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प्रश्न
If \[f\left( 0 \right) = f\left( 1 \right) = 0, f'\left( 1 \right) = 2 \text { and y } = f \left( e^x \right) e^{f \left( x \right)}\] write the value of \[\frac{dy}{dx} \text{ at x } = 0\] ?
उत्तर
\[\text{ We have,} f\left( 0 \right) = f\left( 1 \right) = 0 , f'\left( 1 \right) = 2\]
\[\text { and, } \]
\[y = f\left( e^x \right) e^{f\left( x \right)}\]
\[\Rightarrow \frac{dy}{dx} = \frac{d}{dx}\left[ f\left( e^x \right) \times e^{f\left( x \right)} \right]\]
\[ \Rightarrow \frac{dy}{dx} = f\left( e^x \right)\frac{d}{dx} e^{f\left( x \right)} + e^{f\left( x \right)} \frac{d}{dx}f\left( e^x \right) \left[ \text{Using product rule } \right]\]
\[ \Rightarrow \frac{dy}{dx} = f\left( e^x \right) \times e^{f\left( x \right)} \frac{d}{dx}f\left( x \right) + e^{f\left( x \right)} \times f'\left( e^x \right)\frac{d}{dx}\left( e^x \right)\]
\[ \Rightarrow \frac{dy}{dx} = f\left( e^x \right) \times e^{f\left( x \right)} \times f'\left( x \right) + e^{f\left( x \right)} \times f'\left( e^x \right) \times e^x \]
\[\text{ Putting x } = 0, \text{ we get }, \]
\[\frac{dy}{dx} = f\left( e^0 \right) \times e^{f\left( 0 \right)} \times f'\left( 0 \right) + e^{f\left( 0 \right)} \times f'\left( e^0 \right) \times e^0 \]
\[ \Rightarrow \frac{dy}{dx} = f\left( 1 \right) e^{f\left( 0 \right)} \times f'\left( 0 \right) + e^{f\left( 0 \right)} \times f'\left( 1 \right) \times 1\]
\[ \Rightarrow \frac{dy}{dx} = 0 \times e^0 \times f'\left( 0 \right) + e^0 \times 2 \times 1 .........\left[ \because f\left( x \right) = f\left( 1 \right) = 0 \text{ and }f'\left( 1 \right) = 2 \right]\]
\[ \Rightarrow \frac{dy}{dx} = 0 + 1 \times 2 \times 1\]
\[ \Rightarrow \frac{dy}{dx} = 2\]
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