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

Find  of the following:

Question 1. 2x + 3y = sin x

Solution:

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

2 + 3  = cos x

3  = cos x – 2

 = (cosx – 2)/3 

Question 2. 2x + 3y = sin y

Solution:

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

2 + 3  = cos y

(cosy – 3)  = 2

Question 3. ax + by² = cos y

Solution:

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

a + b * 2y() = -sin y * 

(2by + siny)  = -a

Question 4. xy + y² = tan x + y

Solution:

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

(x * + y) + 2y = sec²x + 

(x + 2y – 1) = sec²x – y

Question 5. x² + xy + y² = 100

Solution:

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

2x + (x + y) + 2y = 0

(x + 2y) *  = -(2x + y)

Question 6. x³ + x²y + xy² + y³ = 81

Solution:

Differentiate both sides w.r.t. x

3x²+(x² + y * 2x) + (x * 2y *  + y²) + 3y² *  = 0

(x² + 2xy + 3y²) = -(3x² + 2xy + y²)

Question 7. Sin²y + cos xy = π

Solution:

Differentiate both sides w.r.t. x

2 sin y *  (sin⁡y) – sin(xy) *  xy = 0

2sin y * cosy  – sin(xy)(x *  + y) = 0

(2sin cos y – sin (xy) – x))  = y(xy)

Question 8. sin² x + cos² y = 1

Solution:

 2 sin x *  (sin ⁡x) + 2 cos y *  (cos ⁡y) = 0

2 sin x * cos x + 2 cos y*(-sin y) *  = 0

2 sin x * cos x – 2 cos x – 2 cos y sin y *  = 0

Sin(2x) – sin(2y) –  = 0

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

Solution:

Put x = tanθ          

θ = tan^-1x

y = sin^-1(sin 2θ)          

y = 2θ

y = 2tan^-1x         -(1)

On differentiating eq(1), we get

Question 10. , -1/√3 < x < 1/√3

Solution:

Put x = tanθ         

θ = tan^-1x

y = 

y = tan^-1(tan 3θ)         

y = 3θ

y = 3tan^-1x          -(1)

On differentiating eq(1), we get

Question 11. , 0 < x < 1

Solution:

Put x = tanθ          

θ  = tan^-1 x

y = 

y = cos^-1(cos 2θ)   

y = 2θ 

y = 2tan^-1x         -(1)

On differentiating eq(1), we get

Question 12. , 0 < x < 1

Solution:

Put x = tanθ       

θ = tan^-1x

y = sin^-1(cos 2θ)

y = sin^-1(sin (π/2 – 2θ))         

y = π/2 – 2θ

y = π/2 – 2 tan^-1x  

Question 13. , -1 < x < 1

Solution:

Put x = tanθ     

θ = tan^-1x

y = cos^{-1}()        

y = cos^-1(sin 2θ)

y = cos^-1(cos (π/2 – 2θ))         

y = π/2 – 2θ

y = π/2 – 2tan^-1x  

Question 14. , -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

Question 15. , 0 < x < 1/√2

Solution:

Put x = tanθ

y = sec^-1(1/cos2θ))

y = sec^-1(sec2θ) = 2θ

y = 2cos^-1x

 =