Differentiating Exponential Functions

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.

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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}

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Differentiating e^{kx}

e^{x} differentiates to itself.


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.

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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})}

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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)).


A LevelAQAEdexcelOCR
A Level AQA Edexcel OCR

Example: Differentiating \ln(f(x))

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

[2 marks]

Recall the formula:


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}

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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


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




\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}

Related Topics


The Exponential Function

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