CBSE 2025 · Region 7 · Set 1 · Q23 · 2 marks
Differentiate $\displaystyle \left(\frac{5^{\mathrm{x}}}{\mathrm{x}^{5}}\right)$ with respect to x .If $\displaystyle -2 \mathrm{x}^{2}-5 \mathrm{x} y+y^{3}=76$, then find $\displaystyle \frac{d y}{d \mathrm{x}}$.
Differentiate $\displaystyle \left(\frac{5^{\mathrm{x}}}{\mathrm{x}^{5}}\right)$ with respect to x .
If $\displaystyle -2 \mathrm{x}^{2}-5 \mathrm{x} y+y^{3}=76$, then find $\displaystyle \frac{d y}{d \mathrm{x}}$.
Marking-scheme solution
Let, $\displaystyle y=\frac{5^{\mathrm{x}}}{\mathrm{x}^{5}}=5^{\mathrm{x}} \cdot \mathrm{x}^{-5} \Rightarrow \frac{d y}{d \mathrm{x}}=\left(5^{\mathrm{x}}\right)^{\prime} \cdot \mathrm{x}^{-5}+5^{\mathrm{x}} \cdot\left(\mathrm{x}^{-5}\right)^{\prime}$
\[=\frac{5^{\mathrm{x}}}{\mathrm{x}^{5}} \log 5-\frac{5^{\mathrm{x}+1}}{\mathrm{x}^{6}}
\]
Differentiating $\displaystyle -2 \mathrm{x}^{2}-5 \mathrm{x} y+y^{3}=76$, with respect to ' $\displaystyle \mathrm{x}$ '
\[-4 \mathrm{x}-5 y-5 \mathrm{x} \frac{d y}{d \mathrm{x}}+3 y^{2} \frac{d y}{d \mathrm{x}}=0
\]
\[\Rightarrow \frac{d y}{d \mathrm{x}}=\frac{4 \mathrm{x}+5 y}{3 y^{2}-5 \mathrm{x}}
\]
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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.