CBSE 2023 · Region 3 · Set 1 · Q25 · 2 marks
If $\displaystyle \mathrm{y}=\mathrm{x}^{\frac{1}{\mathrm{x}}}$, then find $\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}$ at $\displaystyle \mathrm{x}=1$.If $\displaystyle \mathrm{x}=\mathrm{a} \sin 2 \mathrm{t}, \mathrm{y}=\mathrm{a}(\cos 2 \mathrm{t}+\log \tan \mathrm{t})$, then find $\displaystyle \frac{d \mathrm{y}}{d \mathrm{x}}$.
If $\displaystyle \mathrm{y}=\mathrm{x}^{\frac{1}{\mathrm{x}}}$, then find $\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}$ at $\displaystyle \mathrm{x}=1$.
If $\displaystyle \mathrm{x}=\mathrm{a} \sin 2 \mathrm{t}, \mathrm{y}=\mathrm{a}(\cos 2 \mathrm{t}+\log \tan \mathrm{t})$, then find $\displaystyle \frac{d \mathrm{y}}{d \mathrm{x}}$.
Marking-scheme solution
(a)
$\displaystyle \mathrm{y}=\mathrm{x}^{1 / \mathrm{x}}$
$$\Rightarrow \log \mathrm{y}=\frac{1}{\mathrm{x}} \log \mathrm{x}
$$$\Rightarrow \frac{1}{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\log \mathrm{x}}{\mathrm{x}^{2}}+\frac{1}{\mathrm{x}^{2}} \Rightarrow \frac{d \mathrm{y}}{d \mathrm{x}}=\mathrm{x}^{\frac{1}{\mathrm{x}}} \frac{(1-\log \mathrm{x})}{\mathrm{x}^{2}}$
$\displaystyle \Rightarrow\left(\frac{d \mathrm{y}}{d \mathrm{x}}\right)_{\mathrm{x}=1}=1$
(b)
$\displaystyle \frac{d \mathrm{x}}{d \mathrm{t}}=2 \mathrm{a} \cos 2 \mathrm{t}$
$$\frac{d \mathrm{y}}{d \mathrm{t}}=$\displaystyle 2$ \mathrm{a}\left(-\sin $\displaystyle 2$ \mathrm{t}+\frac{\sec ^{$\displaystyle 2$} \mathrm{t}}{$\displaystyle 2$ \tan \mathrm{t}}\right)=$\displaystyle 2$ \mathrm{a} \frac{\cos ^{$\displaystyle 2$} $\displaystyle 2$ \mathrm{t}}{\sin $\displaystyle 2$ \mathrm{t}}
$$$\frac{d \mathrm{y}}{d \mathrm{x}}=\cot 2 \mathrm{t}$
Continuity and DifferentiabilityLogarithmic DifferentiationApplyvery_short_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.