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Mathematics · 2025 · 2 marks
CBSE 2025 · Region 1 · Set 1 · Q21
(a) Differentiate 2^cos^2 x w. r. t cos^2 x.
(a)
Differentiate $\displaystyle 2^{\cos ^{2} x}$ w. r. t $\displaystyle \cos ^{2} x$.
OR
(b)
If $\displaystyle \tan ^{-1}\left(x^{2}+\mathrm{y}^{2}\right)=\mathrm{a}^{2}$, then find $\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}$.
Marking-scheme solution
Let $\displaystyle u=2^{\cos ^{2} x} \Rightarrow \frac{d u}{d x}=2^{\cos ^{2} x}(-2 \cos x \sin x) \log 2$
Let $\displaystyle v=\cos ^{2} x \Rightarrow \frac{d v}{d x}=-2 \cos x \sin x$
Now $\displaystyle \frac{d u}{d v}=\frac{\left(\dfrac{d u}{d x}\right)}{\left(\dfrac{d v}{d x}\right)}=2^{\cos ^{2} x} \log 2$
\[\tan ^{-1}\left(x^{2}+\mathrm{y}^{2}\right)=\mathrm{a}^{2} \Rightarrow x^{2}+\mathrm{y}^{2}=\tan \mathrm{a}^{2}
\]
Differentiatebothsides wrt $\displaystyle \boldsymbol{x}$,
\[\begin{aligned}
& 2 x+2 \mathrm{y} \frac{d \mathrm{y}}{d x}=0 \\
& \Rightarrow \frac{d \mathrm{y}}{d x}=-\frac{x}{\mathrm{y}}
\end{aligned}
\]
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CBSE Class 12 Mathematics past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.