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प्रश्न
Differentiate \[x^{x^2 - 3} + \left( x - 3 \right)^{x^2}\] ?
उत्तर
\[\text{ Let y } = x^{x^2 - 3} + \left( x - 3 \right)^{x^2} \]
\[ \text{ Also, let u } = x^{x^2 - 3} \text{ and }v = \left( x - 3 \right)^{x^2} \]
\[ \therefore y = u + v\]
Differentiate both sides with respect to x,
\[\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} . . . \left( i \right)\]
\[\text{ Now, u }= x^{x^2 - 3} \]
\[ \therefore \log u = \log\left( x^{x^2 - 3} \right)\]
\[ \Rightarrow \log u = \left( x^2 - 3 \right) \log x\]
Differentiating with respect to x,
\[\frac{1}{u}\frac{du}{dx} = \log x\frac{d}{dx}\left( x^2 - 3 \right) + \left( x^2 - 3 \right)\frac{d}{dx}\left( \log x \right)\]
\[ \Rightarrow \frac{1}{u}\frac{du}{dx} = \log x\left( 2x \right) + \left( x^2 - 3 \right)\left( \frac{1}{x} \right)\]
\[ \Rightarrow \frac{du}{dx} = x^{x^2 - 3} \left[ \frac{x^2 - 3}{x} + 2x \log x \right]\]
\[\text{ Also, v }= \left( x - 3 \right)^{x^2} \]
\[ \therefore \log v = \log \left( x - 3 \right)^{x^2} \]
\[ \Rightarrow \log v = x^2 \log\left( x - 3 \right)\]
Differentiating both sides with respect to x,
\[\frac{1}{v}\frac{dv}{dx} = \log\left( x - 3 \right)\frac{d}{dx}\left( x^2 \right) + x^2 \frac{d}{dx}\left[ \log\left( x - 3 \right) \right]\]
\[ \Rightarrow \frac{1}{v}\frac{dv}{dx} = \log\left( x - 3 \right) \left( 2x \right) + x^2 \left( \frac{1}{x - 3} \right)\frac{d}{dx}\left( x - 3 \right)\]
\[ \Rightarrow \frac{dv}{dx} = v\left[ 2x \log\left( x - 3 \right) + \frac{x^2}{x - 3} \times 1 \right]\]
\[ \Rightarrow \frac{dv}{dx} = \left( x - 3 \right)^{x^2} \left[ \frac{x^2}{x - 3} + 2x\log\left( x - 3 \right) \right]\]
\[\text{ Substituing the expressions of }\frac{du}{dx}and \frac{dv}{dx}in equation \left( i \right)\]
\[\frac{dy}{dx} = x^{x^2 - 3} \left[ \frac{x^2 - 3}{x} + 2x \log x \right] + \left( x - 3 \right)^{x^2} \left[ \frac{x^2}{x - 3} + 2x \log\left( x - 3 \right) \right]\]
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