Forgetting the Chain Rule

Q: What exactly is a composite function?

It is a function inside another function, for example (2x2−5)3(2x^{2}-5)^{3}(2x2−5)3, where the inner part 2x2−52x^{2}-52x2−5 is raised to a power, or 2x+1=(2x+1)1/2\sqrt{2x+1}=(2x+1)^{1/2}2x+1=(2x+1)1/2. Whenever the thing being raised to a power is not just xxx, you are differentiating a composite and the chain rule applies.

Q: Does \sqrt{2x+1} need the chain rule too?

Yes. Rewrite it as (2x+1)1/2(2x+1)^{1/2}(2x+1)1/2; then dydx=12(2x+1)−1/2⋅2=(2x+1)−1/2\frac{dy}{dx}=\tfrac{1}{2}(2x+1)^{-1/2}\cdot 2=(2x+1)^{-1/2}dxdy=21(2x+1)−1/2⋅2=(2x+1)−1/2. The factor of 222 from differentiating the inside is exactly the step that is easy to drop.

The error

When a function is a composite, an inner expression tucked inside a power, such as $(3x+2)^{4}$ , or inside a root like $\sqrt{2x+1}$ , the derivative is not finished once you bring the power down. You still have to multiply by the derivative of the inside.

The most common slip in Add Math differentiation is stopping too early: writing $4(3x+2)^{3}$ and moving on, as though the bracket were a single letter. It is tempting because the power rule feels complete on its own, and for a plain $x^{n}$ it is.

But $(3x+2)$ is not $x$ ; its own rate of change, $3$ , has to come along for the ride. Miss it and every value that follows, gradient, tangent, turning point, is scaled by the wrong factor.

Wrong line, then the fix

Take $y=(3x+2)^{4}$ . The tempting shortcut treats the bracket as if it were $x$ :

Wrong, derivative of the inside is missingMust memorise

\frac{dy}{dx}=4(3x+2)^{3}

The correct working carries the derivative of the inside, $\frac{d}{dx}(3x+2)=3$ , and multiplies:

CorrectMust memorise

\frac{dy}{dx}=4(3x+2)^{3}\cdot 3=12(3x+2)^{3}

The two answers differ by a factor of $3$ . If the inside were simply $x$ , that factor would be $1$ and the shortcut would happen to work, which is exactly why the habit forms and then fails on a real bracket.

The reliable fix

Chain ruleMust memorise

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

1
Spot the composite
Look for an inner expression raised to a power or sitting inside a root; if the bracket is anything other than a single $x$ , the chain rule is needed.
2
Name the inside
Let $u$ be the inner expression, so $y$ becomes a simple power of $u$ .
3
Differentiate the outside
Find $\frac{dy}{du}$ with the power rule, leaving $u$ intact.
4
Differentiate the inside
Find $\frac{du}{dx}$ , this is the piece students drop.
5
Multiply and rewrite
Compute $\frac{dy}{du}\cdot\frac{du}{dx}$ , then replace $u$ by the original expression.

One line to remember it

Power down, then times the inside

Every bracket leaves a fingerprint. If nothing on the outside of your bracket changed, you have forgotten to multiply by the derivative of the inside.

Practice where the error hides

Q1[4 marks]

Given $y=(2x^{2}-5)^{3}$ , find $\frac{dy}{dx}$ , and hence the gradient of the curve at $x=1$ .

Show worked solution

This is a composite: an inner expression $2x^{2}-5$ raised to the power $3$ . Let $u=2x^{2}-5$ , so $y=u^{3}$ .

\frac{dy}{du}=3u^{2}=3(2x^{2}-5)^{2},\qquad \frac{du}{dx}=4x

Multiply the two parts using the chain rule:

\frac{dy}{dx}=3(2x^{2}-5)^{2}\cdot 4x=12x(2x^{2}-5)^{2}

Now substitute $x=1$ . First the inside: $2(1)^{2}-5=2-5=-3$ , so $(2x^{2}-5)^{2}=(-3)^{2}=9$ .

\left.\frac{dy}{dx}\right|_{x=1}=12(1)(9)=108

The gradient of the curve at $x=1$ is $108$ . The dropped factor $4x$ is what separates this from the wrong answer $3(2x^{2}-5)^{2}$ , which at $x=1$ would have given only $27$ .

How one-to-one teaching helps

Our teachers turn the chain rule into a habit you cannot skip: name the inside as $u$ every single time, so the $\frac{du}{dx}$ step is written before you are allowed to finish. In a one-to-one lesson we watch the exact line where you would normally stop and rebuild it with you until the extra factor appears automatically.

Because the paper awards method marks, showing $\frac{dy}{du}$ and $\frac{du}{dx}$ separately protects your score even under time pressure. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

To drill the chain rule, message us on WhatsApp to arrange a one-hour paid class from RM50/hr at the teacher's rate.

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Frequently asked questions

What exactly is a composite function?

It is a function inside another function, for example $(2x^{2}-5)^{3}$ , where the inner part $2x^{2}-5$ is raised to a power, or $\sqrt{2x+1}=(2x+1)^{1/2}$ . Whenever the thing being raised to a power is not just $x$ , you are differentiating a composite and the chain rule applies.

How do I know I have forgotten the chain rule?

Check whether your derivative includes the derivative of the inside. If your answer looks exactly like the power rule applied to a plain $x$ , nothing multiplied on from the bracket's contents, you have almost certainly stopped one step early.

Do I still get marks if I miss the inside derivative?

Under analytic marking you can earn method marks for correctly bringing the power down and reducing it, but you lose the accuracy marks because the final expression is wrong by a constant or variable factor. Writing $\frac{dy}{du}$ and $\frac{du}{dx}$ as separate lines makes your method visible and easy to credit.

Does

\sqrt{2x+1}

need the chain rule too?

Yes. Rewrite it as $(2x+1)^{1/2}$ ; then $\frac{dy}{dx}=\tfrac{1}{2}(2x+1)^{-1/2}\cdot 2=(2x+1)^{-1/2}$ .

The factor of $2$ from differentiating the inside is exactly the step that is easy to drop.

Source:SRC-DSKP-EN

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

Forgetting the chain rule

The error

Wrong line, then the fix

The reliable fix

One line to remember it

Practice where the error hides

How one-to-one teaching helps

Frequently asked questions

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