NCERT Solutions Class 12 Maths Chapter 5 Exercise 5.3

NCERT Solutions for Class 12 Chapter 5 Exercise 5.3

Mathematics is a science that deals with the Logic of Quantities, Shapes, and Arrangements. It is like a practice that goes everywhere around students. Therefore, Mathematics is a subject that students learn since their childhood.

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Chapter – 5 in Class 12 Mathematics is Continuity and Differentiability. The chapter includes the following topics – Continuity, Differentiability, Exponential and Logarithmic Functions, Logarithmic Differentiation, Derivatives of Functions in Parametric Forms, Second Order Derivatives and Mean Value Theorem. Hence, the chapter covers a wide range of topics with very important concepts.

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NCERT Solutions for Class 12th Maths Chapter 5 Continuity and Differentiability (Ex 5.3) Exercise 5.3

Students can download the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.3 from Extramarks’ website. These solutions are easily accessible and are in PDF format.

Access NCERT Solutions for Mathematics Chapter 5 – Continuity and Differentiability

Important Points

The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.3 comprises the following topics –

  • Derivatives of Implicit Functions
  • Derivatives of Inverse Trigonometric Functions

Some important concepts of the Exercise 5.3 Class 12th are –

  • Sum, difference, product, and quotient of continuous functions are continuous.
  • Every differentiable function is continuous, but the converse is not true.
  • Rolle’s Theorem: If f : [a, b] → R is continuous on [a, b] and differentiable on (a, b) such that f(a) = f(b), then there exists some c in (a, b) such that f ′(c) = 0.
  • Mean Value Theorem: If f : [a, b] → R is continuous on [a, b] and differentiable on (a, b). Then there exists some c in (a, b) such that f'(c) = (f(b) – f(a))/(b-a).

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Look for the NCERT Solutions for Class 12th Maths Chapter 5 Exercise 5.3

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NCERT Solutions for Class 12 Maths Chapter 5 Other Exercises

Chapter 5 – Continuity and Differentiability Exercises
Exercise 5.1 10 Short Questions and 24 Long Questions
Exercise 5.3
9 short Questions and 6 long Questions
Exercise 5.4
5 Short Questions and 5 Long Questions
Exercise 5.5
4 Short Questions and 14 Long Questions
Exercise 5.6
1 Short Question and 1 Long Question
Exercise 5.7
10 Short Questions and 7 Long Questions
Exercise 5.8
Questions with Solutions


Find dy/dx in the following:
2x + 3y = sinx


Given:  2x + 3y = sinxDifferentiating w.r.t. x, we getddx(2x + 3y)=ddxsinx2ddxx+3dydx=cosx        2+3dydx=cosx        dydx=cosx23


Find dy/dx in the following:
2x + 3y = siny


Given:  2x + 3y = sinyDifferentiating w.r.t. x, we getddx(2x + 3y)=ddxsiny2ddxx+3dydx=cosydydx        2+3dydx=cosydydx3dydxcosydydx=2  dydx(3cosy)=2    dydx=2(3cosy)    dydx=2(cosy3)


Find dydx in the following: ax + by2 = cosy


Given:  ax + by2 = cosyDifferentiating w.r.t. x, we getddx(ax + by2)=ddxcosyaddxx+2bydydx=sinydydx        a+2bydydx=sinydydx2bydydx+sinydydx=a  dydx(2by+siny)=a    dydx=a(2by+siny)    dydx=a(2by+siny)


Find dydx in the following:xy+y2=tanx+y


Given:  xy + y2=tanx+yDifferentiating w.r.t. x, we get  ddx(xy + y2)=ddx(tanx+y)[xdydx+yddxx]+ddxy2=sec2x+dydx[By Product Rule]            xdydx+y+2ydydx=sec2x+dydx        xdydx+2ydydxdydx=sec2xy              (x+2y1)dydx=sec2xy                  dydx=sec2xy(x+2y1)


Find dydx in the following x3+ x2y + xy2+ y3= 81


Given:  x3+ x2y + xy2+ y3= 81Differentiating w.r.t. x, we get                  ddx(x3+ x2y + xy2+ y3)=ddx81  ddxx3+ddxx2y+ddxxy2+ddxy3=03x2+(x2dydx+yddxx2)+(xddxy2+y2ddxx)+3y2dydx=0[By product rule]        3x2+x2dydx+y×2x+(2xydydx+y2×1)+3y2dydx=0                      3x2+x2dydx+2xy+2xydydx+y2+3y2dydx=0              (x2+2xy+3y2)dydx=3x22xyy2      dydx=(3x2+2xy+y2)(x2+2xy+3y2)


Find dydx in the followingx2+ xy + y2= 100


Given:  x2+ xy + y2=100Differentiating w.r.t. x, we get      ddx(x2+ xy + y2)=ddx100  ddxx2+ddxxy+ddxy2=02x+(xdydx+yddxx)+2ydydx=0[By product rule]      2x+xdydx+y×1+2ydydx=0    dydx(x+2y)=2xydydx=2x+yx+2y




Given:  sin2y+cosxy=KDifferentiating w.r.t. x, we get                      ddxsin2y+cosxy=ddxK                      ddxsin2y+ddxcosxy=0              2sinyddxsinysinxyddxxy=0Using chain rule2sinycosydydxsinxyxdydx+yddxx=0By product rule          sin2ydydxxsinxydydxysinxy=0            dydxsin2yxsinxy=ysinxy        dydx=ysinxysin2yxsinxy


Find dydx in the followingsin2x + cos2y = 1


Given:  sin2x + cos2y=1Differentiating w.r.t. x, we get                            ddx(sin2x + cos2y)=ddx1                    ddxsin2x+ddxcos2y=02sinxddxsinx+2cosyddx(cosy)=0[Using chain rule]    2sinxcosx+2cosy(siny)dydx=0                sin2xsin2ydydx=0        sin2ydydx=sin2x    dydx=sin2xsin2y


Find dydx in the followingy=sin1(2x1+x2)


The given relationship is  y=sin1(2x1+x2)  siny=2x1+x2Differentiating both sides w.r.t. x, we get    ddxsiny=ddx(2x1+x2)      cosydydx=(1+x2)ddx2x2xddx(1+x2)(1+x2)2[By Quotient Rule]        1sin2ydydx=(1+x2)×22x(0+2x)(1+x2)2      1(2x1+x2)2dydx=2+2x24x2(1+x2)2          [Putting value of siny](1+x2)24x2(1+x2)2dydx=22x2(1+x2)2      (1x2)2(1+x2)2dydx=22x2(1+x2)2            (1x2)(1+x2)dydx=22x2(1+x2)2        dydx=22x2(1+x2)(1x2)      dydx=2(1x2)(1+x2)(1x2)      dydx=2(1+x2)


Find dydx in the following:y=tan1(3xx31+3x2),13<x<13


We​​ have y=tan1(3xx313x2)    Putting x=tanθ, we get          y=tan1(3tanθtan3θ13tan2θ)    =tan1(tan3θ)Since,13<x<13,thenx=tanθ  13<tanθ<13π6<θ<π6π2<3θ<π2    y=3θ[π2<3θ<π2]=3tan1x[x=tanθθ=tan1x]Differentiating both sides w.r.t. x, we get  dydx=3ddxtan1x=3×11+x2=31+x2ddxtan1(3xx313x2)=31+x2


Find dydx in the following: y=cos1(1x21+x2),    0<x<1


We​​ have y=cos1(1x21+x2),    0<x< 1    Putting x=tanθ, we get          y=cos1(1tan2θ1+tan2θ)    =cos1(cos2θ)Since,0<  x <  1,thenx=tanθ          0<tanθ<1    0<θ<π4    0<2θ<π2    y=2θ[0<2θ<π2]=2tan1x[x=tanθθ=tan1x]Differentiating both sides w.r.t. x, we getdydx=2ddxtan1x=2×11+x2=21+x2ddxcos1(1tan2θ1+tan2θ)=21+x2


Find dydx in the following:y=sin1(1x21+x2),    0<x<1


We​​ have y=sin1(1x21+x2),    0<x< 1    Putting x=tanθ, we get          y=sin1(1tan2θ1+tan2θ)    =sin1(cos2θ)    =sin1{sin(π22θ)}Since,0<  x <  1,then  x=tanθ          0<tanθ<1    0<θ<π4    0<2θ<π2      0<π22θ<π2    y=π22θ[0<π22θ<π2]=π22tan1x[x=tanθθ=tan1x]Differentiating both sides w.r.t. x, we getdydx=02ddxtan1x=2×11+x2=21+x2Thus,ddxsin1(1x21+x2)=21+x2


Find dydx in the following:y=cos1(2x1+x2),    1<x<1


We​​ have y=cos1(2x1+x2),    1<x<1    Putting x=tanθ, we get          y=cos1(2tanθ1+tan2θ)    =cos1(sin2θ)    =cos1{cos(π22θ)}Since,1<x<1,then  x=tanθ          1<tanθ<1  π4<θ<π4  π2<2θ<π2      π2>2θ>π2      π>π22θ>0      0<π22θ<π        y=π22θ=π22tan1x[x=tanθθ=tan1x]Differentiating both sides w.r.t. x, we getdydx=ddxπ22ddxtan1x=02×11+x2ddxcos1(2x1+x2)=21+x2.


Find dydx in the followingy=sin1(2x1x2),    12<x<12


We​​ have y=sin1(2x1x2),    12< x<12    Putting x=sinθ, we get          y=sin1(2sinθ1sin2θ)    =sin1(2sinθcosθ)    =sin1(sin2θ)Since,12< x<12,then  x=sinθ        12<sinθ<12      π4<θ<π4      π2<2θ<π2      y=2θ=2sin1x[x=sinθθ=sin1x]Differentiating w.r.t. x, we get  dydx=ddx(2sin1x)=21x2Thus,ddx{sin1(2x1x2)}=21x2


Find dy/dx in the following

y = sec1(12x21),    0<x<12


We have y=sec1(12x21),     0< x < 12    =cos1(2x21)Putting x=cosθ, ​we gety=cos1(2cos2θ1)    =cos1(cos2θ)Since,0<x <12        0<cosθ  <12[x=cosθ]        3π2<θ<7π4        0<θ<π4        0<2θ<π2y=2θ      =2cos1x[x=cosθθ=cos1x]Differentiating w.r.t. x, we get  dydx=2ddxcos1x=2(11x2)=21x2Thus,ddxsec1(12x21)=21x2.

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