CBSE 2026 · Region 3 · Set 1 · Q30 · 3 marks
If $\displaystyle (\sin \mathrm{x})^{\mathrm{y}}=\mathrm{y}^{\cos \mathrm{x}}$, then find $\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}$.
Marking-scheme solution
$\displaystyle (\sin x)^{y}=y^{\cos x} \Rightarrow y \log \sin x=\cos x \log y$
On diff. wrt x both sides, we get
$\displaystyle \dfrac{y \cos x}{\sin x}+\log \sin x \dfrac{dy}{dx}=\dfrac{\cos x}{y} \dfrac{dy}{dx}-\log y \cdot \sin x$
$\displaystyle \Rightarrow \dfrac{dy}{dx}=\dfrac{y\left(y \cos x+\log y \cdot \sin^{2} x\right)}{\sin x(\cos x-y \log \sin x)}$ or $\displaystyle \dfrac{(y \cot x+\sin x \cdot \log y) y}{\cos x-y \log \sin x}$
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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.