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Question
If \[f\left( 1 \right) = 4, f'\left( 1 \right) = 2\] find the value of the derivative of \[\log \left( f\left( e^x \right) \right)\] w.r. to x at the point x = 0 ?
Solution
\[\text { We have, } f\left( 1 \right) = 4 \text { and }f'\left( 1 \right) = 2\]
\[\text {Let y }= \log\left\{ f\left( e^x \right) \right\}\]
\[\Rightarrow \frac{dy}{dx} = \frac{d}{dx}\left[ \log\left\{ f\left( e^x \right) \right\} \right]\]
\[ \Rightarrow \frac{dy}{dx} = \frac{1}{f\left( e^x \right)} \times \frac{d}{dx}\left\{ f\left( e^x \right) \right\}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{1}{f\left( e^x \right)} \times f'\left( e^x \right) \times \frac{d}{dx}\left( e^x \right)\]
\[ \Rightarrow \frac{dy}{dx} = \frac{e^x f'\left( e^x \right)}{f\left( e^x \right)}\]
\[\text { Putting x } = 0, \text { we get }, \]
\[\frac{dy}{dx} = \frac{e^0 f'\left( e^0 \right)}{f\left( e^0 \right)}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{1f'\left( 1 \right)}{f\left( 1 \right)}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{2}{4} \left[ \because f'\left( 1 \right) = 2 \text { and }f\left( 1 \right) = 4 \right]\]
\[ \Rightarrow \frac{dy}{dx} = \frac{1}{2}\]
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