G4 - Product and quotient rules | AQA A-Level Mathematics

A product or quotient has two changing pieces. Learn how their changes combine, choose a convenient derivative rule and keep the original domain when an expression simplifies.

Differentiating a product

If $y=x^2e^x$ , why is its derivative not just $2xe^x$ ? The polynomial changes, but so does the exponential. We need to account for both changes.

You already know

The power rule gives $(x^n)'=nx^{n-1}$ , and $(e^x)'=e^x$ . Differentiate sums term by term. These rules are not in the formula booklet: learn them.

For a concrete warning, $x\times x=x^2$ has derivative $2x$ , whereas multiplying the two derivatives gives only $1$ . In a product, each factor's change contributes while the other factor is held fixed.

Product rule

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

\frac{d}{dx}(uv)=u'v+uv'

Not in the formula booklet: learn it.

Here is why it works. Put $p(x)=u(x)v(x)$ , add and subtract $u(x+h)v(x)$ , then divide by $h$ :

\begin{aligned} \frac{p(x+h)-p(x)}h &=u(x+h)\frac{v(x+h)-v(x)}h +v(x)\frac{u(x+h)-u(x)}h \end{aligned}

As $h\to0$ , differentiability makes $u(x+h)\to u(x)$ . Taking the two derivative limits gives $uv'+vu'$ . The definition of a derivative by this limit is in the formula booklet; the product rule derived from it is not.

Worked example 1: a polynomial times an exponential

Question. Find $dy/dx$ for $y=x^2e^x$ , in factorised form.

Plan. Differentiate each factor in turn, then factorise the sum.

\begin{aligned} u=x^2,&\quad v=e^x && \text{choose the factors}\\ u'=2x,&\quad v'=e^x && \text{differentiate separately}\\ y'&=2xe^x+x^2e^x && \text{product rule}\\ &=xe^x(x+2) && \text{common factors} \end{aligned}

Why this rule? Multiplication is the last operation, and both factors vary with $x$ .

Answer. $dy/dx=xe^x(x+2)$ , for all real $x$ .

Check. At $x=0$ , the derivative is $0$ ; the difference quotient there is $he^h\to0$ .

Keep the polynomial factor and change only the second factor from exponential to sine. This changes its derivative, but the two-contribution structure stays the same. Trigonometric calculus uses radians; $(\sin x)'=\cos x$ is a recall result, not supplied.

Worked example 2: a polynomial times sine

Question. Find $dy/dx$ for $y=x^2\sin x$ , with $x$ in radians.

Plan. Apply the product rule to the polynomial and trigonometric factors.

\begin{aligned} u'=2x,&\quad v'=\cos x && \text{differentiate the factors}\\ y'&=2x\sin x+x^2\cos x && \text{product rule} \end{aligned}

Why keep the sine in the first term? That term differentiates only $x^2$ ; the second term accounts for the sine changing.

Answer. $dy/dx=2x\sin x+x^2\cos x$ .

Check. At $x=\pi/2$ , this gives $\pi$ , since the cosine contribution is zero.

Sometimes the two contributions cancel completely. On $x>0$ , both $x$ and $1/x$ vary, but their product is $1$ : the product-rule terms are $1/x+x(-1/x^2)=0$ . Two moving pieces can produce a constant result.

Watch out

Wrong: $(x\sin x)'=1\times\cos x=\cos x$ . Multiplying derivatives omits the unchanged factors. Right: $(x\sin x)'=\sin x+x\cos x$ .

For three factors, grouping the first two and applying the rule twice gives $(uvw)'=u'vw+uv'w+uvw'$ . This extension is not supplied: derive it by differentiating exactly one factor in each term. Now change only the polynomial power in example 2.

Differentiating a quotient

Changing a denominator affects the whole quotient, so a quotient derivative is not a quotient of derivatives.

For instance, $(x^2+1)/(x-1)$ has a changing numerator and denominator and is undefined at $x=1$ . To derive its general rule, write $y=u/v$ as $u=vy$ , on an interval where $v\ne0$ . The product rule gives

\begin{aligned} u'&=v'y+vy' && \text{differentiate }u=vy\\ vy'&=u'-v'\frac uv && \text{substitute }y=u/v\\ y'&=\frac{vu'-uv'}{v^2} && \text{divide by }v \end{aligned}

This derivation explains both the subtraction and the squared denominator.

Quotient rule

For differentiable $u(x)$ , $v(x)$ , with $v(x)\ne0$ ,

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

In the formula booklet.

Choose the complete numerator and denominator as your two functions. Keep brackets until the subtraction has been dealt with.

Worked example 3: keeping the subtraction intact

Question. Differentiate $y=(x^2+1)/(x-1)$ , simplifying the numerator.

Plan. Use the quotient rule and retain the original excluded input.

\begin{aligned} u=x^2+1,&\quad v=x-1 && \text{whole numerator and denominator}\\ u'=2x,&\quad v'=1 && \text{differentiate}\\ y'&=\frac{2x(x-1)-(x^2+1)}{(x-1)^2} && \text{quotient rule}\\ &=\frac{x^2-2x-1}{(x-1)^2} && \text{expand the subtraction} \end{aligned}

Why keep the bracket? The minus sign applies to both terms of $x^2+1$ .

Answer. $dy/dx=(x^2-2x-1)/(x-1)^2$ , for $x\ne1$ .

Check. Division gives $y=x+1+2/(x-1)$ . Its derivative $1-2/(x-1)^2$ is the same expression. The reciprocal-power derivative used in this check is a recall result, not supplied.

Change only the constant in the denominator. The subtraction rule stays identical, but the excluded input and the coefficient of $x$ change.

Worked example 4: a shifted denominator

Question. Differentiate $y=(x^2+1)/(x-2)$ , simplifying the numerator.

Plan. Use the same quotient structure with denominator $x-2$ .

\begin{aligned} y'&=\frac{2x(x-2)-(x^2+1)}{(x-2)^2} && \text{quotient rule}\\ &=\frac{x^2-4x-1}{(x-2)^2} && \text{collect terms} \end{aligned}

Why square $x-2$ ? The quotient rule squares the whole original denominator, not its derivative.

Answer. $dy/dx=(x^2-4x-1)/(x-2)^2$ , for $x\ne2$ .

Check. At $x=0$ , the derivative is $-1/4$ . Division gives $y=x+2+5/(x-2)$ , whose derivative gives the same value.

The order of subtraction is worth keeping visible even when the derivatives are simple.

Watch out

Wrong: $y'=[2x(x-2)-x^2+1]/(x-2)^2$ . Here $-(x^2+1)$ has become $-x^2+1$ . Right: subtract both terms, giving numerator $x^2-4x-1$ .

Try the same structure with one more change to the denominator's constant. State the restriction alongside your answer.

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G4 - Product and quotient rules

Differentiating a product

Differentiating a quotient

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