# Differentiating Exponential Functions

A LevelAQAEdexcelOCREdexcel 2022

## Differentiating Exponential Functions

In this section, we’ll be looking at how to differentiate equations of the form $\textcolor{red}{a}^{\textcolor{orange}{k}\textcolor{blue}{x}}$, $e^{\textcolor{orange}{k}\textcolor{blue}{x}}$ and $e^{f(\textcolor{blue}{x})}$. We’ll also look at the derivative of logarithmic functions.

Make sure you are happy with the following topics before continuing.

A Level

## 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

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

A Level

## 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.

A Level

## 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

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

By using the Chain Rule, we can see that

$\dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}} = \dfrac{d\textcolor{limegreen}{y}}{d\textcolor{red}{u}} \times \dfrac{d\textcolor{red}{u}}{d\textcolor{blue}{x}}$

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

A Level

## 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 $\dfrac{d\textcolor{blue}{x}}{d\textcolor{limegreen}{y}} = e^\textcolor{limegreen}{y} = \textcolor{blue}{x}$, by extension.

So, by finding the reciprocal, we have

$\dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}} = \dfrac{1}{\textcolor{blue}{x}}$

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

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

A Level
A Level

## Example: Differentiating $\ln(f(x))$

Find the derivative of $\ln(x^{2}+3x-8)$

[2 marks]

Recall the formula:

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

In this example: $f(x)=x^{2}+3x-8$, so $f'(x)=2x+3$.

Hence, our answer is $\dfrac{2x+3}{x^{2}+3x-8}$

A Level

## Example Questions

For $y = 3^{2x}$, we have

$\dfrac{dy}{dx} = 2 \times 3^{2x}\ln 3$

$= 9\ln9 \equiv 19.775$ when $x = 1$

Then, the straight line equation of the tangent $y = mx + c$ is given by

$9 = 19.775 + c$

meaning

$c = -10.775$

So, the equation of the tangent line is

$y = 19.775x - 10.775$

Setting $u = x^2 - 3x + 1$ and $g(u) = e^u$, and using the chain rule, we have

$f'(x) = g'(u) \times u'$

$= e^u \times (2x - 3)$

$= (2x - 3)e^{x^2 - 3x + 1}$

Set $u = x^3 + x - 3$ and $y = \ln u$, and use the chain rule:

$\dfrac{du}{dx} = 3x^2 + 1$

and

$\dfrac{dy}{du} = \dfrac{1}{u}$

Therefore, we have

$\dfrac{dy}{dx} = (3x^2 + 1)\dfrac{1}{x^3 + x - 3} = \dfrac{3x^2 + 1}{x^3 + x - 3}$

A Level

A Level

A Level

A Level

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