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Class 12 NCERT Solutions- Mathematics Part I – Chapter 5 Continuity And Differentiability – Exercise 5.3

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  • Last Updated : 09 Mar, 2021

Find \frac{dy}{dx}  of the following:

Question 1. 2x + 3y = sin x

Solution:

On differentiating both sides w.r.t. x, we get

2 + 3 \frac{dy}{dx}  = cos x

\frac{dy}{dx}  = cos x – 2

\frac{dy}{dx}  = (cosx – 2)/3 

Question 2. 2x + 3y = sin y

Solution:

On differentiating both sides w.r.t. x, we get

2 + 3 \frac{dy}{dx}  = cos y\frac{dy}{dx}

(cosy – 3) \frac{dy}{dx}  = 2

\frac{dy}{dx} = \frac{2}{(cos⁡y - 3)}

Question 3. ax + by2 = cos y

Solution:

On differentiating both sides w.r.t. x, we get

a + b * 2y(\frac{dy}{dx} ) = -sin y * \frac{dy}{dx}

(2by + siny) \frac{dy}{dx}  = -a

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

Question 4. xy + y2 = tan x + y

Solution:

On differentiating both sides w.r.t. x, we get

(x *\frac{dy}{dx}  + y) + 2y\frac{dy}{dx}  = sec2x + \frac{dy}{dx}

(x + 2y – 1)\frac{dy}{dx}  = sec2x – y

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

Question 5. x2 + xy + y2 = 100

Solution:

On differentiating both sides w.r.t. x, we get

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

(x + 2y) * \frac{dy}{dx}  = -(2x + y)

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

Question 6. x3 + x2y + xy2 + y3 = 81

Solution:

Differentiate both sides w.r.t. x

3x2+(x2\frac{dy}{dx}  + y * 2x) + (x * 2y * \frac{dy}{dx}  + y2) + 3y2 \frac{dy}{dx}  = 0

(x2 + 2xy + 3y2)\frac{dy}{dx}  = -(3x2 + 2xy + y2)

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

Question 7. Sin2y + cos xy = π

Solution:

Differentiate both sides w.r.t. x

2 sin y * \frac{dy}{dx}  (sin⁡y) – sin(xy) * \frac{dy}{dx}  xy = 0

2sin y * cosy \frac{dy}{dx}  – sin(xy)(x * \frac{dy}{dx}  + y) = 0

(2sin cos y – sin (xy) – x)) \frac{dy}{dx}  = y(xy)

\frac{dy}{dx} = \frac{y sin⁡(xy)}{2 sin⁡ y cos⁡ y - x sin⁡ (xy)}

\frac{dy}{dx} = \frac{ysin(xy)}{sin2y - x sinxy}

Question 8. sin2 x + cos2 y = 1

Solution:

 2 sin x * \frac{dy}{dx}  (sin ⁡x) + 2 cos y * \frac{dy}{dx}  (cos ⁡y) = 0

2 sin x * cos x + 2 cos y*(-sin y) * \frac{dy}{dx}  = 0

2 sin x * cos x – 2 cos x – 2 cos y sin y * \frac{dy}{dx}  = 0

Sin(2x) – sin(2y) – \frac{dy}{dx}  = 0

\frac{dy}{dx} = \frac{sin(2x)}{sin(2y)}

Question 9. y = sin-1(\frac{2x}{(1 + x2)}

Solution:

Put x = tanθ          

θ = tan-1x

y = sin^{-1}\frac{2tanθ}{1+tan^2θ}

y = sin-1(sin 2θ)          

y = 2θ

y = 2tan-1x         -(1)

On differentiating eq(1), we get

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

Question 10. Y=tan^{-1}\frac{(3x-x^3)}{(1-3x^2 )}  , -1/√3 < x < 1/√3

Solution:

Put x = tanθ         

θ = tan-1x

y = tan^{-1}(\frac{3tanθ - tan^3θ}{1-3tan^2θ})

y = tan-1(tan 3θ)         

y = 3θ

y = 3tan-1x          -(1)

On differentiating eq(1), we get

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

Question 11. y = cos^{-1}(\frac{1-x^2}{1+x^2})  , 0 < x < 1

Solution:

Put x = tanθ          

θ  = tan-1 x

y = cos^{-1}(\frac{1-tan^2θ}{1+tan^2θ})

y = cos-1(cos 2θ)   

y = 2θ 

y = 2tan-1x         -(1)

On differentiating eq(1), we get

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

Question 12. y = sin^{-1}(\frac{1-x^2}{1+x^2})  , 0 < x < 1

Solution:

Put x = tanθ       

θ = tan-1x

y = sin^{-1}(\frac{1 - tan^2θ}{1 + tan^2θ})

y = sin-1(cos 2θ)

y = sin-1(sin (π/2 – 2θ))         

y = π/2 – 2θ

y = π/2 – 2 tan-1x  

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

Question 13. y = cos^{-1}(\frac{2x}{1+x^2})  , -1 < x < 1

Solution:

Put x = tanθ     

θ = tan-1x

y = cos^{-1}(\frac{2tanθ}{1 + tan^2θ} )        

y = cos-1(sin 2θ)

y = cos-1(cos (π/2 – 2θ))         

y = π/2 – 2θ

y = π/2 – 2tan-1x  

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

Question 14. y = sin^{-1}(2x\sqrt{1-x^2})  , -1/√2 < x < 1/√2

Solution:

Put x = sinθ     

θ =  sin-1 x

y = sin-1(2sinθ√(1 – sin2θ))

y = sin-1(sin 2θ) = 2θ

y = 2sin-1x

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

Question 15. y = sec^{-1}(\frac{1}{2x^2-1})  , 0 < x < 1/√2

Solution:

Put x = tanθ

y = sec^{-1}(\frac{1}{2 cos^2 - 1})         

y = sec-1(1/cos2θ))

y = sec-1(sec2θ) = 2θ

y = 2cos-1x

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


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