CBSE 2024 · Region 4 · Set 2 · Q24 · 2 marks
If $\displaystyle \mathrm{y}=\cos ^{3}\left(\sec ^{2} 2 \mathrm{t}\right)$, find $\displaystyle \frac{\mathrm{dy}}{\mathrm{dt}}$.If $\displaystyle x^{\mathrm{y}}=\mathrm{e}^{x-\mathrm{y}}$, prove that $\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\log x}{(1+\log x)^{2}}$.
If $\displaystyle \mathrm{y}=\cos ^{3}\left(\sec ^{2} 2 \mathrm{t}\right)$, find $\displaystyle \frac{\mathrm{dy}}{\mathrm{dt}}$.
If $\displaystyle x^{\mathrm{y}}=\mathrm{e}^{x-\mathrm{y}}$, prove that $\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\log x}{(1+\log x)^{2}}$.
Marking-scheme solution
$$\begin{aligned}
& \mathrm{y}=\cos ^{3}\left(\sec ^{2} 2 \mathrm{t}\right) \\
& \Rightarrow \frac{d \mathrm{y}}{d \mathrm{t}}=3 \cos ^{2}\left(\sec ^{2} 2 \mathrm{t}\right)\left[-\sin \left(\sec ^{2} 2 \mathrm{t}\right)\right] \times \frac{d\left(\sec ^{2} 2 \mathrm{t}\right)}{d \mathrm{t}} \\
& \Rightarrow \frac{d \mathrm{y}}{d \mathrm{t}}=-3 \cos ^{2}\left(\sec ^{2} 2 \mathrm{t}\right) \cdot \sin \left(\sec ^{2} 2 \mathrm{t}\right) \times 2 \sec 2 \mathrm{t} \cdot \sec 2 \mathrm{t} \tan 2 \mathrm{t} \cdot 2 \\
& \therefore \frac{d \mathrm{y}}{d \mathrm{t}}=-12 \cos ^{2}\left(\sec ^{2} 2 \mathrm{t}\right) \times \sin \left(\sec ^{2} 2 \mathrm{t}\right) \times \sec ^{2} 2 \mathrm{t} \times \tan 2 \mathrm{t} .
\end{aligned}
\begin{aligned}
& A s, x^{\mathrm{y}}=\mathrm{e}^{x-\mathrm{y}} \Rightarrow \log \left(x^{\mathrm{y}}\right)=\log \left(\mathrm{e}^{x-\mathrm{y}}\right) \\
& \Rightarrow \mathrm{y} \log x=(x-\mathrm{y}) \Rightarrow \mathrm{y}=\frac{x}{1+\log x}
\end{aligned}
Now, Differentiating both the sides wrt $\displaystyle x$
\frac{d \mathrm{y}}{d x}=\frac{(\log x+1) \cdot 1-x\left(\dfrac{1}{x}\right)}{(\log x+1)^{2}}=\frac{\log x}{(1+\log x)^{2}}
$$
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CBSE Class 12 Mathematics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.