Quotient Rule

When a Denominator Varies

The Product Rule separates the changes in two varying factors. A quotient presents a related challenge: both numerator and denominator can change, and a changing denominator affects the value of the entire expression. The quotient rule gives the resulting derivative in terms of the values and derivatives of the numerator and denominator at the point.

The denominator condition is essential. If $g(a)\ne0$, continuity of $g$ at $a$ ensures that $g$ remains nonzero sufficiently close to $a$. Thus $f/g$ is defined on a neighborhood of $a$. We will first find the derivative of the reciprocal $1/g$ at $a$, then combine it with the Product Rule. Throughout, differentiability is assumed at the point $a$, not throughout a neighborhood.

Theorem (Quotient Rule): Let $I$ be an interval, let $a$ be an interior point of $I$, and let $f,g:I\to\mathbb{R}$ be differentiable at $a$. If $g(a)\ne0$, then $f/g$ is defined on a neighborhood of $a$, is differentiable at $a$, and $$ \left(\frac{f}{g}\right)'(a) =\frac{f'(a)g(a)-f(a)g'(a)}{(g(a))^2}. $$

The Derivative of a Reciprocal

The reciprocal rule is the key step. The fact that $g(a)\ne0$ gives more than a defined value at $a$: differentiability of $g$ implies continuity there, so values of $g$ close to $a$ remain separated from zero. The reciprocal is therefore defined near $a$, and its difference quotient can be computed exactly.

Theorem (Derivative of a Reciprocal): Let $I$ be an interval, let $a$ be an interior point of $I$, and let $g:I\to\mathbb{R}$ be differentiable at $a$. If $g(a)\ne0$, then $1/g$ is defined on a neighborhood of $a$ and is differentiable at $a$, with $$ \left(\frac{1}{g}\right)'(a)=-\frac{g'(a)}{(g(a))^2}. $$

Proof. Differentiability of $g$ at $a$ implies continuity at $a$, by the theorem that differentiability implies continuity. Set $c=|g(a)|$, which is positive. By continuity, there is a $\delta_1>0$ such that, for $x\in I$ with $|x-a|<\delta_1$,

|g(x)-g(a)|<\frac{c}{2}.

The triangle inequality then gives

|g(x)|\ge |g(a)|-|g(x)-g(a)|>\frac{c}{2}>0.

Consequently, $1/g(x)$ is defined for every such $x$. Because $a$ is an interior point of $I$, we can also restrict to a sufficiently small neighborhood contained in $I$. For nonzero $h$ small enough that $a+h\in I$ and $|h|<\delta_1$, algebra gives

\frac{\dfrac{1}{g(a+h)}-\dfrac{1}{g(a)}}{h} = -\frac{g(a+h)-g(a)}{h\,g(a+h)g(a)}.

The difference quotient $(g(a+h)-g(a))/h$ tends to $g'(a)$. Continuity gives $g(a+h)\to g(a)$, and both $g(a)$ and the nearby values $g(a+h)$ are nonzero. Therefore the right-hand side tends to

-\frac{g'(a)}{g(a)g(a)} =-\frac{g'(a)}{(g(a))^2}.

This is the derivative of $1/g$ at $a$, as claimed. $\square$

This is a pointwise differentiability conclusion: the hypotheses guarantee that $1/g$ is defined near $a$ and differentiable at $a$. They do not say that $g$, or $1/g$, is differentiable at every point of that neighborhood.

Deriving the Quotient Rule

For $x$ near $a$, the quotient can be written as a product: $(f/g)(x)=f(x)(1/g(x))$. The Product Rule applies at $a$, since $f$ and $1/g$ are both differentiable there. Using the reciprocal derivative just proved gives

\begin{aligned} \left(\frac{f}{g}\right)'(a) &=f'(a)\frac{1}{g(a)} +f(a)\left(-\frac{g'(a)}{(g(a))^2}\right)\\ &=\frac{f'(a)g(a)-f(a)g'(a)}{(g(a))^2}. \end{aligned}

Proof of the Quotient Rule. The reciprocal theorem shows that $1/g$ is defined on a neighborhood of $a$ and differentiable at $a$. The Product Rule, applied at $a$ to $f$ and $1/g$, shows that $f(1/g)=f/g$ is differentiable at $a$. Substituting the reciprocal derivative into the Product Rule gives the displayed formula. This proves the Quotient Rule. $\square$

The negative sign in the numerator comes from the reciprocal derivative. The denominator is squared because differentiating $1/g$ produces $1/(g(a))^2$. Both features matter: changing the sign or using just $g(a)$ in the denominator generally gives an incorrect result.

Worked Applications

Worked Example: A Quotient with Two Changing Factors

Let $q(x)=(x^2+4)/(3x-1)$, and find $q'(1)$. The denominator at $1$ is $3(1)-1=2$, so it is nonzero. Set $f(x)=x^2+4$ and $g(x)=3x-1$. Then

f(1)=5,\qquad f'(1)=2,\qquad g(1)=2,\qquad g'(1)=3.

The Quotient Rule gives

q'(1)=\frac{(2)(2)-(5)(3)}{2^2} =\frac{4-15}{4} =-\frac{11}{4}.

To check the calculation, differentiate the quotient algebraically:

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

At $x=1$, the numerator is $2(1)(2)-3(5)=4-15=-11$, and the denominator is $2^2=4$. This gives $-11/4$, in agreement with the rule.

Worked Example: A Zero Numerator at the Point

Consider $r(x)=(x^2-9)/(x+3)$ at $a=3$. The denominator is $3+3=6$, so the quotient is defined near $3$. Let $f(x)=x^2-9$ and $g(x)=x+3$. Their values and derivatives at $3$ are

f(3)=0,\qquad f'(3)=6,\qquad g(3)=6,\qquad g'(3)=1.

Consequently,

r'(3)=\frac{(6)(6)-(0)(1)}{6^2} =\frac{36}{36} =1.

Here the zero numerator removes the term involving $g'(3)$, but it does not make the derivative zero. Indeed, for $x\ne-3$, $(x^2-9)/(x+3)=x-3$. In a neighborhood of $3$, this identity shows directly that $r(x)=x-3$, whose derivative is $1$. The cancellation is valid near $3$, where $x+3\ne0$; it does not define the original quotient at $x=-3$.

Worked Example: A Negative Derivative from the Denominator

Let $s(x)=(2x+1)/(x^2+1)$, and calculate $s'(-1)$. The denominator is $(-1)^2+1=2$, which is nonzero. For $f(x)=2x+1$ and $g(x)=x^2+1$, we have

f(-1)=-1,\qquad f'(-1)=2,\qquad g(-1)=2,\qquad g'(-1)=-2.

Thus the Quotient Rule yields

s'(-1)=\frac{(2)(2)-(-1)(-2)}{2^2} =\frac{4-2}{4} =\frac{1}{2}.

For an independent check, the numerator of the derivative at a general $x$ is $2(x^2+1)-(2x+1)(2x)$. At $x=-1$, this is $2(2)-(-1)(-2)=4-2=2$, and the squared denominator is $4$. The derivative is therefore $2/4=1/2$.

What the Hypotheses Tell Us

The condition $g(a)\ne0$ has two roles. It makes the quotient meaningful at the point, and, together with continuity of $g$, guarantees that the quotient is defined on some neighborhood of the point. A derivative is determined by values close to $a$, so having only a value at $a$ is not enough: the quotient must be defined for nearby inputs as well.

The theorem does not require $g(x)\ne0$ for every $x\in I$. It requires only $g(a)\ne0$, which ensures a neighborhood of $a$ on which the denominator is nonzero. Nor does it require $f$ and $g$ to be differentiable throughout that neighborhood. Their differentiability at $a$, along with the nonzero denominator condition, is sufficient for the conclusion at $a$.

When $f(a)=0$, the formula reduces to

\left(\frac{f}{g}\right)'(a)=\frac{f'(a)}{g(a)}.

This follows by substituting $f(a)=0$ into the Quotient Rule, not by discarding a term without checking its value. Conversely, if $g(a)=0$, the rule cannot be applied at $a$. For instance, $x/x$ is defined away from zero and equals $1$ there, but the quotient $x/x$ itself is not defined at zero. One cannot use the Quotient Rule at zero because its denominator hypothesis fails, even though the simplified expression $1$ has a derivative there. A simplification may define a new extension, but that is a separate question from differentiating the original quotient at a point where it is undefined.

For reliable calculations, identify the numerator and denominator first, check the denominator value at the point, and then substitute into the formula in the order shown. In particular, keep the entire expression $f'(a)g(a)-f(a)g'(a)$ in the numerator and square $g(a)$ only after evaluating it.

Check Your Understanding

Use the hypotheses and formula of the Quotient Rule to answer the following.

Why does $g(a)\ne0$, together with differentiability of $g$ at $a$, ensure that $1/g$ is defined on a neighborhood of $a$?
If $f(a)=4$, $f'(a)=3$, $g(a)=2$, and $g'(a)=1$, what is $(f/g)'(a)$?
State the formula for $(f/g)'(a)$ when $f(a)=0$ and $g(a)\ne0$.
Does the Quotient Rule require $f$ and $g$ to be differentiable at every point near $a$? Explain.
Why can the Quotient Rule not be applied to $x/x$ at $a=0$, even though the quotient equals $1$ whenever $x\ne0$?

What You'll Learn