The Quotient Rule, calculus lesson

Average cost per item, concentration of a solution and many other everyday quantities are one thing divided by another. To know how fast such a ratio changes you need the derivative of a quotient. It comes from the product rule with one extra ingredient, the derivative of a reciprocal.

Key terms

Reciprocal rule

Definition If $g$ is differentiable at $x$ and $g(x) \neq 0$ , then $\frac{1}{g}$ is differentiable at $x$ and

\left(\frac{1}{g}\right)'(x) = -\frac{g'(x)}{g(x)^2}.

Proof. Since $g$ is differentiable at $x$ , it is continuous there. So $g(x + h) \to g(x) \neq 0$ , and in particular $g(x + h) \neq 0$ for all small $h$ . For those $h \neq 0$ ,

\frac{\frac{1}{g(x + h)} - \frac{1}{g(x)}}{h} = -\frac{g(x + h) - g(x)}{h} \cdot \frac{1}{g(x + h) g(x)} \to -\frac{g'(x)}{g(x)^2}.

The first factor tends to $g'(x)$ , and the second tends to $\frac{1}{g(x)^2}$ because $g(x + h) \to g(x)$ .

In plain words When $g$ grows, $\frac{1}{g}$ shrinks, which explains the minus sign. Dividing by $g^2$ accounts for how sensitive $\frac{1}{g}$ is to changes in $g$ .

Example For $y = \frac{1}{x^2 + 1}$ the rule gives

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

valid for all $x$ since $x^2 + 1 > 0$ . At $x = 1$ this is $-\frac{2}{4} = -\frac{1}{2}$ . The rule also proves the power rule for negative integers. For a positive integer $n$ and $x \neq 0$ ,

\frac{d}{dx} \frac{1}{x^n} = -\frac{n x^{n - 1}}{x^{2n}} = -n x^{-n - 1}.

Quotient rule

Definition If $f$ and $g$ are differentiable at $x$ and $g(x) \neq 0$ , then $\frac{f}{g}$ is differentiable at $x$ and

\left(\frac{f}{g}\right)'(x) = \frac{f'(x) g(x) - f(x) g'(x)}{g(x)^2}.

In plain words Derivative of the top times the bottom, minus the top times the derivative of the bottom, all over the bottom squared. The order in the numerator matters, because subtraction is not symmetric.

Example For $y = \frac{x^2}{x + 1}$ with $x \neq -1$ ,

y' = \frac{2x(x + 1) - x^2 \cdot 1}{(x + 1)^2} = \frac{x^2 + 2x}{(x + 1)^2}.

At $x = 1$ the slope is $\frac{3}{4}$ .

Nonzero denominator

Definition The quotient rule needs $g(x) \neq 0$ at the point in question. Where $g(x) = 0$ the quotient $\frac{f}{g}$ is not defined at $x$ , so it has no derivative there, whatever the formula seems to give after simplifying. In the proof, $g(x) \neq 0$ together with continuity also guarantees $g(x + h) \neq 0$ for small $h$ , so every difference quotient is defined.

In plain words You can only ask for a slope where the function exists. Cancelling a factor does not create a value at a point that was never in the domain.

Example Consider

q(x) = \frac{x^2 - 1}{x - 1}.

For $x \neq 1$ it equals $x + 1$ , so $q'(x) = 1$ for $x \neq 1$ . But $q$ is undefined at $1$ , so $q'(1)$ does not exist. If you want a derivative there, you must first take the continuous extension $x + 1$ , which is a different function.

Rational functions

Definition A rational function is a quotient of polynomials

R(x) = \frac{P(x)}{Q(x)},

where $Q$ is not the zero polynomial, with domain $\{x \mid Q(x) \neq 0\}$ . By the quotient rule, $R$ is differentiable at every point of its domain, and

R'(x) = \frac{P'(x) Q(x) - P(x) Q'(x)}{Q(x)^2}

is again a rational function, defined on the same domain.

In plain words Rational functions are smooth everywhere they are defined, and their derivatives are rational functions too.

Example For $R(x) = \frac{3x - 1}{x^2 - 4}$ , defined for $x \neq \pm 2$ ,

R'(x) = \frac{3(x^2 - 4) - (3x - 1)(2x)}{(x^2 - 4)^2} = \frac{-3x^2 + 2x - 12}{(x^2 - 4)^2}.

At $x = 0$ this gives $R'(0) = \frac{-12}{16} = -\frac{3}{4}$ .

Derivation from the product rule

Definition Write $\frac{f}{g} = f \cdot \frac{1}{g}$ . Under the hypotheses of the quotient rule, both factors are differentiable at $x$ , the second by the reciprocal rule. The product rule gives

\left(\frac{f}{g}\right)' = f' \cdot \frac{1}{g} + f \cdot \left(-\frac{g'}{g^2}\right) = \frac{f' g - f g'}{g^2}.

A shortcut often seen is to set $q = \frac{f}{g}$ , differentiate $f = q g$ to get $f' = q' g + q g'$ , and solve for $q'$ . That recovers the same formula, but it quietly assumes $q$ is already known to be differentiable, which is exactly what the reciprocal rule supplies.

In plain words A quotient is a product with a reciprocal in it. Product rule plus reciprocal rule gives the quotient rule.

Example Take $f(x) = x^2 + 1$ and $g(x) = x$ for $x \neq 0$ . Splitting first, $\frac{x^2 + 1}{x} = x + x^{-1}$ , whose derivative is $1 - \frac{1}{x^2}$ . The quotient rule gives the same answer,

\frac{2x \cdot x - (x^2 + 1) \cdot 1}{x^2} = \frac{x^2 - 1}{x^2} = 1 - \frac{1}{x^2}.

Check yourself

Differentiate $y = \frac{2x + 3}{x - 1}$ and state where the result holds.

Show the answer

For every $x \neq 1$ the quotient rule gives

y' = \frac{2(x - 1) - (2x + 3)}{(x - 1)^2} = -\frac{5}{(x - 1)^2}.

Suppose $f(3) = 2$ , $f'(3) = 4$ , $g(3) = -1$ and $g'(3) = 5$ . Find $\left(\frac{f}{g}\right)'(3)$ .

Show the answer

Since $g(3) = -1 \neq 0$ , the quotient rule applies and gives

\left(\frac{f}{g}\right)'(3) = \frac{f'(3) g(3) - f(3) g'(3)}{g(3)^2} = \frac{4 \cdot (-1) - 2 \cdot 5}{1} = -14.

Find the tangent line to $y = \frac{4}{x^2 + 1}$ at $x = 1$ .

Show the answer

By the reciprocal and constant multiple rules $y' = -\frac{8x}{(x^2 + 1)^2}$ , so the slope at $1$ is $-2$ . The point is $(1, 2)$ , and the tangent line is $y = 2 - 2(x - 1)$ , or $y = -2x + 4$ .

The Quotient Rule

Theory, part 47

Practice, part 48

Key terms

Reciprocal rule

Quotient rule

Nonzero denominator

Rational functions

Derivation from the product rule

Check yourself

How this lesson connects

Builds on

Used later in