2.4, 3.4 - Product and quotient rules | Cambridge A-Level Mathematics

Two changing factors contribute to the change in their product. You will learn how to account for both contributions, differentiate fractions without losing the subtraction order, and combine these rules with the chain rule for logarithmic, exponential and trigonometric functions.

The product rule

If $x\times x=x^2$ , why does differentiating the two factors separately give $1$ , rather than the correct answer $2x$ ? Multiplying derivatives misses how each changing factor interacts with the other factor.

For a concrete picture in algebra, let the factors $u$ and $v$ change by small amounts $\Delta u$ and $\Delta v$ . Their product changes by

(u+\Delta u)(v+\Delta v)-uv =v\Delta u+u\Delta v+\Delta u\Delta v.

Divide by the input change and take the limit. For differentiable functions, the final small-product contribution tends to zero. The other two contributions remain.

The product rule

If $u$ and $v$ are differentiable functions of $x$ ,

\frac{\mathrm d}{\mathrm dx}(uv) =v\frac{\mathrm du}{\mathrm dx}+u\frac{\mathrm dv}{\mathrm dx}.

In the formula booklet.

For $x\times x$ , this gives $x(1)+x(1)=2x$ , answering the opening puzzle. Each term differentiates one factor and leaves the other unchanged.

Worked example 1: a linear factor times an exponential

Question. Differentiate $y=xe^x$ .

Plan. Differentiate each factor once in the product rule, then factor out the common exponential.

\begin{aligned} u&=x,&u'&=1\\ v&=e^x,&v'&=e^x\\ y'&=e^x(1)+x(e^x)\\ &=e^x(1+x) \end{aligned}

Answer. $y'=e^x(1+x)$ .

Check. At $x=0$ , the difference quotient from the original function is $he^h/h=e^h\to1$ , agreeing with $y'(0)=1$ .

Change only the algebraic factor from $x$ to $x^2$ . Both contributions are still needed.

Worked example 2: a squared factor times an exponential

Question. Differentiate $y=x^2e^x$ .

Plan. Use the product rule, then collect the common factor $xe^x$ .

\begin{aligned} y'&=e^x(2x)+x^2(e^x)\\ &=xe^x(x+2) \end{aligned}

Answer. $y'=xe^x(x+2)$ .

Check. At $x=1$ , a central difference with step $0.001$ gives approximately $8.15485$ ; the derivative gives $3e\approx8.15485$ .

You may expand a polynomial product first if that is shorter. The product rule matters most when expansion cannot remove the product, as with $x^2e^x$ or $x\sin x$ .

Watch out

$(uv)'\ne u'v'$ . For $x^2e^x$ , the wrong expression $2xe^x$ counts the change in $x^2$ but misses the changing exponential. The missing term is $x^2e^x$ .

Change the algebraic power once more.

The quotient rule

A fraction changes when its numerator changes and when its denominator changes. Increasing a positive numerator raises the fraction, but increasing a positive denominator lowers it. The subtraction in the quotient rule reflects those competing effects.

To see the rule, write $y=u/v$ as $u=yv$ , where $v\ne0$ . The product rule gives $u'=y'v+yv'$ . Substitute $y=u/v$ and solve for $y'$ .

The quotient rule

For differentiable $u$ and $v$ , with $v\ne0$ ,

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

In the formula booklet.

Keep brackets around both numerator contributions until the subtraction has been performed.

Worked example 3: a linear denominator

Question. Differentiate $y=(x^2+1)/(x+2)$ .

Plan. Identify numerator and denominator, apply $vu'-uv'$ , then simplify the numerator.

\begin{aligned} u&=x^2+1,&u'&=2x\\ v&=x+2,&v'&=1\\ y'&=\frac{(x+2)(2x)-(x^2+1)(1)}{(x+2)^2}\\ &=\frac{x^2+4x-1}{(x+2)^2} \end{aligned}

Answer. $y'=(x^2+4x-1)/(x+2)^2$ , for $x\ne-2$ .

Check. Division gives $y=x-2+5/(x+2)$ . Differentiating this gives $1-5/(x+2)^2$ , the same result.

Keep the numerator but square the $x$ in the denominator. Its derivative is now $2x$ , and more terms cancel.

Worked example 4: a quadratic denominator

Question. Differentiate $y=(x^2+1)/(x^2+2)$ .

Plan. Apply the quotient rule before cancelling any terms.

\begin{aligned} y'&=\frac{(x^2+2)(2x)-(x^2+1)(2x)}{(x^2+2)^2}\\ &=\frac{2x[(x^2+2)-(x^2+1)]}{(x^2+2)^2}\\ &=\frac{2x}{(x^2+2)^2} \end{aligned}

Answer. $y'=2x/(x^2+2)^2$ , for every real $x$ , since $x^2+2>0$ .

Check. Rewrite $y=1-1/(x^2+2)$ . The chain rule gives $y'=2x/(x^2+2)^2$ , independently confirming the subtraction sign.

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2.4, 3.4 - Product and quotient rules

The product rule

The quotient rule

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