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If f(1)=2: f^{\prime}(1)=6 , find the value of the derivative of \log f\left(e^x\right) w.r.t. x at the point x=0.

Option: 1

\frac{6}{5}


Option: 2

3


Option: 3

\frac{1}{3}


Option: 4

0


Answers (1)

best_answer

Let y=\log f\left(e^x\right)

Then take derivative of the given term,

\frac{d y}{d x}=\frac{1}{f\left(e^x\right)} \cdot \frac{d}{d x}\left(f\left(e^x\right)\right)

\begin{aligned} & =\frac{1}{f\left(e^x\right)} \cdot f^{\prime}\left(e^x\right) \cdot \frac{d}{d x}\left(e^x\right) \\ & =\frac{1}{f\left(e^x\right)} \cdot f^{\prime}\left(e^x\right) \cdot e^x \\ & \therefore \frac{d y}{d x}{ }_{x=0}=\frac{1}{f\left(e^0\right)} \cdot f^{\prime}\left(e^0\right) \cdot e^0 \\ & =\frac{1}{f(1)} \cdot f^{\prime}(1) \cdot 1 \\ & =\frac{f^{\prime}(1)}{f(1)} \end{aligned}

Now, Putting the value of f'(1) and f(1) in the obtained value of \frac{d y}{d x}_{x=0},

\begin{aligned} \frac{d y}{d x}_{x=0} & =\frac{f^{\prime}(1)}{f(1)} \\ & =\frac{6}{2} \\ & =3 \end{aligned}

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Pankaj

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