CBSE 2026 · Region 4 · Set 1 · Q28 · 3 marks
If $\displaystyle \mathrm{xy}=\mathrm{e}^{\mathrm{x}-\mathrm{y}}$, then find $\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}$.Differentiate $\displaystyle \tan ^{-1}\left(\frac{\sqrt{1+\mathrm{x}^{2}}-\sqrt{1-\mathrm{x}^{2}}}{\sqrt{1+\mathrm{x}^{2}}+\sqrt{1-\mathrm{x}^{2}}}\right)$ with respect to $\displaystyle \cos ^{-1} \mathrm{x}^{2}$.
If $\displaystyle \mathrm{xy}=\mathrm{e}^{\mathrm{x}-\mathrm{y}}$, then find $\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}$.
Differentiate $\displaystyle \tan ^{-1}\left(\frac{\sqrt{1+\mathrm{x}^{2}}-\sqrt{1-\mathrm{x}^{2}}}{\sqrt{1+\mathrm{x}^{2}}+\sqrt{1-\mathrm{x}^{2}}}\right)$ with respect to $\displaystyle \cos ^{-1} \mathrm{x}^{2}$.
Marking-scheme solution
$\displaystyle x y=e^{x-y} \Rightarrow x \dfrac{dy}{dx}+y=e^{x-y}\left(1-\dfrac{dy}{dx}\right)$
$\displaystyle \Rightarrow \left(x+e^{x-y}\right) \dfrac{dy}{dx}=e^{x-y}-y \Rightarrow \dfrac{dy}{dx}=\dfrac{e^{x-y}-y}{x+e^{x-y}}$ or $\displaystyle \dfrac{y(x-1)}{x(y+1)}$
Alternative method:
$\displaystyle x y=e^{x-y} \Rightarrow \log x+\log y=(x-y)$
$\displaystyle \Rightarrow \dfrac{1}{x}+\dfrac{1}{y} \dfrac{dy}{dx}=1-\dfrac{dy}{dx}$
$\displaystyle \Rightarrow \dfrac{dy}{dx}=\dfrac{1-\dfrac{1}{x}}{\dfrac{1}{y}+1}=\dfrac{y(x-1)}{x(y+1)}$
Let $\displaystyle y=\tan^{-1}\left(\dfrac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}\right)$ & put $\displaystyle x^{2}=\cos \theta$
$\displaystyle y=\tan^{-1}\left(\dfrac{\sqrt{1+\cos \theta}-\sqrt{1-\cos \theta}}{\sqrt{1+\cos \theta}+\sqrt{1-\cos \theta}}\right)$
$\displaystyle y=\tan^{-1}\left(\dfrac{\cos\left(\dfrac{\theta}{2}\right)-\sin\left(\dfrac{\theta}{2}\right)}{\cos\left(\dfrac{\theta}{2}\right)+\sin\left(\dfrac{\theta}{2}\right)}\right)=\tan^{-1}\left(\tan\left(\dfrac{\pi}{4}-\dfrac{\theta}{2}\right)\right)$
Thus, $\displaystyle y=\left(\dfrac{\pi}{4}-\dfrac{\theta}{2}\right) \Rightarrow \dfrac{dy}{d\theta}=\dfrac{-1}{2}$
Continuity and DifferentiabilityLogarithmic DifferentiationApplynumericmedium
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CBSE Class 12 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.