CBSE 2025 · Region 6 · Set 1 · Q22 · 2 marks
Differentiate $\displaystyle \sqrt{\mathrm{e}^{\sqrt{2 \mathrm{x}}}}$ with respect to $\displaystyle \mathrm{e}^{\sqrt{2 \mathrm{x}}}$ for $\displaystyle \mathrm{x}>0$.If $\displaystyle (\mathrm{x})^{y}=(y)^{\mathrm{x}}$, then find $\displaystyle \frac{d y}{d \mathrm{x}}$.
Differentiate $\displaystyle \sqrt{\mathrm{e}^{\sqrt{2 \mathrm{x}}}}$ with respect to $\displaystyle \mathrm{e}^{\sqrt{2 \mathrm{x}}}$ for $\displaystyle \mathrm{x}>0$.
If $\displaystyle (\mathrm{x})^{y}=(y)^{\mathrm{x}}$, then find $\displaystyle \frac{d y}{d \mathrm{x}}$.
Marking-scheme solution
(a)
Let $\displaystyle \mathrm{u}=\sqrt{\mathrm{e}^{\sqrt{2 \mathrm{x}}}}$ and $\displaystyle \mathrm{v}=\mathrm{e}^{\sqrt{2 \mathrm{x}}}$
Derivative of $\displaystyle \sqrt{\mathrm{v}}$ w.r.t. $\displaystyle \mathrm{v}=\frac{1}{2 \sqrt{\mathrm{v}}}$.
Continuity and DifferentiabilityLogarithmic DifferentiationApplyvery_short_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.