AP Inter 2nd Year Maths Exercise 5c Solutions

Referring to the AP Inter 2nd Year Maths Study Material Chapter 5 Continuity and Differentiability Exercise 5c Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Continuity and Differentiability Solutions Exercise 5c

I.

Question 1.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of 2x + 3y = sin x
Solution:
Given that 2x + 3y = sin x.
Differentiating both sides w.r.t x, we have
\(\frac{\mathrm{d}}{\mathrm{dx}}\) (2x) + \(\frac{\mathrm{d}}{\mathrm{dx}}\) (3y) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) sin x
∴ 2 + 3 \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = cos x
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\cos x-2}{3}\)
⇒ 3\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = cos x – 2

Question 2.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of 2x + 3y = sin y
Solution:
Given that 2x + 3y = sin y ⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (2x) + \(\frac{\mathrm{d}}{\mathrm{dx}}\) (3y)
= \(\frac{\mathrm{d}}{\mathrm{dx}}\) sin y
∴ 2 + 3\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = cos y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) – 3\(\frac{\mathrm{dy}}{\mathrm{dx}}\)
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) (cos y – ) = 2
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(=\frac{2}{\cos y-3}\)

Question 3.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of ax + by² = cos y
Solution:
Given that ax + by² = cos y ⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (ax) + \(\frac{\mathrm{d}}{\mathrm{dx}}\) (by²) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) (cos y)
∴ a + b.2y \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = -sin y\(\frac{\mathrm{d}}{\mathrm{dx}}\)
⇒ 2by \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + sin y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = -a
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) (2by + sin y) = -a ⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{-a}{2 b y+\sin y}\)

Question 4.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of xy + y² = tan x + y
Solution:
Given that xy + y² = tan x + y. ⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\)(xy) + \(\frac{\mathrm{d}}{\mathrm{dx}}\)y² = \(\frac{\mathrm{d}}{\mathrm{dx}}\)tan x + \(\frac{\mathrm{d}}{\mathrm{dx}}\)y
(x\(\frac{\mathrm{d}}{\mathrm{dx}}\)y + y\(\frac{\mathrm{d}}{\mathrm{dx}}\)x) + 2y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = sec²x + \(\frac{\mathrm{dy}}{\mathrm{dx}}\) [∵ From Product Rule]
⇒ x\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y + 2y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = sec² x + \(\frac{\mathrm{dy}}{\mathrm{dx}}\) ⇒ x\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + 2y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) – \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = sec² x – y
⇒ (x + 2y – 1)\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = sec²x – y
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\sec ^2 x-y}{x+2 y-1}\)

Question 5.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of x² + xy + y² = 100
Solution:
Given that ² + xy + y² = 100
⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\)x² + \(\frac{\mathrm{d}}{\mathrm{dx}}\)xy + \(\frac{\mathrm{d}}{\mathrm{dx}}\)y² = \(\frac{\mathrm{d}}{\mathrm{dx}}\)(100)
∴ 2x + (x\(\frac{\mathrm{d}}{\mathrm{dx}}\)y + y\(\frac{\mathrm{d}}{\mathrm{dx}}\)x) + 2y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0
⇒ 2x + x\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y + 2y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0
⇒ (x + 2y)\(\frac{\mathrm{dy}}{\mathrm{dx}}\) – 2x – y
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{(2 x+y)}{x+2 y}\)

Question 6.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of x³ + x²y + xy² + y³ = 81
Solution:
Given that x³ + x²y + xy² + y³ = 81
⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\)x³ + \(\frac{\mathrm{d}}{\mathrm{dx}}\)x²y + \(\frac{\mathrm{d}}{\mathrm{dx}}\)xy² + \(\frac{\mathrm{d}}{\mathrm{dx}}\)y³ = 81
∴ 3x² + x² \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y.\(\frac{\mathrm{d}}{\mathrm{dx}}\)x²) + x\(\frac{\mathrm{d}}{\mathrm{dx}}\)y² + y² \(\frac{\mathrm{d}}{\mathrm{dx}}\) (x) = 3y²\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0
⇒ 3x² + x² \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y2x + x2y \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y² . 1 + 3y² \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0
⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (x² +2xy + 3y²) = -3x² – 2xy – y²
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\left(3 x^2+2 x y+y^2\right)}{x^2+2 x y+3 y^2}\)

Question 7.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of sin²y + cos xy = k
Solution:
Given that sin² y + cos xy = k
⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (sin y)² + \(\frac{\mathrm{d}}{\mathrm{dx}}\)cos xy = \(\frac{\mathrm{d}}{\mathrm{dx}}\)(k)
∴ 2(sin.y)\(\frac{\mathrm{d}}{\mathrm{dx}}\)sin y – sin xy\(\frac{\mathrm{d}}{\mathrm{dx}}\)(xy) = 0
⇒ 2sin y cos y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) – sin xy(x\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y.1) = 0
⇒ sin2y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) – x sin xy\(\frac{\mathrm{dy}}{\mathrm{dx}}\) – y sin xy = 0
⇒ (sin2y – x sin xy)\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = y sin xy
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{y \sin x y}{\sin 2 y-x \sin x y}\)

Question 8.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of sin²x + cos² y = 1
Solution:
Given that sin² x + cos² y = 1
⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (sin x)² + \(\frac{\mathrm{d}}{\mathrm{dx}}\) (cos y)² = \(\frac{\mathrm{d}}{\mathrm{dx}}\) (1)
∴ 2(sin x)\(\frac{\mathrm{d}}{\mathrm{dx}}\) sin x + 2(cos y)\(\frac{\mathrm{d}}{\mathrm{dx}}\) cos y = 0
⇒ 2 sin x cos x + 2cos y(-sin y\(\frac{\mathrm{dy}}{\mathrm{dx}}\)) = 0
⇒ 2 sin x cos x – 2 sin y cos y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0
⇒ sin 2x – sin 2y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = o
⇒ sin 2y \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = sin 2x
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\sin 2 x}{\sin 2 y}\)

II.

Question 1.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) in y = sin^-1\(\left(\frac{2 x}{1+x^2}\right)\)
Solution:
Given that y = sin^-1\(\left(\frac{2 x}{1+x^2}\right)\)
To simplify the given inverse t-function, we put x = tanθ.
∴ y = sin^-1\(\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)\) = sin^-1(sin 2θ) = 2θ
⇒ y = 2tan^-1x (∵ x tanθ ⇒ θ = tan^-1x)
∴ \(\frac{d y}{d x}=2 \frac{1}{1+x^2}=\frac{2}{1+x^2}\)

Question 2.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) in y = tan^-1\(\left(\frac{3 x-x^3}{1-3 x^2}\right)\), \(-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}\)
Solution:
Given y = tan^-1\(\left(\frac{3 x-x^3}{1-3 x^2}\right)\), \(-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}\)
Put x = tanθ
∴ y = tan^-1\(\left(\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right)\) = tan^-1(tan 3θ) = 3θ
⇒ y = 3 tan^-1x (∵ x = tan θ ⇒ θ = tan^-1x)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 3 . \(\frac{1}{1+x^2}=\frac{3}{1+x^2}\)

Question 3.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) in y = cos^-1\(\left(\frac{1-x^2}{1+x^2}\right)\), 0 < x < 1
Solution:
Given that y = cos^-1\(\left(\frac{1-x^2}{1+x^2}\right)\), 0 < x < 1
Put x = tanθ
∴ y = cos^-1\(\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)\) = cos^-1(cos 2θ)
y = 2θ = 2tan^-1x (∵ x = tan θ ⇒ θ = tan^-1x)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 2 . \(\frac{1}{1+x^2}=\frac{2}{1+x^2}\)

Question 4.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of y = sin^-1\(\left(\frac{1-x^2}{1+x^2}\right)\), 0 < x < 1
Solution:
Given that y = sin^-1\(\left(\frac{1-x^2}{1+x^2}\right)\)
Put x = tanθ
∴ y = sin^-1\(\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)\) = sin^-1(cos 2θ)
= sin^-1 sin(\(\frac{\pi}{2}\) – 2θ) = \(\frac{\pi}{2}\) – 2θ
⇒ y = \(\frac{\pi}{2}\) – 2 tan^-1x (∵ x = tan θ ⇒ θ = tan^-1x)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0 – 2\(\frac{1}{1+x^2}=\frac{-2}{1+x^2}\)

Question 5.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of y = cos^-1\(\left(\frac{2 x}{1+x^2}\right)\), -1 < x < 1
Solution:
Given that y = cos^-1\(\left(\frac{2 x}{1+x^2}\right)\)
Put x = tanθ
∴ y = cos^-1\(\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)\) = cos^-1(sin 2θ)
= cos^-1 sin(\(\frac{\pi}{2}\) – 2θ) = \(\frac{\pi}{2}\) – 2θ
⇒ y = \(\frac{\pi}{2}\) – 2 tan^-1x (∵ x = tan θ ⇒ θ = tan^-1x)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0 – 2\(\frac{1}{1+x^2}=\frac{-2}{1+x^2}\)

Question 6.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of y = sin^-1(2x\(\sqrt{1-\mathrm{x}^2}\)), \(-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}\)
Solution:
Given that y = sin^-1(2x\(\sqrt{1-\mathrm{x}^2}\)). Put x = sinθ
∴ y = sin^-1(2 sinθ\(1-\sin ^2 \theta\)) = sin^-1(2sinθ\(\cos ^2 \theta\)) = sin^-1(2 sinθ cosθ)
⇒ y = sin^-1 (sin 2θ) = 2θ = 2 sin^-1 x [∵ x = sin θ ⇒ θ = sin^-1x)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{2}{\sqrt{1-x^2}}\)

Question 7.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of y = sec^-1\(\left(\frac{1}{2 x^2-1}\right),\), 0 < x < \(\frac{1}{\sqrt{2}}\)
Solution:
Put x = cosθ then 2x² – 1 = 2cos²θ – 1 = cos 2θ) = 2θ
∴ y = sec^-1\(\left(\frac{1}{\cos 2 \theta}\right)\) = sec^-1 (sec 2θ) = 2θ = 2 cos^-1
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) (2cos^-1x)
= 2\(\left(\frac{-1}{\sqrt{1-x^2}}\right)=\frac{-2}{\sqrt{1-x^2}}\)