Question:
If $f(1)=4, f^{\prime}(1)=2$, find the value of the derivative of $\log \left(f\left(e^{x}\right)\right)$ with respect to $x$ at the point $x=0$.
Solution:
Using the Chain Rule of Differentiation, derivative of $\log \left(f\left(e^{x}\right)\right)$ w.r.t. $x$ is $\frac{1}{f\left(e^{x}\right)} \cdot f^{\prime}\left(e^{x}\right)$
So, the value of the derivative at $x=0$ is
$\frac{1}{f\left(e^{0}\right)} \cdot f^{\prime}\left(e^{0}\right)=\frac{1}{f(1)} \cdot f^{\prime}(1)$
$=\frac{1}{4} \cdot 2$
$=\frac{1}{2}$
So, the value of the derivative at $x=0$ is $0.5$ (Ans)
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.