Question:
If $f^{\prime}(1)=2$ and $y=f\left(\log _{e} x\right)$, find $\cdot \frac{d y}{d x} \cdot$ at $x=e$
Solution:
$y=f\left(\log _{e} x\right)$
Using the Chain Rule of Differentiation,
$\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{f}^{\prime}\left(\log _{\mathrm{e}} \mathrm{x}\right) \cdot \frac{1}{\mathrm{x}}$
So, at $\mathrm{x}=\mathrm{e}$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{f}^{\prime}\left(\log _{\mathrm{e}} \mathrm{e}\right) \cdot \frac{1}{\mathrm{e}}$
$=f^{\prime}(1) \cdot \frac{1}{e}$
$=\frac{2}{e}($ Ans $)$