Proof of the Quotient Rule

A Direct Route to the Quotient Rule

The Quotient Rule was stated and derived in the previous tutorial by writing a quotient as a product with a reciprocal. Here we follow a different route: start with the difference quotient of the quotient itself and simplify it exactly. This makes the source of each term in the formula visible, and it provides a useful model for proofs in which an algebraic rearrangement exposes limits already known to exist.

Let $f$ and $g$ be differentiable at an interior point $a$, with $g(a)\ne0$. Differentiability implies continuity, so $g(x)$ stays nonzero for all $x$ sufficiently close to $a$. Thus the quotient is defined near $a$, as required to form its difference quotient. The key is to combine the two fractions before taking a limit.

Identity (Quotient Increment Identity): Suppose $g(a)\ne0$ and $g(a+h)\ne0$. Then $$ \frac{\dfrac{f(a+h)}{g(a+h)}-\dfrac{f(a)}{g(a)}}{h} = \frac{g(a)\dfrac{f(a+h)-f(a)}{h} -f(a)\dfrac{g(a+h)-g(a)}{h}} {g(a+h)g(a)} $$ whenever $h\ne0$.

Proof. Put the two fractions in the numerator over the common denominator $g(a+h)g(a)$. Their numerator becomes $f(a+h)g(a)-f(a)g(a+h)$. Add and subtract $f(a)g(a)$:

\begin{aligned} f(a+h)g(a)-f(a)g(a+h) &=f(a+h)g(a)-f(a)g(a)+f(a)g(a)-f(a)g(a+h)\\ &=g(a)\bigl(f(a+h)-f(a)\bigr) -f(a)\bigl(g(a+h)-g(a)\bigr). \end{aligned}

Divide this expression by $h\,g(a+h)g(a)$, which is permitted because $h\ne0$ and both denominator values are nonzero. Separating the two difference quotients gives the stated identity. $\square$

Proof from the Difference Quotient

The identity is exact; it is not an approximation. It separates the change in the numerator function from the change in the denominator function. The two difference quotients then have limits given by the derivatives of $f$ and $g$.

Theorem (Quotient Rule, Direct Proof): 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 differentiable at $a$, and $$ \left(\frac{f}{g}\right)'(a) =\frac{f'(a)g(a)-f(a)g'(a)}{(g(a))^2}. $$

Proof. Write $G=g(a)$, so $G\ne0$. Since $g$ is differentiable at $a$, it is continuous there. In particular, for $x$ sufficiently close to $a$, continuity gives $|g(x)-G|<|G|/2$. The triangle inequality yields

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

Thus $g(a+h)\ne0$ for all sufficiently small $h$, and the quotient increment identity applies. As $h\to0$, its two difference quotients satisfy

\frac{f(a+h)-f(a)}{h}\longrightarrow f'(a), \qquad \frac{g(a+h)-g(a)}{h}\longrightarrow g'(a).

Also, continuity of $g$ gives $g(a+h)\to G$. The denominator in the identity therefore tends to $G^2$, which is nonzero. Taking the limit on the right-hand side gives

\begin{aligned} \lim_{h\to0} \frac{g(a)\dfrac{f(a+h)-f(a)}{h} -f(a)\dfrac{g(a+h)-g(a)}{h}} {g(a+h)g(a)} &=\frac{Gf'(a)-f(a)g'(a)}{G^2}\\ &=\frac{f'(a)g(a)-f(a)g'(a)}{(g(a))^2}. \end{aligned}

The expression on the left before taking the limit is the difference quotient of $f/g$ at $a$. Since its limit exists and has the displayed value, $f/g$ is differentiable at $a$ with that derivative. This proves the formula directly. $\square$

This proof does not require $f$ and $g$ to be differentiable throughout a neighborhood of $a$. Their derivatives at $a$ provide the limits of the two difference quotients; continuity of $g$ at $a$ supplies the needed nonzero denominator nearby.

Reading the Formula from the Algebra

The numerator $f'(a)g(a)-f(a)g'(a)$ comes from the two terms in the exact identity. The change in $f$ contributes $g(a)f'(a)$, while the change in $g$ contributes $-f(a)g'(a)$. The minus sign is already present before any limit is taken. The denominator is $(g(a))^2$ because the common denominator $g(a+h)g(a)$ tends to $g(a)g(a)$.

The denominator condition is doing more than making the formula's final denominator nonzero. It ensures that the quotient exists at the point and that its difference quotient can be formed for all sufficiently small nonzero increments. If $g(a)=0$, this proof stops before the limit calculation: the original quotient may not even be defined at $a$.

Worked Example: A Quadratic Numerator

Let $q(x)=(2x^2-1)/(x+2)$, and find $q'(1)$. Set $f(x)=2x^2-1$ and $g(x)=x+2$. At $a=1$, the denominator is $g(1)=3$, so the quotient rule applies. The needed values are

f(1)=1,\qquad f'(1)=4,\qquad g(1)=3,\qquad g'(1)=1.

Substitution gives

q'(1)=\frac{(4)(3)-(1)(1)}{3^2} =\frac{12-1}{9} =\frac{11}{9}.

For a check, differentiating the quotient expression gives $q'(x)=[4x(x+2)-(2x^2-1)]/(x+2)^2$. At $x=1$, the numerator is $4(1)(3)-(2(1)^2-1)=12-1=11$, and the denominator is $3^2=9$. The result is again $11/9$.

Worked Example: A Negative Denominator Value

Consider $r(x)=(x^3+2)/(2x-1)$ at $a=0$. The denominator is $g(0)=-1$, which is nonzero; it need not be positive. For $f(x)=x^3+2$ and $g(x)=2x-1$,

f(0)=2,\qquad f'(0)=0,\qquad g(0)=-1,\qquad g'(0)=2.

The quotient rule gives

r'(0)=\frac{(0)(-1)-(2)(2)}{(-1)^2} =\frac{-4}{1} =-4.

The squared denominator is positive even though $g(0)$ is negative. Direct differentiation gives numerator $3x^2(2x-1)-2(x^3+2)$ over $(2x-1)^2$. At $x=0$, that numerator is $0-2(2)=-4$, and the denominator is $(-1)^2=1$, confirming the result.

Worked Example: The Numerator Vanishes at the Point

Let $s(x)=(x^2-4)/(x-1)$, and calculate $s'(2)$. Here $f(2)=0$, while $g(2)=1$, so the denominator condition holds. The values are

f(2)=0,\qquad f'(2)=4,\qquad g(2)=1,\qquad g'(2)=1.

Therefore,

s'(2)=\frac{(4)(1)-(0)(1)}{1^2}=4.

The zero value of $f(2)$ eliminates the term involving $g'(2)$, but not the term involving $f'(2)$. To check, for $x\ne1$ we can factor $x^2-4=(x-2)(x+2)$, but that does not cancel $x-1$. Instead, the direct derivative numerator is $2x(x-1)-(x^2-4)$. At $x=2$, this is $2(2)(1)-0=4$, and the denominator is $1$, as required.

Common Errors and Why the Proof Is Useful

A direct proof is especially helpful for checking the sign and the squared denominator. One common mistake is to write $f(a)g'(a)-f'(a)g(a)$, reversing the order of the two terms. The quotient increment identity shows why that order is fixed: the change in the denominator is subtracted. Another is to use $g(a)$, rather than $(g(a))^2$, below the fraction. The identity's denominator is a product of two values of $g$, and both tend to $g(a)$.

It is also important to distinguish a nonzero denominator at $a$ from a denominator that is nonzero everywhere on $I$. Only the first is assumed. Continuity ensures nonvanishing on some neighborhood of $a$, which is precisely what a derivative calculation at $a$ needs. Zeros farther away do not prevent this pointwise argument.

Finally, a formal rearrangement should preserve the conditions under which it is valid. In the quotient increment identity, division by $g(a+h)g(a)$ is allowed only when both factors are nonzero. The value $g(a)\ne0$ is given, and continuity supplies the same property for nearby $a+h$. The proof would be incomplete if it took the limit of a quotient without first ensuring that its denominator is nonzero near the point.

The broader technique is to begin with the defining difference quotient, combine terms exactly, and then identify familiar limits. Here the exact identity isolates the increments of $f$ and $g$, allowing differentiability at one point to finish the argument. This approach is a useful complement to the reciprocal-and-product proof: it shows how the rule follows directly from the definition of the derivative.

Check Your Understanding

Use the direct difference-quotient proof and the quotient rule to answer the following.

In the quotient increment identity, what algebraic step produces the negative sign in the term involving the increment of $g$?
Why does the denominator $g(a+h)g(a)$ tend to a nonzero limit under the hypotheses of the Quotient Rule?
If $f(a)=5$, $f'(a)=2$, $g(a)=-2$, and $g'(a)=3$, calculate $(f/g)'(a)$.
Does the quotient rule require $g$ to be nonzero on all of $I$? Explain what the proof actually needs.
Why is the quotient rule not applicable at $a=0$ to the expression $x/x$, even though it equals $1$ when $x\ne0$?

What You'll Learn