Mc12, Mc13 - Product and quotient rules | OCR A-Level Mathematics B (MEI)

A derivative must account for every changing factor. Learn the product rule in Mc12 and the quotient rule in Mc13, then choose between rules and simpler algebra. Equivalent expressions share derivatives only where they describe the same function.

Why a product has two derivative terms

How can two non-constant factors multiply to a constant? Take $x\times x^{-1}=1$ , with $x\ne0$ . Multiplying their derivatives would give $-x^{-2}$ , although the product's derivative is zero. The two changes must cancel, rather than multiply.

You already know

You can differentiate powers, exponentials, logarithms and radian trigonometric functions. Function notation $f'(x)$ and differential notation $\mathrm dy/\mathrm dx$ express the same derivative.

For differentiable $u,v$ , insert a useful middle term into the difference quotient:

\frac{u(x+h)v(x+h)-u(x)v(x)}h =u(x+h)\frac{v(x+h)-v(x)}h+v(x)\frac{u(x+h)-u(x)}h

The limit is $uv'+vu'$ , since differentiability gives continuity as well as the two derivative limits. For the opening product, $x(-x^{-2})+x^{-1}(1)=0$ : the two contributions cancel exactly.

Product rule

When both functions are differentiable,

[f(x)g(x)]'=f(x)g'(x)+f'(x)g(x)

Equivalently, for $y=uv$ , $\mathrm dy/\mathrm dx=u\,\mathrm dv/\mathrm dx+v\,\mathrm du/\mathrm dx$ .

Not in the formula sheet: learn it.

Start with a logarithm, whose real domain must remain positive.

Worked example 1: a product with a logarithm

Question. Determine the derivative of $f(x)=x\ln x$ , stating the domain.

Plan. Retain each factor once while differentiating the other.

\begin{aligned} x&>0 &&\text{logarithm domain}\\ f'(x)&=x\left(\frac1x\right)+(1)\ln x &&\text{product rule}\\ &=1+\ln x &&\text{simplify} \end{aligned}

Why this step? The derivative of one factor is weighted by the current size of the other.

Answer. $f'(x)=1+\ln x$ , for $x>0$ .

Check. At one, $f(1)=0$ but $f'(1)=1$ ; the graph can cross an axis with nonzero gradient.

Change only the algebraic factor's power, from one to two.

Worked example 2: changing the retained factor

Question. Determine the derivative of $g(x)=x^2\ln x$ .

Plan. Keep the same logarithmic derivative and update the power term.

\begin{aligned} g'(x)&=x^2\frac1x+(2x)\ln x &&\text{product rule}\\ &=x+2x\ln x &&\text{simplify} \end{aligned}

Why this step? Changing a factor changes both its retained value and the term in which it is differentiated.

Answer. $g'(x)=x+2x\ln x$ , on $x>0$ .

Check. At one, the gradient is one again.

The simplest counterexample to the false rule is $x\times x=x^2$ : its derivative is $2x$ , not one.

Watch out

Wrong: $[fg]'=f'g'$ . Correct: $[fg]'=fg'+f'g$ . Two terms record two contributions.

Keep the logarithm but change the power again.

Applying the product rule to exponential and trig factors

A trigonometric product keeps both the product structure and the input rate. Write down the two factors before trying to tidy the result.

Worked example 3: a line times a cosine

Question. For $f(x)=x\cos(2x)$ , determine $f'(x)$ and its exact value at $\pi/4$ .

Plan. Differentiate the cosine with its radian input factor, then use the product rule.

\begin{aligned} f'(x)&=x[-2\sin(2x)]+(1)\cos(2x) &&\text{product rule}\\ &=\cos(2x)-2x\sin(2x) &&\text{reorder}\\ f'(\pi/4)&=0-2(\pi/4)=-\pi/2 &&\text{exact values} \end{aligned}

Why this step? The factor $x$ is retained in the term where cosine changes.

Answer. $f'(x)=\cos(2x)-2x\sin(2x)$ , with specified gradient $-\pi/2$ .

Check. $f'(0)=1$ ; multiplying the two derivatives would wrongly give zero.

Change only the cosine frequency from two to four.

Worked example 4: a faster oscillation

Question. For $g(x)=x\cos(4x)$ , determine $g'(x)$ and $g'(\pi/4)$ .

Plan. Change the cosine's input factor and keep both product terms.

\begin{aligned} g'(x)&=\cos(4x)-4x\sin(4x) &&\text{product rule}\\ g'(\pi/4)&=\cos\pi-\pi\sin\pi=-1 &&\text{evaluate} \end{aligned}

Why this step? A changed frequency alters the argument at the specified point as well as the derivative coefficient.

Answer. The derivative is as above and the specified gradient is minus one.

Check. At zero, $g'(0)=1$ , as before.

Exponential products can be factored usefully: $(x^2\mathrm e^{-x})'=x\mathrm e^{-x}(2-x)$ . Since the exponential is positive, its stationary points are $(0,0)$ and $(2,4\mathrm e^{-2})$ . Likewise $(x\mathrm e^{2x})'=(1+2x)\mathrm e^{2x}$ , giving gradient one at zero. Three factors can be grouped into two and differentiated again; simple polynomial products may be quicker to expand.

Exam tip

An unsimplified correct derivative can be a complete answer unless a particular form is requested. Simplify when it reveals zeros, signs or a needed later result.

Try the same product with a new frequency.

The quotient rule and its denominator condition

A changing denominator matters even when the numerator stays small. For $y=u/v$ , rewrite the relation as $yv=u$ . The product rule gives $y'v+yv'=u'$ , and substitution of $y=u/v$ gives the following rule.

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Mc12, Mc13 - Product and quotient rules

Why a product has two derivative terms

Applying the product rule to exponential and trig factors

The quotient rule and its denominator condition

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