Differentiating Trig Functions

A LevelAQAEdexcelOCR

Differentiating Trig Functions Revision

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.

A LevelAQAEdexcelOCR

Simple Trig Results

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

A LevelAQAEdexcelOCR
MME Logo

Your 2024 Revision Partner


Open TikTok

General Trig Differentiation

Below is a table of three derivative results.

A LevelAQAEdexcelOCR

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)

A LevelAQAEdexcelOCR

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

A LevelAQAEdexcelOCR

Differentiating Trig Functions 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

Additional Resources


Exam Tips Cheat Sheet

A Level

Formula Booklet

A Level

Differentiating Trig Functions Worksheet and Example Questions

Site Logo


A Level

You May Also Like...

MME Learning Portal

Online exams, practice questions and revision videos for every GCSE level 9-1 topic! No fees, no trial period, just totally free access to the UK’s best GCSE maths revision platform.

View Product

Related Topics


Inverse Trig Functions

A Level

Reciprocal Trig Functions

A Level