# Math Methods – Differentiation 4/5 – Composite Functions

### Math Methods – Differentiation 4/5 – Composite Functions

This article has been written by Stephane Biggs, a Years 7 – 12, VCE Physics, Further Maths, Math Methods and VCE Specialist Maths Tutor at Learnmate.If you’re interested in private tutoring from Stephane then please check out his page here.

Functions of functions (or composite functions) are probably the most difficult expressions to differentiate, but with a bit of extra help and practice, it can be made easy! Before tackling this subject, make sure you are familiar with finding derivatives of sin, cos, polynomials, roots, logs and exponentials. Previous articles that I have written on LearnMate are dedicated to finding derivatives of such expressions. You must also have studied composite functions before learning to differentiate them!

Examples of “functions of functions”, also known as “composite functions”

Sometimes you’ll be given an expression which you can break into 2 parts which you can individually differentiate, but not both together. For example, look at the expression 5sin(x2).

You know how to differentiate 5sin(x) (the answer is 5cos(x)) and you know how to differentiate x(the answer is 2x). If we take 2 functions f(x) and g(x) and say that f(x)=5sin(x) and g(x)= x2 , then we can write 5sin(x2) as what we call a composite function f(g(x)). In the next subheading we will see how to differentiate such composite functions.

Differentiating composite functions

Once you have broken an expression into 2 parts f(x) and g(x), you will NEED to REMEMBER the following formula :

Derivative of f(g(x)) = f’(g(x) . g’(x)

So the derivative of 5sin(x2) is 5cos(x2) . 2x which simplifies to 10x cos(x2). The key to differentiating composite functions is to break them into appropriate functions f(x) and g(x). That can sometimes be tricky. For example, what if you were asked to differentiate (2x-3)7 ? You could just expand, but that would take a very, very long time! Rather notice that if f(x) = x7 and g(x) = 2x-3, then f(g(x)) is (2x-3)7 . Applying the formula would give us the derivative 7(2x-3)6 . 2 which simplifies to 14(2x-3)6.

Much simpler than expanding! Take the time to analyze and understand each step, and then try the following exercises. Get in touch if you need help.

Exercise – Find the derivatives of the following expressions

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Math Methods – Differentiation 4/5 – Composite Functions