Skip to content
spmaddmath.com.my
Tuition

Study

SyllabusFormulasMethodsExam & PapersTools
LocationsPricingBlogOur TeachersContact
EN

Calculus Clinic · Differentiation

Forgetting the inner derivative in the chain rule

For y=[f(x)]ny=[f(x)]^{n}, the chain rule gives dydx=n[f(x)]n1f(x)\frac{dy}{dx}=n[f(x)]^{n-1}\cdot f'(x). The error is stopping at n[f(x)]n1n[f(x)]^{n-1} and forgetting the ×f(x)\times f'(x), the derivative of the inside.

The error

The chain rule differentiates a function of a function: an outer operation wrapped around an inner expression. For a power like (3x2+1)4(3x^{2}+1)^{4}, the outer job is the fourth power and the inner expression is 3x2+13x^{2}+1.

The rule says bring the power down, reduce it by one, and then multiply by the derivative of the inside. The tempting slip in Add Math is to do the first part perfectly, 4(3x2+1)34(3x^{2}+1)^{3}, and stop there, forgetting the ×6x\times 6x.

It happens because the power-rule half feels like the whole answer, and the inside looks like it was already dealt with when you copied it down. But without the inner derivative the working silently ignores how fast the inside itself is changing.

When the inside is just xx the missing factor is 11 and no harm is done, which is exactly why the habit of leaving it off survives, until the inside is anything more than xx, and every value that follows is wrong.

Wrong line, then the fix

Take y=(3x2+1)4y=(3x^{2}+1)^{4}. Bringing the power down but forgetting the inside gives:

Wrong, the inner derivative is missingMust memorise
dydx=4(3x2+1)3\frac{dy}{dx}=4(3x^{2}+1)^{3}

The inner expression is 3x2+13x^{2}+1, whose derivative is 6x6x. The chain rule multiplies by it:

CorrectMust memorise
dydx=4(3x2+1)36x=24x(3x2+1)3\frac{dy}{dx}=4(3x^{2}+1)^{3}\cdot 6x=24x(3x^{2}+1)^{3}

The wrong line is missing the whole factor 6x6x, so it is off by that much at every value of xx except where 6x=16x=1. A quick way to notice: after a chain-rule power, your answer should almost always carry an extra term out front that came from the inside.

The reliable fix

Chain ruleMust memorise
dydx=dydududx\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}
  1. 1

    Name the inside

    Let uu be the inner expression, so yy becomes a simple power of uu, e.g. y=u4y=u^{4} with u=3x2+1u=3x^{2}+1.

  2. 2

    Differentiate the outside

    Find dydu\frac{dy}{du}, treating uu as a single block: dydu=4u3\frac{dy}{du}=4u^{3}.

  3. 3

    Differentiate the inside

    Find dudx\frac{du}{dx} separately: dudx=6x\frac{du}{dx}=6x. This is the factor that gets forgotten.

  4. 4

    Multiply, never add

    Combine as dydududx\frac{dy}{du}\cdot\frac{du}{dx}, then rewrite uu back in terms of xx.

  5. 5

    Check the front factor

    Confirm your answer carries the inner derivative as a factor. If the inside was not simply xx, that factor must be there.

One line to remember it

Peel the outside, then times the derivative of the inside

Differentiate the outer power first, then multiply by the derivative of whatever is inside the bracket. No inner derivative in your answer means you stopped one step too early.

Practice where the error hides

Q1[4 marks]

Given y=(2x35)3y=(2x^{3}-5)^{3}, find dydx\frac{dy}{dx}, and hence the gradient of the curve at x=1x=1.

Show worked solution

Let the inside be u=2x35u=2x^{3}-5, so y=u3y=u^{3}. Then dydu=3u2\frac{dy}{du}=3u^{2} and the inner derivative is dudx=6x2\frac{du}{dx}=6x^{2}.

dydx=dydududx=3u26x2=3(2x35)26x2=18x2(2x35)2\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=3u^{2}\cdot 6x^{2}=3(2x^{3}-5)^{2}\cdot 6x^{2}=18x^{2}(2x^{3}-5)^{2}

At x=1x=1, the inside is 2(1)35=32(1)^{3}-5=-3, so (2x35)2=(3)2=9(2x^{3}-5)^{2}=(-3)^{2}=9 and 18x2=18(1)=1818x^{2}=18(1)=18:

dydxx=1=18(1)9=162\left.\frac{dy}{dx}\right|_{x=1}=18(1)\cdot 9=162

So the gradient at x=1x=1 is 162162. Forgetting the inner derivative would have left 3(2x35)2=3(9)=273(2x^{3}-5)^{2}=3(9)=27, a value six times too small, because the missing factor 6x26x^{2} equals 66 at x=1x=1.

How one-to-one teaching helps

Our teachers make the inner derivative a fixed second step, not an optional extra: name the inside, differentiate the outside, then write ×dudx\times\frac{du}{dx} on the same line before simplifying. In a one-to-one lesson we watch for the exact moment your pen wants to stop after the power, and we build the reflex to ask "derivative of the inside?"

every time. Because the paper awards method marks, showing dydu\frac{dy}{du} and dudx\frac{du}{dx} separately keeps your working easy to credit.

Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English. To lock in 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

When can I safely skip the inner derivative?

Only when the inside is simply xx, because then dudx=1\frac{du}{dx}=1 and multiplying by it changes nothing. For (x)4=x4(x)^{4}=x^{4} the answer is 4x34x^{3}.

The moment the inside is anything more, like 3x2+13x^{2}+1, the inner derivative must be included.

How do I spot that a question needs the chain rule?

Look for a bracket raised to a power, a root, or any function wrapped around an expression that is not just xx, such as (2x1)5(2x-1)^{5}, 4x+3\sqrt{4x+3}, or (x2+1)2(x^{2}+1)^{-2}. Whenever there is an inside more complex than xx, differentiate it and multiply.

Do I add or multiply the inner derivative?

Multiply. The chain rule is dydududx\frac{dy}{du}\cdot\frac{du}{dx}, a product.

Adding the inner derivative instead of multiplying by it is a separate error that also gives the wrong answer.

Does forgetting the inner derivative cost marks?

Yes. Under analytic marking you may keep a method mark for bringing the power down correctly, but you lose the accuracy marks because the derivative is incomplete.

Writing dudx\frac{du}{dx} as its own line makes the full method visible and creditable.

Source:SRC-DSKP-EN

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

Ready to get started?

Book a Trial Classfrom RM50/hr · One-hour paid trial · Same-day reply
Book a Trial ClassOne-hour paid trial · Same-day reply