Answer
\[\begin{align}
& \mathbf{a}.\text{ }f\left( g\left( x \right) \right)={{e}^{g\left( x \right)}},\text{ with }g\left( x \right)=kx \\
& \mathbf{b}.\frac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]=k{{e}^{kx}} \\
\end{align}\]
Work Step by Step
\[\begin{align}
& \mathbf{a}. \\
& \text{Let }f\left( x \right)={{e}^{kx}} \\
& \text{We can write }{{e}^{kx}}\text{ as }{{e}^{g\left( x \right)}}\text{ with }g\left( x \right)=kx,\text{ then} \\
& f\left( g\left( x \right) \right)={{e}^{g\left( x \right)}},\text{ with }g\left( x \right)=kx \\
& \\
& \mathbf{b}. \\
& \frac{d}{dx}\left[ {{e}^{kx}} \right]=\frac{d}{dx}\left[ f\left( g\left( x \right) \right) \right] \\
& \text{By the chain rule} \\
& \frac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]=f'\left( g\left( x \right) \right)g'\left( x \right) \\
& \text{Where} \\
& \frac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]={{e}^{kx}}\frac{d}{dx}\left[ kx \right] \\
& \frac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]={{e}^{kx}}\left( k \right) \\
& \frac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]=k{{e}^{kx}} \\
\end{align}\]
