Logarithmic Differentiation PYQs
Practice Class 12 CBSE Board Previous Year Questions (2014-2026)
Q1
2024
00:00
If $y = (\sin x)^{\cos x} + (\cos x)^{\sin x}$, find $dy/dx$. 5 Marks
Let $y = u + v$. $u = (\sin x)^{\cos x} \Rightarrow \log u = \cos x \log(\sin x)$.
$\frac{1}{u}\frac{du}{dx} = -\sin x \log(\sin x) + \cos x \frac{\cos x}{\sin x}$.
$\frac{du}{dx} = (\sin x)^{\cos x} [\cos x \cot x - \sin x \log(\sin x)]$.
$v = (\cos x)^{\sin x} \Rightarrow \log v = \sin x \log(\cos x)$.
$\frac{1}{v}\frac{dv}{dx} = \cos x \log(\cos x) + \sin x \frac{-\sin x}{\cos x}$.
$\frac{dv}{dx} = (\cos x)^{\sin x} [\cos x \log(\cos x) - \sin x \tan x]$.
Sum of $du/dx$ and $dv/dx$
Q2
2023
00:00
If $y = x^{\sin x} + (\sin x)^x$, find $dy/dx$. 5 Marks
Let $y = u + v$. $u = x^{\sin x} \Rightarrow \frac{du}{dx} = x^{\sin x} [\cos x \log x + \frac{\sin x}{x}]$.
$v = (\sin x)^x \Rightarrow \frac{dv}{dx} = (\sin x)^x [\log(\sin x) + x \cot x]$.
Sum of $du/dx$ and $dv/dx$
Q3
2022 (Term 2)
00:00
If $y = x^x + x^{1/x}$, find $dy/dx$. 5 Marks
$du/dx = x^x (1 + \log x)$.
$dv/dx = x^{1/x} \left(\frac{1-\log x}{x^2}\right)$.
Sum of derivatives
Q4
2019
00:00
If $y = (\sin x)^x + \sin^{-1}(\sqrt{x})$, find $dy/dx$. 4 Marks
$du/dx = (\sin x)^x [x \cot x + \log(\sin x)]$.
$dv/dx = \frac{1}{\sqrt{1-(\sqrt{x})^2}} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{x}\sqrt{1-x}}$.
Sum of derivatives
Q5
2015
00:00
If $y = \frac{x\cos^{-1}x}{\sqrt{1-x^2}} - \log\sqrt{1-x^2}$, prove $dy/dx = \frac{\cos^{-1}x}{(1-x^2)^{3/2}}$. 5 Marks
Differentiate using quotient rule and chain rule.
$\frac{dy}{dx} = \frac{\sqrt{1-x^2}(\cos^{-1}x - \frac{x}{\sqrt{1-x^2}}) - x\cos^{-1}x(\frac{-x}{\sqrt{1-x^2}})}{1-x^2} - \frac{1}{\sqrt{1-x^2}} \cdot \frac{-x}{\sqrt{1-x^2}}$.
Simplifying leads to the result.
Proved
Q6
2021
00:00
If $y = (\cos x)^x + x^{\cos x}$, find $dy/dx$. 5 Marks
Let $u = (\cos x)^x \Rightarrow \frac{du}{dx} = (\cos x)^x [\log(\cos x) - x \tan x]$.
Let $v = x^{\cos x} \Rightarrow \frac{dv}{dx} = x^{\cos x} [\frac{\cos x}{x} - \sin x \log x]$.
Sum of derivatives
Q7
2020
00:00
If $y = (x \cos x)^x + (x \sin x)^{1/x}$, find $dy/dx$. 5 Marks
Let $u = (x \cos x)^x \Rightarrow \log u = x [\log x + \log(\cos x)]$.
$\frac{du}{dx} = (x \cos x)^x [\log(x \cos x) + 1 - x \tan x]$.
Let $v = (x \sin x)^{1/x} \Rightarrow \log v = \frac{1}{x} [\log x + \log(\sin x)]$.
$\frac{dv}{dx} = (x \sin x)^{1/x} [\frac{x \cot x + 1 - \log(x \sin x)}{x^2}]$.
Sum of derivatives
Q8
2018
00:00
If $y = x^{\cos x} + (\cos x)^{\sin x}$, find $dy/dx$. 5 Marks
$du/dx = x^{\cos x} [\frac{\cos x}{x} - \sin x \log x]$.
$dv/dx = (\cos x)^{\sin x} [\cos x \log(\cos x) - \sin x \tan x]$.
Sum of derivatives
Q9
2017
00:00
If $y = (\sin x)^x + \sin^{-1}\sqrt{x}$, find $dy/dx$. 4 Marks
$du/dx = (\sin x)^x [\log(\sin x) + x \cot x]$.
$dv/dx = \frac{1}{2\sqrt{x-x^2}}$.
Sum of derivatives
Q10
2016
00:00
Find $dy/dx$ if $y = \sin^{-1}\left[\frac{6x - 4\sqrt{1-4x^2}}{5}\right]$. 3 Marks
$y = \sin^{-1}\left[\frac{3}{5}(2x) - \frac{4}{5}\sqrt{1-(2x)^2}\right]$.
Let $2x = \sin \alpha$ and $3/5 = \cos \beta \Rightarrow 4/5 = \sin \beta$.
$y = \sin^{-1}(\sin \alpha \cos \beta - \cos \alpha \sin \beta) = \sin^{-1}(\sin(\alpha - \beta))$.
$y = \alpha - \beta = \sin^{-1}(2x) - \cos^{-1}(3/5)$.
$\frac{dy}{dx} = \frac{1}{\sqrt{1-4x^2}} \cdot 2 - 0 = \frac{2}{\sqrt{1-4x^2}}$.
$\frac{2}{\sqrt{1-4x^2}}$