9B - The quotient rule

Can you differentiate a fraction by dividing the derivatives of its numerator and denominator? You will use the quotient rule, choose when algebra gives a shorter route, and evaluate derivatives while keeping the original function's domain.

Differentiating a quotient

For $y=\frac{x}{x+1}$ , dividing the two derivatives would give $1/1=1$ . But writing $y=1-\frac1{x+1}$ gives the derivative $\frac1{(x+1)^2}$ , which varies with $x$ . What accounts for the changing denominator?

Write a quotient as a product: $y=u v^{-1}$ . The product and chain rules give

\begin{aligned} \frac{dy}{dx}&=u'v^{-1}-uv^{-2}v'\\ &=\frac{u'v-uv'}{v^2} &&\text{use a common denominator} \end{aligned}

The minus sign comes from differentiating the reciprocal. This explains both the order of the numerator terms and the squared denominator.

Quotient rule

If $u(x)$ and $v(x)$ are differentiable at the point considered and $v(x)\ne0$ , then

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

Differentiate the numerator and multiply by the denominator; subtract the numerator times the derivative of the denominator; divide by the denominator squared.

On the formulae sheet.

Choose which expression is $u$ and which is $v$ , then write their derivatives before substituting into the rule.

Worked example 1: a quotient of linear expressions

Question. Differentiate $y=\frac{2x+1}{x-2}$ .

Plan. Use $u=2x+1$ , $v=x-2$ , and keep the numerator subtraction in brackets.

\begin{aligned} u'&=2 &v'&=1\\ \frac{dy}{dx}&=\frac{2(x-2)-(2x+1)(1)}{(x-2)^2} &&\text{apply the quotient rule}\\ &=\frac{2x-4-2x-1}{(x-2)^2} &&\text{subtract the whole second term}\\ &=-\frac5{(x-2)^2} &&\text{collect terms} \end{aligned}

Answer. $\frac{dy}{dx}=-\frac5{(x-2)^2}$ , for $x\ne2$ .

Check. Rewriting the original as $2+5(x-2)^{-1}$ and using the chain rule gives $-5(x-2)^{-2}$ , the same result by another route.

Keep the numerator and change only the denominator to a quadratic. Its derivative must change too.

Worked example 2: a quadratic denominator

Question. Differentiate $y=\frac{2x+1}{x^2-2}$ .

Plan. Use $v'=2x$ , then simplify the full numerator before considering cancellation.

\begin{aligned} \frac{dy}{dx}&=\frac{2(x^2-2)-(2x+1)(2x)}{(x^2-2)^2} &&\text{use }u'v-uv'\\ &=\frac{2x^2-4-4x^2-2x}{(x^2-2)^2} &&\text{expand both products}\\ &=-\frac{2(x^2+x+2)}{(x^2-2)^2} &&\text{collect and factor} \end{aligned}

Answer. $\frac{dy}{dx}=-\frac{2(x^2+x+2)}{(x^2-2)^2}$ , for $x\ne\pm\sqrt2$ .

Check. At $x=1$ , the derivative is $-8$ . The gradient calculated directly from the original function at $0.9999$ and $1.0001$ is approximately $-8.0000005$ , consistent with that value.

A common factor may be cancelled only when it multiplies the whole numerator and denominator.

Watch out

Reversing $u'v-uv'$ reverses the derivative's sign. Also, the denominator is $v^2$ , not $(v')^2$ : in the second example it is $(x^2-2)^2$ , not $4x^2$ .

Apply the same rule to a new quadratic denominator.

Choosing quotient or product form

The quotient rule is reliable, but simplifying a fraction first may reveal a shorter differentiation problem.

When the denominator is a single power of $x$ , dividing each numerator term by it often gives a sum of powers. When a denominator is a bracket that does not cancel, either the quotient rule or the product form $uv^{-1}$ works. The derivation in the first part shows why these methods agree.

Worked example 3: cancelling a common factor

Question. Differentiate $y=\frac{x^2+3x}{x}$ , where $x\ne0$ .

Plan. Factor the numerator and cancel the common factor before differentiating.

\begin{aligned} y&=\frac{x(x+3)}x=x+3 &&\text{valid because }x\ne0\\ \frac{dy}{dx}&=1 &&\text{differentiate the simplified expression} \end{aligned}

Answer. $\frac{dy}{dx}=1$ , for $x\ne0$ .

Check. The quotient rule gives

\frac{(2x+3)x-(x^2+3x)}{x^2} =\frac{x^2}{x^2}=1,

on the same domain. The two methods agree, but cancellation needed less work.

Keep the numerator and square the denominator. Cancellation now leaves a reciprocal term rather than a linear function.

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9B - The quotient rule

Differentiating a quotient

Choosing quotient or product form

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