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.

BoardCBSE
TextbookNCERT
ClassClass 12
SubjectMaths
ChapterChapter 5
Chapter NameContinuity and Differentiability
ExerciseEx 5.3
Number of Questions Solved15
CategoryNCERT Solutions

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 } “><br><img loading=

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
2quad sinquad yfrac { dquad 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 { 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 } } <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 }(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<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
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 13.1

Ex 5.3 Class 12 Maths Question 14.
y=sin^{ -1 }left( 2xsqrt { 1-{ x }^{ 2 } } right) ,-frac { 1 }{ sqrt { 2 } } <x<frac { 1 }{ sqrt { 2 } }
Solution:
y=sin^{ -1 }left( 2xsqrt { 1-{ x }^{ 2 } } right) ,-frac { 1 }{ sqrt { 2 } } <x<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<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 }{ cos2theta } right)
y=sec^{ -1 }(sec2theta )=2theta ,quad y=2{ cos }^{ -1 }x
therefore frac { dy }{ dx } =frac { -2 }{ sqrt { { 1-x }^{ 2 } } }

All Chapter NCERT Solutions For Class12 Maths

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All Subject NCERT Solutions For Class12

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