Pure 7.4 - Product, quotient and chain rules | Edexcel A-Level Mathematics

Differentiate expressions built from products, quotients and nested functions by reading their structure first. Use the same chain-rule idea to handle reciprocal trigonometric functions, inverse relations and connected rates of change.

Choose the rule from the structure

If a balloon's radius grows steadily, does its volume grow steadily too? A sphere of radius $r$ has volume $V=\frac43\pi r^3$ . Doubling the radius multiplies volume by $8$ , so equal increases in radius should not give equal increases in volume. We will connect the two rates precisely.

You already know

Differentiate sums term by term, but first identify what is added. Standard rules give $(x^n)'=nx^{n-1}$ , $(e^{kx})'=ke^{kx}$ , $(\sin x)'=\cos x$ and $(\cos x)'=-\sin x$ , with radian input.

Two functions multiplied need a product rule; one divided by another needs a quotient rule; one inside another needs a chain rule. More than one may appear in the same question.

Composite function

A composite applies one function to the output of another. For $y=(3x+1)^5$ , the inner function is $u=3x+1$ and the outer function is $y=u^5$ . The expression $x^5+3x+1$ is a sum instead.

Look at the outermost operation before looking inside its pieces.

Expression	Outermost operation	What also happens inside
$x^2\sin x$	product	standard derivatives of the factors
$e^{3x}/x$	quotient	a linear exponential input
$(3x+1)^5$	composition	a linear inner derivative
$x^2(3x+1)^5$	product	chain rule for the second factor

For a product, multiplying the derivatives is generally wrong. The simple check $x\cdot x=x^2$ already disproves that guess: the true derivative is $2x$ , whereas $1\cdot1=1$ .

Key idea

Choose the rule for the outermost operation, then differentiate its pieces using any further rules they need.

The same expression can sometimes be rewritten to use a different valid route; what matters is that every dependence is included.

Product and quotient rules

A product changes when either factor changes. If the factors change by $\Delta u$ and $\Delta v$ , its change is $u\Delta v+v\Delta u+\Delta u\Delta v$ . In the derivative limit, the last term contributes zero, leaving two terms.

Product rule

For differentiable functions $u(x),v(x)$ ,

\frac{\mathrm{d}}{\mathrm{d}x}(uv) =u\frac{\mathrm{d}v}{\mathrm{d}x}+v\frac{\mathrm{d}u}{\mathrm{d}x}

Not in the formula booklet: learn it.

Differentiate one factor while leaving the other alone, then swap roles and add.

Worked example 1: a power times sine

Question. Differentiate $y=2x^4\sin x$ .

Plan. Set $u=2x^4$ , $v=\sin x$ , and form the two product-rule terms.

\begin{aligned} u'&=8x^3,\qquad v'=\cos x && \text{differentiate the factors}\\ y'&=2x^4\cos x+8x^3\sin x && \text{product rule}\\ &=2x^3(x\cos x+4\sin x) && \text{optional factorisation} \end{aligned}

Why this step? Neither factor is constant, so both changing contributions are required.

Answer. $y'=2x^4\cos x+8x^3\sin x$ .

Check. Near zero, $\sin x$ behaves like $x$ , so $y$ behaves like $2x^5$ and the derivative behaves like $10x^4$ , as the result does.

Change only the trigonometric factor from sine to cosine.

Worked example 2: a power times cosine

Question. Differentiate $y=2x^4\cos x$ .

Plan. Keep the power factor and replace the trig derivative by $-\sin x$ .

\begin{aligned} u'&=8x^3,\qquad v'=-\sin x && \text{local derivatives}\\ y'&=-2x^4\sin x+8x^3\cos x && \text{two product contributions} \end{aligned}

Why this step? The negative sine term comes from differentiating cosine; it does not change the sign of the other term.

Answer. $y'=8x^3\cos x-2x^4\sin x$ .

Check. Near zero, $y$ behaves like $2x^4$ , so its leading derivative is $8x^3$ , matching the result.

The same method gives $(x^2\cos3x)'=2x\cos3x-3x^2\sin3x$ : the extra $3$ comes from the input of cosine.

Watch out

Wrong: $(2x^4\sin x)'=8x^3\cos x$ . That multiplies the derivatives and loses both true product-rule terms.

Change only the power in the second example.

Quotient rule

A quotient can be treated as a product with a reciprocal. Writing $y=u/v$ as $yv=u$ , then differentiating the product gives $y'v+yv'=u'$ . Substitute $y=u/v$ and solve for $y'$ .

Quotient rule

For differentiable $u,v$ , where $v\ne0$ ,

\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac uv\right) =\frac{vu'-uv'}{v^2}

In the formula booklet.

The numerator's order matters. It is denominator times derivative of numerator, minus numerator times derivative of denominator.

Worked example 3: an exponential quotient

Question. Differentiate $y=e^{3x}/x$ , for $x\ne0$ .

Plan. Identify $u=e^{3x}$ , $v=x$ , then use the quotient rule before factoring.

\begin{aligned} u'&=3e^{3x},\qquad v'=1 && \text{differentiate each piece}\\ y'&=\frac{x(3e^{3x})-e^{3x}}{x^2} && \text{quotient rule}\\ &=\frac{e^{3x}(3x-1)}{x^2} && \text{factor the numerator} \end{aligned}

Why this step? Keeping the unsimplified numerator makes its subtraction order visible.

Answer. $y'=e^{3x}(3x-1)/x^2$ .

Check. Rewriting as $e^{3x}x^{-1}$ gives $3e^{3x}x^{-1}-e^{3x}x^{-2}$ , the same result.

Change only the denominator's power from $1$ to $4$ .

Worked example 4: a higher denominator power

Question. Differentiate $y=e^{3x}/x^4$ , for $x\ne0$ .

Plan. Use $v=x^4$ , so its derivative is $4x^3$ , and simplify common factors afterwards.

\begin{aligned} y'&=\frac{x^4(3e^{3x})-e^{3x}(4x^3)}{x^8} && \text{square the whole denominator}\\ &=\frac{e^{3x}x^3(3x-4)}{x^8} =\frac{e^{3x}(3x-4)}{x^5} && \text{cancel common factors} \end{aligned}

Why this step? Squaring $x^4$ gives $x^8$ , even though cancellation later reduces the power.

Answer. $y'=e^{3x}(3x-4)/x^5$ , still with $x\ne0$ .

Check. The product route $e^{3x}x^{-4}$ gives $3e^{3x}x^{-4}-4e^{3x}x^{-5}$ .

With a different denominator, the same order applies:

\left(\frac{e^{2x}}{x^2+1}\right)' =\frac{2e^{2x}(x^2+1)-2xe^{2x}}{(x^2+1)^2} =\frac{2e^{2x}(x^2-x+1)}{(x^2+1)^2}

Here $x^2+1>0$ , so every real input is allowed.

Watch out

Wrong: reverse $vu'-uv'$ , or leave the denominator as $v$ . Write $vu'-uv'$ over $v^2$ before inserting expressions.

Change only the denominator power once more.

Chain rule for composites

For $y=(\cos x)^2$ , changing $x$ first changes $\cos x$ , which then changes its square. The rates multiply. This is a dependency chain, not ordinary cancellation of fractions.

Chain rule

If $y=f(u)$ , $u=g(x)$ , with the functions differentiable at the relevant inputs,

\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{\mathrm{d}y}{\mathrm{d}u}\frac{\mathrm{d}u}{\mathrm{d}x}

Not in the formula booklet: learn it.

Differentiate the outside while leaving its input unchanged, then multiply by the derivative of that input.

Worked example 5: a squared trig function

Question. Differentiate $y=\cos^2x$ .

Plan. Use $u=\cos x$ , $y=u^2$ .

\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}u}&=2u,\qquad \frac{\mathrm{d}u}{\mathrm{d}x}=-\sin x && \text{two linked rates}\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=2\cos x(-\sin x)=-2\sin x\cos x && \text{multiply} \end{aligned}

Why this step? The outside is a square, so its input remains $\cos x$ until the outside derivative is taken.

Answer. $-2\sin x\cos x$ .

Check. $\cos^2x=(1+\cos2x)/2$ gives derivative $-\sin2x$ , which is equivalent.

Change only the inner frequency from $1$ to $2$ .

Worked example 6: adding one input layer

Question. Differentiate $y=\cos^2(2x)$ .

Plan. Follow the layers square, cosine, then linear input.

\begin{aligned} y'&=2\cos(2x)\,[-\sin(2x)]\,2 && \text{one derivative per layer}\\ &=-4\sin(2x)\cos(2x) && \text{collect the multipliers} \end{aligned}

Why this step? The final $2$ is the rate of the innermost input; omitting it halves the answer.

Answer. $-4\sin(2x)\cos(2x)$ .

Check. The identity form $(1+\cos4x)/2$ differentiates to $-2\sin4x$ , the same result.

Changing the trig function does not change the dependency logic:

(\tan^2(2x))'=2\tan(2x)\,[2\sec^2(2x)] =4\tan(2x)\sec^2(2x)

where $\cos2x\ne0$ . Likewise, $((1+e^{2x})^4)'=4(1+e^{2x})^3(2e^{2x})=8e^{2x}(1+e^{2x})^3$ . In the combined form $x^2(3x+1)^5$ , product rule gives $2x(3x+1)^5+15x^2(3x+1)^4$ . Work from outside inward rather than trying to remember a separate rule for every appearance.

Watch out

Wrong: $(\cos^2x)'=(-\sin x)^2$ . Differentiation does not pass through the square that way. Differentiate the outside square first: $2\cos x(-\sin x)$ .

Change the inner frequency again.

Reciprocal trigonometric derivatives

The reciprocal identities $\sec x=1/\cos x$ , $\operatorname{cosec}x=1/\sin x$ and $\cot x=\cos x/\sin x$ let you derive the next rules instead of guessing their signs.

For example, chain rule gives

(\sec x)'=-(\cos x)^{-2}(-\sin x)=\sec x\tan x

and

(\operatorname{cosec}x)'=-(\sin x)^{-2}\cos x =-\operatorname{cosec}x\cot x

Quotient rule gives

(\cot x)'=\frac{-\sin^2x-\cos^2x}{\sin^2x} =-\operatorname{cosec}^2x

Reciprocal trigonometric derivatives

For constant $k$ , radian input, and wherever the original functions are defined,

\begin{aligned} (\sec kx)'&=k\sec(kx)\tan(kx)\\ (\operatorname{cosec}kx)'&=-k\operatorname{cosec}(kx)\cot(kx)\\ (\cot kx)'&=-k\operatorname{cosec}^2(kx) \end{aligned}

In the formula booklet.

Secant excludes zeros of cosine; cosecant and cotangent exclude zeros of sine. A printed formula does not remove those restrictions.

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Last Accessed

Pure Mathematics (Papers 1 and 2)

Statistics (Paper 3, Section A)

Mechanics (Paper 3, Section B)

Pure 7.4 - Product, quotient and chain rules

Choose the rule from the structure

Product and quotient rules

Quotient rule

Chain rule for composites

Reciprocal trigonometric derivatives

Continue this lesson.

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