NCERT Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)Exercise 5.3

Jolly September 06, 2022

NCERT Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)Exercise 5.3

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-5 (Continuity And Differentiability)Exercise 5.3 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)Exercise 5.3

Exercise 5.3

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-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

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

Chapter-5 (Continuity And Differentiability)

Exercise 5.1
Exercise 5.2
Exercise 5.4
Exercise 5.5
Exercise 5.6
Exercise 5.7
Exercise 5.8