CBSE 2025 · Region 7 · Set 3 · Q28 · 3 marks
Differentiate $\displaystyle \log \left(\mathrm{x}^{\mathrm{x}}+\operatorname{cosec}^{2} \mathrm{x}\right)$ with respect to x .
Marking-scheme solution
\[\begin{aligned}
\frac{d}{d \mathrm{x}} \log \left(\mathrm{x}^{\mathrm{x}}+\operatorname{cosec}^{2} \mathrm{x}\right) & =\frac{1}{\mathrm{x}^{\mathrm{x}}+\operatorname{cosec}^{2} \mathrm{x}} \frac{d}{d \mathrm{x}}\left(e^{\mathrm{x} \log \mathrm{x}}+\operatorname{cosec}^{2} \mathrm{x}\right) \quad\left(\because \mathrm{x}^{\mathrm{x}}=e^{\mathrm{x} \log \mathrm{x}}\right) \\
& =\frac{1}{\mathrm{x}^{\mathrm{x}}+\operatorname{cosec}^{2} \mathrm{x}}\left[e^{\mathrm{x} \log \mathrm{x}}(1+\log \mathrm{x})-2 \operatorname{cosec}^{2} \mathrm{x} \cot \mathrm{x}\right] \\
& =\frac{1}{\mathrm{x}^{\mathrm{x}}+\operatorname{cosec}^{2} \mathrm{x}}\left[\mathrm{x}^{\mathrm{x}}(1+\log \mathrm{x})-2 \operatorname{cosec}^{2} \mathrm{x} \cot \mathrm{x}\right]
\end{aligned}
\]
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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.