Differentiation Rules PYQs
Practice Class 12 CBSE Board Previous Year Questions (2014-2026)
Q1
2024 (Set 65/2/1)
00:00
If $x = a(\theta + \sin\theta)$ and $y = a(1 - \cos\theta)$, find $\frac{dy}{dx}$ at $\theta = \frac{\pi}{2}$. 2 Marks
$\frac{dx}{d\theta} = a(1 + \cos\theta)$.
$\frac{dy}{d\theta} = a(0 - (-\sin\theta)) = a \sin\theta$.
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a \sin\theta}{a(1 + \cos\theta)} = \frac{\sin\theta}{1 + \cos\theta}$.
Using half-angle formulas: $\frac{dy}{dx} = \frac{2\sin(\theta/2)\cos(\theta/2)}{2\cos^2(\theta/2)} = \tan(\theta/2)$.
At $\theta = \frac{\pi}{2}$, $\frac{dy}{dx} = \tan(\pi/4) = 1$.
1
Q2
2017
00:00
Differentiate $\sin^{-1} \frac{2x}{1+x^2}$ with respect to $\tan^{-1} x$. 4 Marks
Let $u = \sin^{-1} \frac{2x}{1+x^2}$. Let $x = \tan \theta$.
$u = \sin^{-1} \frac{2\tan\theta}{1+\tan^2\theta} = \sin^{-1}(\sin 2\theta) = 2\theta = 2\tan^{-1} x$.
$\frac{du}{dx} = \frac{2}{1+x^2}$.
Let $v = \tan^{-1} x$. $\frac{dv}{dx} = \frac{1}{1+x^2}$.
$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{2/(1+x^2)}{1/(1+x^2)} = 2$.
2
Q3
2018
00:00
If $x = a \sin 2t (1 + \cos 2t)$ and $y = b \cos 2t (1 - \cos 2t)$, find $\frac{dy}{dx}$ at $t = \frac{\pi}{4}$. 4 Marks
$\frac{dx}{dt} = a [2\cos 2t (1 + \cos 2t) + \sin 2t (-2\sin 2t)] = 2a [\cos 2t + \cos^2 2t - \sin^2 2t] = 2a [\cos 2t + \cos 4t]$.
$\frac{dy}{dt} = b [-2\sin 2t (1 - \cos 2t) + \cos 2t (2\sin 2t)] = 2b [-\sin 2t + \sin 2t \cos 2t + \sin 2t \cos 2t] = 2b [-\sin 2t + \sin 4t]$.
At $t = \pi/4$: $2t = \pi/2, 4t = \pi$.
$\frac{dx}{dt} = 2a [\cos(\pi/2) + \cos(\pi)] = 2a [0 - 1] = -2a$.
$\frac{dy}{dt} = 2b [-\sin(\pi/2) + \sin(\pi)] = 2b [-1 + 0] = -2b$.
$\frac{dy}{dx} = \frac{-2b}{-2a} = \frac{b}{a}$.
b/a
Q4
2024 Board
00:00
If $x\sqrt{1+y} + y\sqrt{1+x} = 0$, for $-1 < x < 1$, prove that $\frac{dy}{dx} = -\frac{1}{(1+x)^2}$. 4 Marks
$x\sqrt{1+y} = -y\sqrt{1+x}$. Squaring both sides:
$x^2(1+y) = y^2(1+x) \Rightarrow x^2 + x^2y = y^2 + y^2x$.
$(x^2 - y^2) + (x^2y - y^2x) = 0 \Rightarrow (x-y)(x+y) + xy(x-y) = 0$.
$(x-y)(x+y+xy) = 0$. Since $x \neq y$, we have $x+y+xy=0$.
$y(1+x) = -x \Rightarrow y = -\frac{x}{1+x}$.
$\frac{dy}{dx} = -\frac{(1+x)(1) - x(1)}{(1+x)^2} = -\frac{1}{(1+x)^2}$.
Proved