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

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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

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

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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Last Updated : 09 Mar, 2021

Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.2

Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.4

Chapter 1 - Relations and Functions

Chapter 2 - Inverse Trigonometric Functions

Chapter 3 - Matrices

Chapter 4 - Determinants

Chapter 5 - Continuity and Differentiability

Chapter 6 - Applications of Derivatives

Chapter 1 - Relations and Functions

Chapter 2 - Inverse Trigonometric Functions

Chapter 3 - Matrices

Chapter 4 - Determinants

Chapter 5 - Continuity and Differentiability

Chapter 6 - Applications of Derivatives

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

Question 2. 2x + 3y = sin y

Question 3. ax + by2 = cos y

Question 4. xy + y2 = tan x + y

Question 5. x2 + xy + y2 = 100

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

Question 7. Sin2y + cos xy = π

Question 8. sin2 x + cos2 y = 1

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

Question 10. , -1/√3 < x < 1/√3

Question 11. , 0 < x < 1

Question 12. , 0 < x < 1

Question 13. , -1 < x < 1

Question 14. , -1/√2 < x < 1/√2

Question 15. , 0 < x < 1/√2

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Find $\frac{dy}{dx}$ of the following:

Question 3. ax + by²= cos y

Question 4. xy + y² = tan x + y

Question 5. x² + xy + y² = 100

Question 6. x³+ x²y + xy² + y³ = 81

Question 7. Sin²y + cos xy = π

Question 8. sin² x + cos² y = 1

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

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

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

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

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

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

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