Chain Rule: How to Differentiate

What the chain rule is for

In Add Math, many expressions are built by placing one function inside another. $(2x+3)^{4}$ , $\sqrt{5x-1}$ and $\frac{1}{(3x+2)^{2}}$ are all composite functions: there is an 'inside' expression and an 'outside' operation wrapped around it.

The chain rule is the tool we use to differentiate these.

Trying to differentiate them term by term does not work, because the inside is not a simple $x$ . The chain rule tells us to differentiate the outside and the inside separately, then multiply the two results together.

It appears constantly in the Differentiation chapter and feeds directly into tangents, normals, rates of change and maximum–minimum problems, so mastering it early makes the whole of Form 5 calculus far smoother.

When to reach for it

Reach for the chain rule whenever you see a function 'wrapped around' another expression rather than a plain power of $x$ . The clearest signal is a bracket raised to a power, such as $(4x-1)^{5}$ ; a root, such as $\sqrt{2x+7}$ , which is really a power $\frac{1}{2}$ ; or a reciprocal like $\frac{3}{(x^{2}+1)}$ , which is a negative power.

If you can point to an 'inside' part whose derivative is not simply $1$ , you need the chain rule. A quick test: ask yourself 'could I differentiate this if the inside were just $x$ ?'

If yes, the chain rule turns that easy derivative into the real one by multiplying by the derivative of the inside.

The method, step by step

Chain ruleMust memorise

\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}

1
Spot the composite
Identify the inner expression, the part 'inside' the bracket, root or power. Call it $u$ .
2
Rewrite in terms of u
Express $y$ using $u$ , for example $y=u^{4}$ when $y=(2x+3)^{4}$ .
3
Differentiate the outside
Find $\frac{dy}{du}$ , treating $u$ as the variable.
4
Differentiate the inside
Find $\frac{du}{dx}$ by differentiating the inner expression.
5
Multiply
Apply $\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}$ .
6
Substitute back
Replace $u$ with the original expression in $x$ and simplify.

Worked example

Q1[4 marks]

Given $y=(2x+3)^{4}$ , find $\frac{dy}{dx}$ and hence the gradient of the curve at the point where $x=-1$ .

Show worked solution

Identify the inside expression and let $u=2x+3$ . Then $y=u^{4}$ .

Differentiate the outside with respect to $u$ : $\frac{dy}{du}=4u^{3}$ .

Differentiate the inside with respect to $x$ : $\frac{du}{dx}=2$ .

Apply the chain rule:

\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}=4u^{3}\times 2=8u^{3}

Substitute $u=2x+3$ back in: $\frac{dy}{dx}=8(2x+3)^{3}$ .

For the gradient at $x=-1$ , substitute: $2(-1)+3=1$ , so $\frac{dy}{dx}=8(1)^{3}=8$ .

The gradient of the curve at $x=-1$ is $8$ .

Common mistakes to avoid

Differentiating only the outside and forgetting to multiply by $\frac{du}{dx}$ , the single most common slip.
Reducing the power correctly but mishandling the inside, e.g. writing the derivative of $2x+3$ as $2x$ instead of $2$ .
Bringing the power down but forgetting to subtract one, leaving $(2x+3)^{4}$ instead of $(2x+3)^{3}$ .
Leaving the final answer in terms of $u$ instead of substituting the $x$ -expression back in.
Reaching for the product rule when a coefficient like the $2$ in $2x+3$ simply belongs inside the chain rule.

How one-to-one teaching helps

The step students most often miss is the multiplication by $\frac{du}{dx}$ : they differentiate the bracket, feel finished, and lose easy marks. In a one-to-one lesson our teachers watch you work line by line and catch that exact moment, so the habit of always differentiating the inside becomes automatic.

Because your working is shown step by step, a clean $u$ -substitution keeps every line clear and easy to credit. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

If you would like to see how we teach the chain rule, message us on WhatsApp to arrange a one-hour paid class from RM50/hr at the teacher's rate.

Get 1-to-1 help.

Book a Trial Class

Frequently asked questions

What is a composite function in Add Math?

A composite function is a function placed inside another function, such as $(2x+3)^{4}$ or $\sqrt{5x-1}$ . There is an inner expression and an outer operation.

The chain rule is the method for differentiating this kind of expression, because you cannot differentiate the inside and outside in a single step.

Do I need to memorise the chain rule formula?

Yes. The chain rule, $\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}$ , is a technique you are expected to know and apply, so commit it to memory and practise the $u$ -substitution until it feels automatic.

How do I know when to use the chain rule instead of the product or quotient rule?

Use the chain rule when one function is wrapped around another, a bracket to a power, a root, or a reciprocal. Use the product rule when two functions are multiplied, and the quotient rule when one is divided by another.

Many harder questions combine them, so identify the outermost structure first.

What comes after finding the derivative with the chain rule?

Once you have $\frac{dy}{dx}$ , you can find the gradient at a point, the equation of a tangent or normal, or set $\frac{dy}{dx}=0$ to locate turning points. The chain rule is a building block for most of the Differentiation chapter, so a secure method here pays off across many questions.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

How to differentiate using the chain rule