Differentiating Exponential Functions

Differentiating $a^{kx}$

We can prove the result by Implicit Differentiation, but we’re not worried about that just yet.

For any real values of $textcolor{red}{a}$ and $textcolor{orange}{k}$, $f(textcolor{blue}{x}) = textcolor{red}{a}^{textcolor{orange}{k}textcolor{blue}{x}}$ gives

$f'(textcolor{blue}{x}) = textcolor{orange}{k}textcolor{red}{a}^{textcolor{orange}{k}textcolor{blue}{x}}ln textcolor{red}{a}$

Differentiating $e^{kx}$

$e^{x}$ differentiates to itself.

$dfrac{d}{dx}(e^{x})=e^{x}$

For $f(textcolor{blue}{x}) = e^{textcolor{orange}{k}textcolor{blue}{x}}$ with real values of $textcolor{orange}{k}$, the derivative is given by

$f'(textcolor{blue}{x}) = textcolor{orange}{k}e^{textcolor{orange}{k}textcolor{blue}{x}}$

This is actually an extension of the last definition. Set $textcolor{red}{a} = e$, to give $f'(textcolor{blue}{x}) = textcolor{orange}{k}e^{textcolor{orange}{k}textcolor{blue}{x}}ln e$.

$ln e = 1$, so this part cancels from the expression to give the correct result.

Differentiating $e^{f(x)}$

By extension, we can use the Chain Rule to determine the derivative here.

So, set $textcolor{red}{u} = f(textcolor{blue}{x})$ and $textcolor{limegreen}{y} = e^textcolor{red}{u} = e^{f(textcolor{blue}{x})}$.

Then we have

$dfrac{dtextcolor{red}{u}}{dtextcolor{blue}{x}} = f'(textcolor{blue}{x})$ and $dfrac{dtextcolor{limegreen}{y}}{dtextcolor{red}{u}} = e^textcolor{red}{u}$

By using the Chain Rule, we can see that

$dfrac{dtextcolor{limegreen}{y}}{dtextcolor{blue}{x}} = dfrac{dtextcolor{limegreen}{y}}{dtextcolor{red}{u}} times dfrac{dtextcolor{red}{u}}{dtextcolor{blue}{x}}$

$= f'(textcolor{blue}{x})e^{f(textcolor{blue}{x})}$

Differentiating $ln x$

This one requires a little trick, but the result is pretty obvious once proven.

Begin with $textcolor{limegreen}{y} = ln textcolor{blue}{x}$, and rearrange to get $e^textcolor{limegreen}{y} = textcolor{blue}{x}$.

We have $dfrac{dtextcolor{blue}{x}}{dtextcolor{limegreen}{y}} = e^textcolor{limegreen}{y} = textcolor{blue}{x}$, by extension.

So, by finding the reciprocal, we have

$dfrac{dtextcolor{limegreen}{y}}{dtextcolor{blue}{x}} = dfrac{1}{textcolor{blue}{x}}$

Furthermore, we can use the chain rule to get a general expression for the derivative of $ln(f(x))$.

$dfrac{d}{dx}(ln(f(x)))=dfrac{f'(x)}{f(x)}$