CBSE 2025 · Region 5 · Set 1 · Q34 · 5 marks
For a positive constant 'a', differentiate $\displaystyle \mathrm{a}^{\mathrm{t}+\frac{1}{\mathrm{t}}}$ with respect to $\displaystyle \left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)^{\mathrm{a}}$, where t is a non-zero real number.Find $\displaystyle \frac{d \mathrm{y}}{d \mathrm{x}}$ if $\displaystyle \mathrm{y}^{\mathrm{x}}+\mathrm{x}^{\mathrm{y}}+\mathrm{x}^{\mathrm{x}}=\mathrm{a}^{b}$, where $\displaystyle \mathrm{a}$ and $\displaystyle b$ are constants.
For a positive constant 'a', differentiate $\displaystyle \mathrm{a}^{\mathrm{t}+\frac{1}{\mathrm{t}}}$ with respect to $\displaystyle \left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)^{\mathrm{a}}$, where t is a non-zero real number.
Find $\displaystyle \frac{d \mathrm{y}}{d \mathrm{x}}$ if $\displaystyle \mathrm{y}^{\mathrm{x}}+\mathrm{x}^{\mathrm{y}}+\mathrm{x}^{\mathrm{x}}=\mathrm{a}^{b}$, where $\displaystyle \mathrm{a}$ and $\displaystyle b$ are constants.
Marking-scheme solution
Let $\displaystyle u=\mathrm{a}^{\mathrm{t}+\frac{1}{\mathrm{t}}} \Rightarrow \frac{d u}{d \mathrm{t}}=\mathrm{a}^{\mathrm{t}+\frac{1}{\mathrm{t}}} \cdot \log \mathrm{a} .\left(1-\frac{1}{\mathrm{t}^{2}}\right)$
\[\begin{gathered}
v=\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)^{\mathrm{a}} \Rightarrow \frac{d v}{d \mathrm{t}}=\mathrm{a}\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)^{\mathrm{a}-1} \cdot\left(1-\frac{1}{\mathrm{t}^{2}}\right) \\
\frac{d u}{d v}=\frac{d u / d \mathrm{t}}{d v / d \mathrm{t}}=\frac{\mathrm{a}^{\mathrm{t}+\frac{1}{\mathrm{t}}} \cdot \log \mathrm{a}}{\mathrm{a}\left(\mathrm{t}+\dfrac{1}{\mathrm{t}}\right)^{\mathrm{a}-1}}
\end{gathered}
\]
Let $\displaystyle \mathbf{u}=\mathbf{y}^{\mathbf{x}}, \mathbf{v}=\mathbf{x}^{\mathbf{y}}$ and $\displaystyle \mathbf{w}=\mathbf{x}^{\mathbf{x}}$
\[\Rightarrow \frac{d u}{d \mathrm{x}}+\frac{d v}{d \mathrm{x}}+\frac{d w}{d \mathrm{x}}=0
\]
\[u=\mathrm{y}^{\mathrm{x}} \Rightarrow \log u=\mathrm{x} \cdot \log \mathrm{y} \Rightarrow \frac{1}{u} \cdot \frac{d u}{d \mathrm{x}}=\frac{\mathrm{x}}{\mathrm{y}} \cdot \frac{d \mathrm{y}}{d \mathrm{x}}+\log \mathrm{y}
\]
\[\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{y}^{\mathrm{x}}\left(\frac{\mathrm{x}}{\mathrm{y}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}+\log \mathrm{y}\right)=\mathrm{x} \mathrm{y}^{\mathrm{x}-1} \frac{d \mathrm{y}}{d \mathrm{x}}+\mathrm{y}^{\mathrm{x}} \log \mathrm{y}
\]
\[\Rightarrow \frac{d w}{d \mathrm{x}}=\mathrm{x}^{\mathrm{x}} \cdot(1+\log \mathrm{x})
\]
∴ From (i), we get
\[\begin{array}{r}
\mathrm{x} \mathrm{y}^{\mathrm{x}-1} \cdot \frac{d \mathrm{y}}{d \mathrm{x}}+\mathrm{y}^{\mathrm{x}} \cdot \log \mathrm{y}+\mathrm{y} \mathrm{x}^{\mathrm{y}-1}+\mathrm{x}^{\mathrm{y}} \cdot \log \mathrm{x} \cdot \frac{d \mathrm{y}}{d \mathrm{x}}+\mathrm{x}^{\mathrm{x}} \cdot(1+\operatorname{lo} \\
\Rightarrow \frac{d \mathrm{y}}{d \mathrm{x}}=-\frac{\mathrm{x}^{\mathrm{x}} \cdot(1+\log \mathrm{x})+\mathrm{y}^{\mathrm{x}} \cdot \log \mathrm{y}+\mathrm{y} \mathrm{x}^{\mathrm{y}-1}}{\mathrm{x} \cdot \mathrm{y}^{\mathrm{x}-1}+\mathrm{x}^{\mathrm{y}} \cdot \log \mathrm{x}}
\end{array}
\]
Continuity and DifferentiabilityLogarithmic DifferentiationApplylong_answerhard
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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.