Continuity and Differentiability Class 12 Ncert Solutions 5.3

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

class 12 continuity & Differentiability ex.5.3

NCERT Solutions for Class 12 Maths Exercise 5.3 of Chapter 5 Continuity and Differentiability are available here for doing your homework , class assignments, and preparation of your forthcoming exams. NCERT solutions of maths are the best tools for achieving excellence in the maths subject.NCERT Solutions for Class 12 Maths Exercise 5.3 of Chapter 5 Continuity and Differentiability are created by an expert maths teacher who has experience of 25 years in teaching maths subjects.

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability

Exercise 5.1-Continuity and Differentiability

Exercise 5.2-Continuity and Differentiability

Exercise 5.3-Continuity and Differentiability

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

Find dy/dx in the following exercise 1 to 15.

Q1.2x + 3y = sinx

Ans.Differentiating the given function 2x + 3y = sinx with respect to x

$\frac{d}{dx}\left ( 2x \right )+\frac{d}{dx}\left ( 3y \right )=\frac{d}{dx}\left ( sinx \right )$

$2+3\frac{dy}{dx}=cosx$

$3\frac{dy}{dx}=cosx-2$

$\frac{dy}{dx}=\frac{cosx-2}{3}$

Q2. 2x + 3y = siny

Ans.Differentiating the given function 2x + 3y = sinx with respect to x

$\frac{d}{dx}\left ( 2x \right )+\frac{d}{dx}\left ( 3y \right )=\frac{d}{dx}\left ( siny \right )$

$2+3\frac{dy}{dx}=cosy\frac{dy}{dx}$

$3\frac{dy}{dx}-cosy\frac{dy}{dx}=-2$

$\frac{dy}{dx}\left ( 3-cosy \right )=-2$

$\frac{dy}{dx}=\frac{-2}{\left ( 3-cosy \right )}$

$\frac{dy}{dx}=\frac{2}{ cosy-3 }$

Q3. ax + bx² = cosy

Ans. Given function is ax + bx² = cosy

Derivate function with respect to x , we have

$\frac{d}{dx}(ax) + \frac{d}{dx}(by^{2})=\frac{d}{dx}cosy$

$a+b.2.y\frac{dy}{dx}=-siny\frac{dy}{dx}$

$2by\frac{dy}{dx}+siny\frac{dy}{dx}=-a$

$\frac{dy}{dx}\left (2by+siny \right )=-a$

$\frac{dy}{dx}=\frac{-a}{2by+siny}$

Q4. xy + y² = tan x + y

Ans. Given function is xy + y² = tan x + y

Derivation function will respect to x, we have

$\frac{d}{dx}(xy) + \frac{d}{dx}(y^{2})=\frac{d}{dx} tanx + \frac{d}{dx}y$

$x\frac{dy}{dx} + y\frac{d}{dx}x+2y\frac{dy}{dx} = sec^{2}x + \frac{dy}{dx}$

$x\frac{dy}{dx} + y+2y\frac{dy}{dx} = sec^{2}x + \frac{dy}{dx}$

$x\frac{dy}{dx} +2y\frac{dy}{dx}-\frac{dy}{dx} = sec^{2}x -y$

$\frac{dy}{dx}\left ( x+2y-1 \right ) = sec^{2}x -y$

$\frac{dy}{dx} = \frac{sec^{2}y-y}{ x+2y-1 }$

Q5. x² + xy + y² = 100

Ans.The given function is x² + xy + y² = 100

Differentiating the given function with respect to x

$\frac{d}{dx}\left ( x^{2} \right )+\frac{d}{dx}\left ( xy \right )+\frac{d}{dx}\left ( y^{2} \right )$

$2x+x\frac{dy}{dx}+y\frac{d}{dx}\left ( x \right )+2y\frac{dy}{dx}=0$

$2x+x\frac{dy}{dx}+y+2y\frac{dy}{dx}=0$

$\frac{dy}{dx}\left ( x+2y \right )=-2x-y$

$\frac{dy}{dx}=\frac{-2x-y}{2y+x}$

Q6.x³ + x²y + y²x + y³ = 81

Ans.The given function is x³ + x²y + y²x + y³ = 81

Differentiating the given function with respect to x

$\frac{d}{dx}\left ( x^{3} \right )+\frac{d}{dx}\left ( x^{2} y\right )+\frac{d}{dx}\left ( y^{2} x\right )+\frac{d}{dx}\left ( y^{3} \right )=0$

$3x^{2}+x^{2}\frac{dy}{dx}+y\frac{d}{dx}\left ( x^{2} \right )+y^{2}\frac{dx}{dx}+x\frac{d}{dx}\left ( y^{2} \right )+\frac{d}{dx}\left ( y^{3} \right )=0$

$3x^{2}+x^{2}\frac{dy}{dx}+y.2x+y^{2}+x.2y\frac{dy}{dx}+3y^{2}\frac{dy}{dx}=0$

$x^{2}\frac{dy}{dx} +x.2y\frac{dy}{dx}+3y^{2}\frac{dy}{dx}=-3x^{2}-2yx-y^{2}$

$\frac{dy}{dx}\left ( 3x^{2} +2xy+3y^{2}\right ) =-3x^{2}-2yx-y^{2}$

$\frac{dy}{dx} =-\frac{3x^{2}+2yx+y^{2}}{3x^{2} +2xy+3y^{2}}$

Q7. sin²y + cosxy = π

Ans.The given function is sin²y + cosxy = π

Differentiating the given function with respect to x

$\frac{d}{dx}\left ( siny \right )^{2}+\frac{d}{dx}\left ( cosxy \right )=0$

$2siny\frac{d}{dx}siny-sinxy\frac{d}{dx}\left ( xy \right )=0$

$2siny.cosy\frac{dy}{dx}-sinxy\left ( x\frac{dy}{dx}+y .1\right )=0$

$sin2y\frac{dy}{dx}-sinxy\: . x\frac{dy}{dx}-ysinxy =0$

$\frac{dy}{dx}\left ( sin2y-xsinxy \right )=ysinxy$

$\frac{dy}{dx} =\frac{ysinxy}{ sin2y-xsinxy }$

Q8. sin²x + sin²y = 1

Ans.The given function is sin²x + sin²y = 1

Differentiating the given function with respect to x

$\frac{d}{dx} \left ( sin x\right )^{2}+\frac{d}{dx}\left ( sin y\right )^{2}=1$

$2sinx\frac{d}{dx}sinx+2cosy\frac{d}{dx}cosy = 0$

$2sin\: x\: cos\: x+2cos\: y\left ( -sin\: y \frac{dy}{dx}\right )= 0$

$2sin\: x\: cos\: x-2cos\: y.sin\: y \frac{dy}{dx}= 0$

$2cos\: y.sin\: y \frac{dy}{dx}= 2sin\: x\: cos\: x$

$\frac{dy}{dx}=\frac{2sinx.cosx}{2siny.cosy}$

$\frac{dy}{dx}=\frac{sin2x}{sin2y}$

$Q9.y=sin^{-1}\frac{2x}{1+x^{2}}$

Ans. The given function is

$y=sin^{-1}\frac{2x}{1+x^{2}}$

Simplifying the given function

Let x = tan θ

$y=sin^{-1}\frac{2tan\Theta }{1+tan^{2}\Theta }=sin^{-1}\frac{2sin\Theta /cos\Theta}{sec^{2}\Theta }=sin^{-1} \left ( 2sin\Theta cos\Theta \right ) =sin^{-1}sin2\Theta$

$y=2\Theta$

Putting back

$\Theta =tan^{-1} x$

$y =2tan^{-1} x$

Differentiating it with repect to x

$\frac{dy}{dx} =2\frac{d}{dx}tan^{-1} x$

$\frac{dy}{dx} =2.\frac{1}{1+x^{2}}$

$\frac{dy}{dx} =\frac{2}{1+x^{2}}$

Q10.

$y = tan^{-1}\left ( \frac{3x-x^{3}}{1-3x^{2}} \right ),\frac{-1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}$

Ans. The given function is

$y = tan^{-1}\left ( \frac{3x-x^{3}}{1-3x^{2}} \right )$

First of all simplifying the function

Let x = tanθ

$y = tan^{-1}\left ( \frac{3tan\Theta -tan^{3}\Theta }{1-3tan^{2}\Theta } \right )$

$y = tan^{-1}\left ( tan3\Theta \right )$

y = 3θ

Putting back θ = tan^-1x

y = 3 tan^-1x

Differentiating it with respect to x

$\frac{dy}{dx}=3\frac{d}{dx}\left ( tan^{-1}x \right )$

$=3.\frac{1}{1+x^{2}}$

$\frac{dy}{dx}=\frac{3}{1+x^{2}}$

$Q11.y =cos^{-1}\left ( \frac{1-x^{2}}{1+x^{2}} \right ),0<x<1$

The given function is

$y =cos^{-1}\left ( \frac{1-x^{2}}{1+x^{2}} \right )$

First of all simplifying the given function

Let x = tanθ

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

y =cos^-1(cos2θ)

y = 2θ

Putting back θ =tan^-1x

y = 2tan^-1x

Differentiating it with respect to x

$\frac{dy}{dx}=2\frac{d}{dx}\left ( tan^{-1} x\right )$

$=2.\frac{1}{1+x^{2}}$

$\frac{dy}{dx}=\frac{2}{1+x^{2}}$

$Q12.y=sin^{-1}\left ( \frac{1-x^{2}}{1+x^{2}} \right ),-1<x<1$

Ans. The given function is

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

First of all simplifying the given function

Let x = tanθ

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

y =sin^-1(cos2θ)

y =sin^-1sin(π/2-2θ)

y =π/2 -2θ

Putting back θ =tan^-1x

y = π/2 -2tan^-1x

Differentiating the function with respect to x

$\frac{dy}{dx}=0-2\frac{d}{dx}\left ( tan^{-1}x \right )$

$\frac{dy}{dx}=-\frac{2}{1+x^{2}}$

$Q13.y=cos^{-1}\frac{2x}{1+x^{2}},-1<x<1$

Ans. The given function is

$y=cos^{-1}\frac{2x}{1+x^{2}}$

First of all simplifying the given function

Let x = tanθ

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

y =cos^-1(sin2θ)

y =cos^-1cos(π/2-2θ)

y =π/2 -2θ

Putting back θ =tan^-1x

y = π/2 -2tan^-1x

Differentiating the function with respect to x

$\frac{dy}{dx}=0-2\frac{d}{dx}\left ( tan^{-1}x \right )$

$\frac{dy}{dx}=\frac{-2}{1+x^{2}}$

$Q14.y = sin^{-1}\left ( 2x\sqrt{1-x^{2}} \right ),\frac{-1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$

Ans.The given function is

$y =sin^{-1}\left (2x\sqrt{1-x^{2}} \right )$

First of all simplifying the given function

Let x =sinθ

$y =sin^{-1}\left (2sin\Theta \sqrt{1-sin^{2}\Theta } \right )$

$y =sin^{-1}\left (2sin\Theta \sqrt{cos^{2} \Theta } \right )$

y =sin^-1(2sinθ.cosθ)

y =sin^-1(sin2θ)

y = 2θ

Putting back θ =tan^-1x

y = 2tan^-1x

Differentiating the function with respect to x

$\frac{dy}{dx}=2\frac{d}{dx}\left ( sin^{-1} x\right )$

$\frac{dy}{dx}=\frac{2}{\sqrt{1+x^{2}}}$

$Q15.y = sec^{-1}\left ( \frac{1}{2x^{2}-1} \right ),0<x<\frac{1}{\sqrt{2}}$

Ans.The given function is

$y = sec^{-1}\left ( \frac{1}{2x^{2}-1} \right )$

First of all simplifying the given function

Let x =cosθ

$y = sec^{-1}\left ( \frac{1}{2cos^{2}\Theta -1} \right )$

$y = sec^{-1}\left ( \frac{1}{cos2\Theta } \right )$

y =sec^-1(sec2θ)

y = 2θ

Putting back θ =cos^-1x

y = 2cos^-1x

Differentiating the function with respect to x

$\frac{dy}{dx}=2\frac{d}{dx}\left ( cos^{-1} x\right )$

$\frac{dy}{dx}=2.\frac{-1}{\sqrt{\sqrt{1-x^{2}}}}$

$\frac{dy}{dx}=\frac{-2}{\sqrt{\sqrt{1-x^{2}}}}$

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Continuity and Differentiability Class 12 Ncert Solutions 5.3

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Continuity and Differentiability Class 12 Ncert Solutions 5.3

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

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability

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

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