CBSE 2026 · Region 2 · Set 3 · Q23 · 2 marks
If $\displaystyle x=e^{\sin ^{-1} \mathrm{t}}, y=e^{\cos ^{-1} \mathrm{t}}$, find $\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}$ at $\displaystyle \mathrm{t}=\frac{1}{\sqrt{2}}$
Marking-scheme solution
$\displaystyle x=e^{\sin^{-1} t} \Rightarrow \dfrac{dx}{dt}=e^{\sin^{-1} t} \times \dfrac{1}{\sqrt{1-t^{2}}}$
$\displaystyle \mathrm{y}=e^{\cos^{-1} t} \Rightarrow \dfrac{dy}{dt}=e^{\cos^{-1} t} \times \dfrac{-1}{\sqrt{1-t^{2}}}$
Now, $\displaystyle \dfrac{dy}{dx}=-\dfrac{e^{\cos^{-1} t}}{e^{\sin^{-1} t}}$ or $\displaystyle -e^{\cos^{-1} t-\sin^{-1} t}$
$\displaystyle \left.\dfrac{dy}{dx}\right]_{t=\frac{1}{\sqrt{2}}}=-1$
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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.