NCERT Solutions
Class 12 - Mathematics Part-1 - Chapter 5: CONTINUITY AND DIFFERENTIABILITY
Exercise 5.3

Question. 1

Find \(\dfrac{dy}{dx}\) if \(2x+3y=\sin x\).

Answer:

\(\dfrac{dy}{dx}=\dfrac{\cos x-2}{3}\)

Question. 2

Find \(\dfrac{dy}{dx}\) if \(2x+3y=\sin y\).

Answer:

\(\dfrac{dy}{dx}=\dfrac{2}{\cos y-3}\)

Question. 3

Find \(\dfrac{dy}{dx}\) if \(ax+by^2=\cos y\).

Answer:

\(\dfrac{dy}{dx}=-\dfrac{a}{2by+\sin y}\)

Question. 4

Find \(\dfrac{dy}{dx}\) if \(xy+y^2=\tan x+y\).

Answer:

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

Question. 5

Find \(\dfrac{dy}{dx}\) if \(x^2+xy+y^2=100\).

Answer:

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

Question. 6

Find \(\dfrac{dy}{dx}\) if \(x^3+x^2y+xy^2+y^3=81\).

Answer:

\(\dfrac{dy}{dx}=-\dfrac{3x^2+2xy+y^2}{x^2+2xy+3y^2}\)

Question. 7

Find \(\dfrac{dy}{dx}\) if \(\sin^2 y+\cos(xy)=k\).

Answer:

\(\dfrac{dy}{dx}=\dfrac{y\sin(xy)}{\sin(2y)-x\sin(xy)}\)

Question. 8

Find \(\dfrac{dy}{dx}\) if \(\sin^2 x+\cos^2 y=1\).

Answer:

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

Question. 9

Find \(\dfrac{dy}{dx}\) if \(y=\sin^{-1}\!\left(\dfrac{2x}{1+x^2}\right)\).

Answer:

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

Question. 10

Find \(\dfrac{dy}{dx}\) if \(y=\tan^{-1}\!\left(\dfrac{3x-x^3}{1-3x^2}\right)\), for \(-\dfrac{1}{\sqrt{3}}<x<\dfrac{1}{\sqrt{3}}\).

Answer:

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

Question. 11

Find \(\dfrac{dy}{dx}\) if \(y=\cos^{-1}\!\left(\dfrac{1-x^2}{1+x^2}\right)\), for \(0<x<1\).

Answer:

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

Question. 12

Find \(\dfrac{dy}{dx}\) if \(y=\sin^{-1}\!\left(\dfrac{1-x^2}{1+x^2}\right)\), for \(0<x<1\).

Answer:

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

Question. 13

Find \(\dfrac{dy}{dx}\) if \(y=\cos^{-1}\!\left(\dfrac{2x}{1+x^2}\right)\), for \(-1<x<1\).

Answer:

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

Question. 14

Find \(\dfrac{dy}{dx}\) if \(y=\sin^{-1}\!\left(2x\sqrt{1-x^2}\right)\), for \(-\dfrac{1}{\sqrt{2}}<x<\dfrac{1}{\sqrt{2}}\).

Answer:

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

Question. 15

Find \(\dfrac{dy}{dx}\) if \(y=\sec^{-1}\!\left(\dfrac{1}{2x^2-1}\right)\), for \(0<x<\dfrac{1}{\sqrt{2}}\).

Answer:

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

NCERT Solutions Class 12 – Mathematics Part-1 – Chapter 5: CONTINUITY AND DIFFERENTIABILITY – Exercise 5.3 | Detailed Answers