G4 - Chain rule and composite derivatives | AQA A-Level Mathematics

Identify the inner and outer operations of a composite function, track every layer of change and combine the chain rule with product and quotient rules. Keep the function's domain alongside the algebra.

The two rates in a composite function

Why does differentiating $(2x-3)^5$ produce a factor of $2$ ? Imagine increasing $x$ by $0.01$ : the inner quantity $2x-3$ increases by $0.02$ . The outer fifth power responds to that inner change, not directly to the original one.

You already know

The derivative of $u^5$ with respect to $u$ is $5u^4$ , and a composite $g(f(x))$ applies $f$ first. The power rule is not in the formula booklet: learn it.

If $u=g(x)$ , small changes satisfy $\Delta u\approx g'(x)\Delta x$ . If $y=f(u)$ , then $\Delta y\approx f'(u)\Delta u$ . Combining these local linear changes gives the chain rule. It applies when the inner derivative exists at $x$ and the outer derivative exists at the resulting $u$ .

Chain rule

If $u=g(x)$ and $y=f(u)$ , then

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

Not in the formula booklet: learn it.

The derivative of the outer function is evaluated at the inner function. For example, $f(x)=x^2$ , $g(x)=e^x$ give $gf(x)=e^{x^2}$ , with derivative $2xe^{x^2}$ . In the opposite order, $fg(x)=e^{2x}$ , with derivative $2e^{2x}$ . The exponential derivative and these composite results are recall, not supplied.

Worked example 1: a linear inner function

Question. Find $dy/dx$ for $y=(2x-3)^5$ .

Plan. Differentiate the fifth power with respect to its input, then multiply by the input's derivative.

\begin{aligned} u&=2x-3 && \text{inner function}\\ \frac{dy}{du}&=5u^4,\quad\frac{du}{dx}=2 && \text{two rates}\\ \frac{dy}{dx}&=5(2x-3)^4\times2 && \text{chain rule}\\ &=10(2x-3)^4 && \text{simplify} \end{aligned}

Why multiply by $2$ ? The inner input changes twice as quickly as $x$ .

Answer. $dy/dx=10(2x-3)^4$ , for every real $x$ .

Check. At $x=2$ , the inner value is $1$ , so the outer rate $5$ times the inner rate $2$ gives $10$ .

Change only the inner function from linear to quadratic. Its rate now depends on $x$ , so the extra factor is no longer constant.

Worked example 2: a quadratic inner function

Question. Find $dy/dx$ for $y=(2x^2-3)^5$ , in factorised form.

Plan. Keep the inner quadratic intact while differentiating the outer power.

\begin{aligned} u&=2x^2-3 && \text{inner function}\\ \frac{dy}{du}&=5u^4,\quad\frac{du}{dx}=4x && \text{differentiate each layer}\\ \frac{dy}{dx}&=5(2x^2-3)^4(4x) && \text{multiply the rates}\\ &=20x(2x^2-3)^4 && \text{factorised answer} \end{aligned}

Why leave the bracket intact? The power rule applies to that whole input; expanding creates unnecessary terms.

Answer. $dy/dx=20x(2x^2-3)^4$ .

Check. At $x=0$ , the derivative is $0$ . The original function is even, so its graph has reflection symmetry about the vertical axis, consistent with that horizontal tangent.

A zero inner derivative is allowed in the chain rule: you multiply by zero. Dividing by that derivative would be a different operation with different conditions.

Watch out

Wrong: $[(2x^2-3)^5]'=5(2x^2-3)^4$ . This is only the rate with respect to the bracket. Right: multiply by $4x$ , giving $20x(2x^2-3)^4$ .

Change only the quadratic's coefficient in example 2. Write both layer derivatives before multiplying if that helps you keep track.

Nested powers, roots and logarithms

For several layers, work from the outside in and include a derivative factor from each layer.

In $[\ln(1+x^2)]^3$ , the layers are the cube, logarithm and quadratic. A logarithm requires a positive argument. A square root allows a zero argument as a function value, but its usual finite derivative needs a strictly positive argument.

Composite powers and logarithms

Where the functions and derivatives exist,

\frac{d}{dx}[u(x)]^n=n[u(x)]^{n-1}u'(x)

For $u(x)>0$ ,

\frac{d}{dx}\ln u(x)=\frac{u'(x)}{u(x)}

For negative or fractional powers, check roots and denominators before applying the rule. Not in the formula booklet: learn it.

For example, $y=\sqrt{5-2x}$ is defined on $x\le5/2$ , but $y'=-1/\sqrt{5-2x}$ is finite only on $x<5/2$ . At the endpoint the slope becomes unbounded. This root derivative follows from the recalled power and chain rules; it is not supplied.

Worked example 3: three layers with no exclusions

Question. Differentiate $y=[\ln(1+x^2)]^3$ , stating its real domain.

Plan. Differentiate the cube, logarithm and quadratic in that order.

\begin{aligned} y'&=3[\ln(1+x^2)]^2\frac1{1+x^2}(2x) && \text{one factor per layer}\\ &=\frac{6x[\ln(1+x^2)]^2}{1+x^2} && \text{simplify}\\ 1+x^2&>0 && \text{all real inputs allowed} \end{aligned}

Why keep the logarithm squared? It is the input to the outer cube; differentiating the cube reduces only that outer power.

Answer. $y'=6x[\ln(1+x^2)]^2/(1+x^2)$ , for all real $x$ .

Check. At $x=0$ , the derivative is $0$ . No division by the zero value $\ln1=0$ occurs.

Change only the logarithm's inner quadratic to $4-x^2$ . The differentiation steps are the same, but the domain is now restricted.

Worked example 4: a restricted logarithm

Question. Differentiate $y=[\ln(4-x^2)]^3$ , stating where the derivative applies.

Plan. Find the logarithm's domain, then multiply the three layer rates.

\begin{aligned} 4-x^2&>0\quad\Longrightarrow\quad-2<x<2 && \text{logarithm domain}\\ y'&=3[\ln(4-x^2)]^2\frac1{4-x^2}(-2x) && \text{chain rule}\\ &=-\frac{6x[\ln(4-x^2)]^2}{4-x^2} && \text{simplify} \end{aligned}

Why use the original logarithm to find the domain? The algebraic derivative cannot make a logarithm of a nonpositive number valid.

Answer. $y'=-6x[\ln(4-x^2)]^2/(4-x^2)$ , for $-2<x<2$ .

Check. At $x=\sqrt3$ , the logarithm is $\ln1=0$ , and the derivative is $0$ ; the denominator is $1$ , so this is valid.

The outer cube can take a negative logarithm value. Only the logarithm's argument must be positive; its value need not be positive.

Watch out

Wrong: “The domain requires $\ln(4-x^2)>0$ .” That needlessly excludes valid logarithm values. Right: require $4-x^2>0$ , giving $-2<x<2$ .

Change only the constant inside the logarithm. Check the new endpoints before differentiating.

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G4 - Chain rule and composite derivatives

The two rates in a composite function

Nested powers, roots and logarithms

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