y={ cos }^{ -1 }left( frac { 2x }{ 1+{ x }^{ 2 } } right) ,-1
Solution:
$y={ cos }^{ -1 }left( frac { 2x }{ 1+{ x }^{ 2 } } right) ,-1
put x = tanθ
we get

Ex 5.3 Class 12 Maths Question 14.
$y=sin^{ -1 }left( 2xsqrt { 1-{ x }^{ 2 } } right) ,-frac { 1 }{ sqrt { 2 } }
Solution:
$y=sin^{ -1 }left( 2xsqrt { 1-{ x }^{ 2 } } right) ,-frac { 1 }{ sqrt { 2 } }
put x = tanθ
we get
$y=sin^{ -1 }left( 2sinquad theta sqrt { 1-{ x }^{ 2 } } right)$
$y=sin^{ -1 }left( 2sintheta quad costheta right) quad ={ sin }^{ -1 }(sin2theta )quad =2theta$
$y=2sin^{ -1 }xquad therefore frac { dy }{ dx } =frac { 2 }{ sqrt { { 1-x }^{ 2 } } }$

Ex 5.3 Class 12 Maths Question 15.
$y=sin^{ -1 }left( frac { 1 }{ { 2x }^{ 2 }-1 } right) ,0
Solution:
$y=sin^{ -1 }left( frac { 1 }{ { 2x }^{ 2 }-1 } right) ,0
put x = tanθ
we get
$y=sec^{ -1 }left( frac { 1 }{ { 2cos }^{ 2 }theta -1 } right) ={ sec }^{ -1 }left( frac { 1 }{ cos2theta } right)$
$y=sec^{ -1 }(sec2theta )=2theta ,quad y=2{ cos }^{ -1 }x$
$therefore frac { dy }{ dx } =frac { -2 }{ sqrt { { 1-x }^{ 2 } } }$

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