NCERT Solutions
Class 12 - Mathematics Part-1 - Chapter 5: CONTINUITY AND DIFFERENTIABILITYExercise 5.3
Class 12 - Mathematics Part-1 - Chapter 5: CONTINUITY AND DIFFERENTIABILITY
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:
\(\dfrac{dy}{dx}=-\dfrac{2}{\sqrt{1-x^2}}\)