How Do You Use the Chain Rule for Composite Functions?

Use the chain rule by differentiating the outer function and multiplying by the derivative of the inner function. If $y=f(g(x))$, then

$$\frac{dy}{dx}=f'(g(x))g'(x).$$

This is IB Mathematics: Analysis and Approaches content from SL 5.6: Differentiation rules, examined at both SL and HL on Paper 1 and Paper 2.

The rule expresses how rates of change compose. A change in $x$ first changes the inner expression, and that change then affects the outer function. Multiplication combines these two rates, so every nested layer contributes one derivative factor.

How the Chain Rule Works

A composite function is one function applied inside another. For example, in $y=(3x^2-1)^5$, the outer function is “raise to the fifth power”, while the inner function is $3x^2-1$.

Use these steps:

1. Identify the inner function, $u=g(x)$.
2. Differentiate the outer function with respect to $u$.
3. Multiply by $\frac{du}{dx}$.

Equivalently,

$$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}.$$

The order is important: differentiate from the outside inward, but do not expand unless expansion makes the algebra simpler. Keeping the composite form usually reduces errors and makes the chain-rule structure visible to the examiner.

Worked Example

Let $y=(3x^2-1)^5$. Set $u=3x^2-1$, so .

$$\frac{dy}{du}=5u^4, \qquad \frac{du}{dx}=6x.$$

Therefore,

$$\frac{dy}{dx}=5u^4(6x)=30x(3x^2-1)^4.$$

For several layers, multiply by the derivative of every layer. If $y=\sin((2x+1)^3)$, then

$$\frac{dy}{dx}=6(2x+1)^2\cos((2x+1)^3).$$

Exam Technique

The common misconception is differentiating only the outer function and forgetting (g'(x)). In an IB response, show the outer derivative and inner derivative clearly; this can earn a method mark even if later simplification is incorrect.

For a “Differentiate” command term, the final derivative is required, while clear intermediate working protects method marks. If asked to find a gradient, substitute the stated $x$-value only after obtaining the derivative. A quick substitution check can also reveal a missing constant factor before the final answer.

On Paper 1, differentiate analytically. On Paper 2, technology may check the result but does not replace required working.