Differentiating Trig Functions

A LevelAQAEdexcelOCR

Differentiating Trig Functions

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

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

A LevelAQAEdexcelOCR
https://www.google.com

Simple Trig Results

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

A LevelAQAEdexcelOCR

General Trig Differentiation

Below is a table of three derivative results.

A LevelAQAEdexcelOCR
https://www.google.com

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)=sin⁡(x2−x+1)f(x) = textcolor{blue}{sin} (x^2 – x + 1).

Then we can set u=x2−x+1textcolor{red}{u} = x^2 – x + 1 and f(x)=g(u)=sin⁡uf(x) = g(textcolor{red}{u}) = textcolor{blue}{sin} textcolor{red}{u}.

Then dudx=2x−1dfrac{dtextcolor{red}{u}}{dx} = 2x – 1 and df(x)du=cos⁡udfrac{df(x)}{dtextcolor{red}{u}} = textcolor{limegreen}{cos} u.

So,

df(x)dx=(2x−1)cos⁡(x2−x+1)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)=cos⁡xf(x) = textcolor{limegreen}{cos} x. Then

f′(x)=lim⁡h→0(cos⁡(x+h)−cos⁡xh)f'(x) = limlimits_{textcolor{purple}{h} to 0}left( dfrac{textcolor{limegreen}{cos} (x + textcolor{purple}{h}) – textcolor{limegreen}{cos} x}{textcolor{purple}{h}} right)

=lim⁡h→0(cos⁡xcos⁡h−sin⁡xsin⁡h−cos⁡xh)= limlimits_{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⁡h→0(cos⁡x(cos⁡h−1)−sin⁡xsin⁡hh)= limlimits_{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⁡h→0(−12h2cos⁡x−hsin⁡xh)f'(x) = limlimits_{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⁡h→0(−12hcos⁡x−sin⁡x)= limlimits_{textcolor{purple}{h} to 0}left( dfrac{-1}{2}textcolor{purple}{h}textcolor{limegreen}{cos} x – textcolor{blue}{sin} xright)

=−sin⁡x= -textcolor{blue}{sin} x

A LevelAQAEdexcelOCR

Differentiating Trig Functions Example Questions

Question 1: Give an expression for dydxdfrac{dy}{dx} in terms of yy, when x=tan⁡yx = tan y.

[2 marks]

A Level AQAEdexcelOCR

x=tan⁡yx = tan y gives

dxdy=sec⁡2ydfrac{dx}{dy} = sec ^2 y

Then dydx=cos⁡2ydfrac{dy}{dx} = cos ^2 y.

MME Premium Laptop

Save your answers with

MME Premium

Gold Standard Education

Sign Up Now
Still marking manually? Save your answers with
MME Premium Find out more

Question 2: For tan⁡x2tan x^2, find the derivative with respect to xx.

[2 marks]

A Level AQAEdexcelOCR

Let u=x2u = x^2 and y=tan⁡uy = tan u.

Then

dydx=2x×sec⁡2u=2xcos⁡2x2dfrac{dy}{dx} = 2x times sec ^2 u = dfrac{2x}{cos ^2 x^2}

MME Premium Laptop

Save your answers with

MME Premium

Gold Standard Education

Sign Up Now
Still marking manually? Save your answers with
MME Premium Find out more

Question 3: Prove that the derivative of sin⁡kxsin kx is kcos⁡kxkcos kx, using the first principles technique.

[4 marks]

A Level AQAEdexcelOCR

f′(x)=lim⁡h→0(sin⁡(kx+kh)−sin⁡kxh)f'(x) = limlimits_{h to 0}left( dfrac{sin (kx + kh) – sin kx}{h} right)

 

=lim⁡h→0(sin⁡kxcos⁡kh+sin⁡khcos⁡kx−sin⁡kxh)= limlimits_{h to 0}left( dfrac{sin kx cos kh + sin kh cos kx – sin kx}{h}right)

 

=lim⁡h→0(sin⁡kx(cos⁡kh−1)+sin⁡khcos⁡kxh)= limlimits_{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⁡h→0(−12(kh)2sin⁡kx+khcos⁡kxh)f'(x) = limlimits_{h to 0}left( dfrac{-dfrac{1}{2}(kh)^2 sin kx + khcos kx}{h}right)

 

=lim⁡h→0(−12khsin⁡kx+kcos⁡kx)= limlimits_{h to 0}left( dfrac{-1}{2}khsin kx + kcos kxright)

 

=kcos⁡kx= kcos kx

MME Premium Laptop

Save your answers with

MME Premium

Gold Standard Education

Sign Up Now
Still marking manually? Save your answers with
MME Premium Find out more

Additional Resources

Site Logo

Exam Tips Cheat Sheet

A Level
Open
Site Logo

Formula Booklet

A Level
Open

Specification Points Covered

G1 – Understand and use the derivative of f(x)f(x) as the gradient of the tangent to the graph of y=f(x)y=f(x) at a general point (x,y)(x,y); the gradient of the tangent as a limit; interpretation as a rate of change; sketching the gradient function for a given curve; second derivatives; differentiation from first principles for small positive integer powers of xx and for sin⁡(x)sin(x) and cos⁡(x)cos(x)
G2 – Differentiate ekxe^{kx} and akxa^{kx}, sin⁡(kx)sin(kx), cos⁡(kx)cos(kx), tan⁡(kx)tan(kx) and related sums, differences and constant multiples

Differentiating Trig Functions Worksheet and Example Questions

Site Logo

Differentiation

A Level
Exam Questions Mark Scheme
https://www.google.com

Related Topics

Site Logo

Inverse Trig Functions

A Level
Revise
Site Logo

Reciprocal Trig Functions

A Level
Revise