Differentiating Trig Functions

Differentiating Trig Functions

A LevelAQAEdexcelOCRAQA 2022

Differentiating Trig Functions

We’re now going to look at how to differentiate the simple trig functions – that’s \textcolor{blue}{\sin}, \textcolor{limegreen}{\cos} and \textcolor{red}{\tan}.

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

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Simple Trig Results

Below is a diagram showing the derivative and integral of the basic trigonometric functions, \sin x and \cos x.

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General Trig Differentiation

Below is a table of three derivative results.

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Using the Chain Rule with Trig Functions

We can also use the Chain Rule to differentiate more complex trig functions.

For example, say we have f(x) = \textcolor{blue}{\sin} (x^2 - x + 1).

Then we can set \textcolor{red}{u} = x^2 - x + 1 and f(x) = g(\textcolor{red}{u}) = \textcolor{blue}{\sin} \textcolor{red}{u}.

Then \dfrac{d\textcolor{red}{u}}{dx} = 2x - 1 and \dfrac{df(x)}{d\textcolor{red}{u}} = \textcolor{limegreen}{\cos} u.


\dfrac{df(x)}{dx} = (2x - 1)\textcolor{limegreen}{\cos} (x^2 - x + 1)

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From First Principles

We can also find the derivatives from first principles.

For example, let f(x) = \textcolor{limegreen}{\cos} x. Then

f'(x) = \lim\limits_{\textcolor{purple}{h} \to 0}\left( \dfrac{\textcolor{limegreen}{\cos} (x + \textcolor{purple}{h}) - \textcolor{limegreen}{\cos} x}{\textcolor{purple}{h}} \right)

= \lim\limits_{\textcolor{purple}{h} \to 0}\left( \dfrac{\textcolor{limegreen}{\cos} x \textcolor{limegreen}{\cos} \textcolor{purple}{h} - \textcolor{blue}{\sin} x \textcolor{blue}{\sin} h - \textcolor{limegreen}{\cos} x}{\textcolor{purple}{h}}\right)

= \lim\limits_{\textcolor{purple}{h} \to 0}\left( \dfrac{\textcolor{limegreen}{\cos} x(\textcolor{limegreen}{\cos} \textcolor{purple}{h} - 1) - \textcolor{blue}{\sin} x \textcolor{blue}{\sin} \textcolor{purple}{h}}{\textcolor{purple}{h}}\right)

Using small angle approximations, we have

f'(x) = \lim\limits_{\textcolor{purple}{h} \to 0}\left( \dfrac{-\dfrac{1}{2}\textcolor{purple}{h}^2 \textcolor{limegreen}{\cos} x - \textcolor{purple}{h}\textcolor{blue}{\sin} x}{\textcolor{purple}{h}}\right)

= \lim\limits_{\textcolor{purple}{h} \to 0}\left( \dfrac{-1}{2}\textcolor{purple}{h}\textcolor{limegreen}{\cos} x - \textcolor{blue}{\sin} x\right)

= -\textcolor{blue}{\sin} x

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Example Questions

x = \tan y gives

\dfrac{dx}{dy} = \sec ^2 y

Then \dfrac{dy}{dx} = \cos ^2 y.

Let u = x^2 and y = \tan u.


\dfrac{dy}{dx} = 2x \times \sec ^2 u = \dfrac{2x}{\cos ^2 x^2}

f'(x) = \lim\limits_{h \to 0}\left( \dfrac{\sin (kx + kh) - \sin kx}{h} \right)


= \lim\limits_{h \to 0}\left( \dfrac{\sin kx \cos kh + \sin kh \cos kx - \sin kx}{h}\right)


= \lim\limits_{h \to 0}\left( \dfrac{\sin kx(\cos kh - 1) + \sin kh \cos kx}{h}\right)


Using small angle approximations, we have

f'(x) = \lim\limits_{h \to 0}\left( \dfrac{-\dfrac{1}{2}(kh)^2 \sin kx + kh\cos kx}{h}\right)


= \lim\limits_{h \to 0}\left( \dfrac{-1}{2}kh\sin kx + k\cos kx\right)


= k\cos kx

Related Topics


Inverse Trig Functions

A Level

Reciprocal Trig Functions

A Level

Additional Resources


Exam Tips Cheat Sheet

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Formula Booklet

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Worksheet and Example Questions

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