Implicit Differentiation PYQs
Practice Class 12 CBSE Board Previous Year Questions (2014-2026)
Q1
2024
00:00
If $x^2 + xy + y^2 = 100$, find $dy/dx$. 3 Marks
Differentiating w.r.t $x$: $2x + (x y' + y) + 2y y' = 0$.
$(x + 2y) y' = -(2x + y)$.
$y' = -\frac{2x + y}{x + 2y}$.
$-\frac{2x+y}{x+2y}$
Q2
2023
00:00
Find $dy/dx$ at $(1, \pi/4)$ if $\sin^2 y + \cos xy = k$. 3 Marks
$2\sin y \cos y y' - \sin(xy)(x y' + y) = 0$.
$\sin 2y y' - x \sin(xy) y' - y \sin(xy) = 0$.
$y'(\sin 2y - x \sin(xy)) = y \sin(xy)$.
At $(1, \pi/4)$: $y'(\sin(\pi/2) - 1 \sin(\pi/4)) = (\pi/4) \sin(\pi/4)$.
$y'(1 - 1/\sqrt{2}) = \frac{\pi}{4\sqrt{2}} \Rightarrow y' = \frac{\pi/4\sqrt{2}}{(\sqrt{2}-1)/\sqrt{2}} = \frac{\pi}{4(\sqrt{2}-1)}$.
$\frac{\pi}{4(\sqrt{2}-1)}$
Q3
2020
00:00
Find $dy/dx$ if $x^y + y^x = a^b$. 5 Marks
Let $u = x^y, v = y^x$. $u + v = a^b \Rightarrow \frac{du}{dx} + \frac{dv}{dx} = 0$.
$\frac{du}{dx} = x^y [\frac{y}{x} + \log x \frac{dy}{dx}]$.
$\frac{dv}{dx} = y^x [\frac{x}{y}\frac{dy}{dx} + \log y]$.
$x^y \frac{y}{x} + x^y \log x y' + y^x \frac{x}{y} y' + y^x \log y = 0$.
$y' (x^y \log x + x y^{x-1}) = -(y x^{y-1} + y^x \log y)$.
$y' = -\frac{y x^{y-1} + y^x \log y}{x^y \log x + x y^{x-1}}$.
Ratio of terms
Q4
2018
00:00
If $e^x + e^y = e^{x+y}$, prove that $dy/dx = -e^{y-x}$. 3 Marks
$e^x + e^y \frac{dy}{dx} = e^{x+y}(1 + \frac{dy}{dx})$.
$e^x + e^y y' = e^{x+y} + e^{x+y} y'$.
$y'(e^y - e^{x+y}) = e^{x+y} - e^x$.
Substitute $e^{x+y} = e^x + e^y$:
$y'(e^y - (e^x + e^y)) = (e^x + e^y) - e^x$.
$y'(-e^x) = e^y \Rightarrow y' = -\frac{e^y}{e^x} = -e^{y-x}$.
Proved
Q5
2015 (All India)
00:00
If $\cos y = x \cos(a+y)$, prove $dy/dx = \frac{\cos^2(a+y)}{\sin a}$. 3 Marks
$x = \frac{\cos y}{\cos(a+y)}$.
Differentiate $x$ w.r.t $y$: $\frac{dx}{dy} = \frac{\cos(a+y)(-\sin y) - \cos y (-\sin(a+y))}{\cos^2(a+y)}$.
$\frac{dx}{dy} = \frac{\sin(a+y)\cos y - \cos(a+y)\sin y}{\cos^2(a+y)} = \frac{\sin(a+y-y)}{\cos^2(a+y)} = \frac{\sin a}{\cos^2(a+y)}$.
So $\frac{dy}{dx} = \frac{\cos^2(a+y)}{\sin a}$.
Proved
Q6
2022
00:00
If $ax^2 + 2hxy + by^2 = 0$, find $dy/dx$. 2 Marks
$2ax + 2h(x y' + y) + 2by y' = 0$.
$ax + hx y' + hy + by y' = 0$.
$y'(hx + by) = -(ax + hy)$.
$y' = -\frac{ax + hy}{hx + by}$.
$-\frac{ax+hy}{hx+by}$
Q7
2019
00:00
Find $dy/dx$ if $y^x = x^y$. 3 Marks
$x \log y = y \log x$.
$1 \cdot \log y + x \cdot \frac{1}{y} y' = y' \log x + y \cdot \frac{1}{x}$.
$y'(\frac{x}{y} - \log x) = \frac{y}{x} - \log y$.
$y' = \frac{(y/x) - \log y}{(x/y) - \log x} = \frac{y(y - x \log y)}{x(x - y \log x)}$.
$\frac{y(y - x \log y)}{x(x - y \log x)}$
Q8
2017
00:00
Find $dy/dx$ if $\sin^2 x + \cos^2 y = 1$. 2 Marks
$2\sin x \cos x + 2\cos y (-\sin y) y' = 0$.
$\sin 2x - \sin 2y y' = 0$.
$y' = \frac{\sin 2x}{\sin 2y}$.
$\frac{\sin 2x}{\sin 2y}$
Q9
2016
00:00
If $x \cos(a+y) = \cos y$, prove $dy/dx = \frac{\cos^2(a+y)}{\sin a}$, and hence prove $\sin a \cdot y'' + \sin 2(a+y) \cdot y' = 0$. 5 Marks
$x = \frac{\cos y}{\cos(a+y)} \Rightarrow \frac{dx}{dy} = \frac{\sin a}{\cos^2(a+y)} \Rightarrow y' = \frac{\cos^2(a+y)}{\sin a}$.
Differentiate $y'$ w.r.t $x$:
$y'' = \frac{1}{\sin a} [2\cos(a+y)(-\sin(a+y)) y'] = -\frac{\sin 2(a+y)}{\sin a} y'$.
$\sin a y'' = -\sin 2(a+y) y' \Rightarrow \sin a y'' + \sin 2(a+y) y' = 0$.
Proved
Q10
2025 Board
00:00
If $\tan^{-1}(x^2 + y^2) = a^2$, then find $dy/dx$. 2 Marks
$x^2 + y^2 = \tan(a^2)$.
Differentiating w.r.t $x$: $2x + 2y y' = 0$ (since $\tan(a^2)$ is constant).
$2y y' = -2x \Rightarrow y' = -x/y$.
-x/y