Deriving the Partial Quotient Rule

Why a Quotient Rule Is Needed

A partial derivative measures change in one coordinate while the other coordinates are held fixed. If a function is written as a quotient, both its numerator and denominator may change with the chosen coordinate. The derivative must account for both changes.

The rule resembles the single-variable quotient rule, but the notation must make clear which coordinate is changing. We will derive it from the difference quotient, then apply it in examples where the denominator is constant with respect to the chosen variable.

Key idea: To find a partial derivative of a quotient with respect to \(x\), vary \(x\) and hold \(y\) fixed. To find the partial derivative with respect to \(y\), vary \(y\) and hold \(x\) fixed.

The Partial Quotient Rule

Let \(f\) and \(g\) be functions of \(x\) and \(y\), and define \(q(x,y)=f(x,y)/g(x,y)\) wherever \(g(x,y)\ne 0\). Suppose the relevant partial derivatives of \(f\) and \(g\) exist at a point \((a,b)\), and \(g(a,b)\ne 0\). Then the partial derivatives of the quotient at that point are given by the following rules.

Theorem: At a point where \(g\ne 0\), if the required partial derivatives exist, then \(\displaystyle \frac{\partial}{\partial x}\left(\frac{f}{g}\right)=\frac{g f_x-f g_x}{g^2}\) and \(\displaystyle \frac{\partial}{\partial y}\left(\frac{f}{g}\right)=\frac{g f_y-f g_y}{g^2}\). All functions and partial derivatives on the right are evaluated at the same point. The denominator is squared, and the order of the two numerator terms matters.

Here \(f_x\) means that \(f\) is differentiated with respect to \(x\), holding \(y\) fixed; \(g_x\) is interpreted the same way. For a \(y\)-partial, \(x\) is held fixed instead. The theorem requires a nonzero denominator at the point. Because \(g\) has the relevant partial derivative there, it is continuous along the corresponding coordinate line at that point. Thus, if \(g(a,b)\ne0\), it remains nonzero for sufficiently small changes along that line, so the difference quotient of \(f/g\) is defined near the point in the direction being considered.

Deriving the Rule from a Difference Quotient

We derive the \(x\)-partial formula at \((a,b)\). Write the difference quotient for \(q=f/g\), varying \(x\) by \(h\) while keeping \(y=b\):

\frac{q(a+h,b)-q(a,b)}{h} = \frac{1}{h}\left( \frac{f(a+h,b)}{g(a+h,b)} - \frac{f(a,b)}{g(a,b)} \right).

Combining the fractions gives a common denominator. Adding and subtracting terms in the numerator isolates the changes in \(f\) and \(g\):

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

Now divide each difference in the numerator by \(h\). As \(h\) approaches zero, the difference quotients for \(f\) and \(g\) approach \(f_x(a,b)\) and \(g_x(a,b)\). Also, \(g(a+h,b)\) approaches \(g(a,b)\). The denominator therefore approaches \(g(a,b)^2\), which is nonzero.

\begin{aligned} q_x(a,b) &= \frac{ g(a,b)f_x(a,b)-f(a,b)g_x(a,b)} {g(a,b)^2}. \end{aligned}

This proves the \(x\)-partial formula. For the \(y\)-partial, use the difference quotient that changes \(y\) to \(b+h\) while keeping \(x=a\). The same algebra gives \(g(a,b)f_y(a,b)-f(a,b)g_y(a,b)\) in the numerator and \(g(a,b)^2\) in the denominator.

Why the square appears: Subtracting two fractions introduces the product \(g(a+h,b)g(a,b)\) in the denominator. In the limit, both factors approach \(g(a,b)\), giving \(g(a,b)^2\).

Worked Examples

Worked Example: Differentiate with Respect to \(x\)

Find \(q_x\) for \(\displaystyle q(x,y)=\frac{x^2+2xy}{y+1}\), where \(y\ne-1\). Because this is an \(x\)-partial, treat \(y\) as fixed.

Set \(f(x,y)=x^2+2xy\) and \(g(x,y)=y+1\). Then

f_x=2x+2y,\qquad g_x=0,\qquad g^2=(y+1)^2.

Substitute these expressions into the partial quotient rule:

\begin{aligned} q_x &=\frac{g f_x-f g_x}{g^2}\\ &=\frac{(y+1)(2x+2y)-(x^2+2xy)(0)}{(y+1)^2}\\ &=\frac{2x+2y}{y+1}. \end{aligned}

The denominator condition is \(y+1\ne0\), or \(y\ne-1\). Check the result using \(gq_x+qg_x=f_x\): the left side is \((y+1)\frac{2x+2y}{y+1}+\frac{x^2+2xy}{y+1}(0)=2x+2y\), equal to \(f_x\).

Worked Example: Differentiate with Respect to \(y\)

Find \(q_y\) for \(\displaystyle q(x,y)=\frac{3x+y^2+1}{x+3}\), where \(x\ne-3\). For this partial derivative, hold \(x\) fixed.

Let \(f(x,y)=3x+y^2+1\) and \(g(x,y)=x+3\). The needed quantities are

f_y=2y,\qquad g_y=0,\qquad g^2=(x+3)^2.

Applying the rule gives

\begin{aligned} q_y &=\frac{g f_y-f g_y}{g^2}\\ &=\frac{(x+3)(2y)-(3x+y^2+1)(0)}{(x+3)^2}\\ &=\frac{2y}{x+3}. \end{aligned}

The quotient is defined when \(x\ne-3\). To check, multiply the result by \(g\) and use \(g_y=0\): \((x+3)\frac{2y}{x+3}+\frac{3x+y^2+1}{x+3}(0)=2y=f_y\).

Worked Example: Keep the Chosen Variable Explicit

Find \(q_x\) for \(\displaystyle q(x,y)=\frac{x^2y+4y}{2y-1}\), where \(y\ne\frac12\). Here \(y\) is fixed, so the denominator has zero \(x\)-partial.

Choose \(f(x,y)=x^2y+4y\) and \(g(x,y)=2y-1\). Differentiating only with respect to \(x\) gives

f_x=2xy,\qquad g_x=0,\qquad g^2=(2y-1)^2.

Therefore

\begin{aligned} q_x &=\frac{g f_x-f g_x}{g^2}\\ &=\frac{(2y-1)(2xy)-(x^2y+4y)(0)}{(2y-1)^2}\\ &=\frac{2xy}{2y-1}. \end{aligned}

The condition is \(2y-1\ne0\), so \(y\ne\frac12\). The check is \((2y-1)\frac{2xy}{2y-1}+\frac{x^2y+4y}{2y-1}(0)=2xy=f_x\), as required.

A Product-Rule View

The difference-quotient derivation shows why the formula is valid without assuming in advance that the quotient is differentiable. There is also a useful way to remember its structure. Since \(f=qg\), differentiating this identity with respect to \(x\) using the partial product rule gives \(f_x=q_xg+qg_x\). Solving for \(q_x\) yields \(q_x=(g f_x-fg_x)/g^2\), because \(q=f/g\). This is a memory aid once the quotient is known to have a partial derivative; the difference-quotient argument supplies the direct justification.

Common Mistake: Forgetting the Denominator’s Derivative

A frequent error is to differentiate only the numerator and leave the denominator unchanged. For example, for \(q(x,y)=(x^2+1)/(y+2)\), an incorrect calculation of \(q_x\) might be

\text{Incorrect:}\qquad q_x=\frac{2x}{y+2}.

This answer happens to be correct here because \(g_x=0\). But that shortcut is not the quotient rule; it works only when the denominator is constant with respect to the selected variable. The general rule includes the term \(-f g_x\). Omitting it when \(g_x\ne0\) loses the effect of the changing denominator. The rule also requires \(g\ne0\); substituting a point where the denominator vanishes is not allowed.

Practical check: Before differentiating a quotient, identify the chosen variable and compute both \(f\)'s and \(g\)'s partial derivatives with respect to that variable. A zero derivative of \(g\) must come from holding the other variable fixed, not from assuming every denominator is constant.

When the Rule Matters

The quotient rule is a general tool for partial differentiation, including situations in which both numerator and denominator depend on the chosen coordinate. The essential discipline is to apply the same partial derivative consistently to every term. A denominator may have zero partial derivative if it depends only on the coordinate being held fixed, but it need not be zero in other situations.

Before using the rule, check that the denominator is nonzero at the point of interest and that the required partial derivatives exist. Then keep the order \(g f_x-f g_x\) or \(g f_y-f g_y\) intact, and square the entire denominator. These checks prevent the most common sign, variable, and domain errors.

Check Your Understanding

Answer each question by identifying the differentiated variable and stating any denominator restriction.

For \(q=f/g\), what is the formula for \(q_y\), and which variable is held fixed?
Why does the difference-quotient derivation produce \(g^2\) in the denominator?
In \(\displaystyle q(x,y)=\frac{x+y}{y-4}\), which partial derivative treats the denominator as constant, and why?
What condition must hold at a point before the quotient rule can be applied there?
Explain why the difference-quotient derivation is a direct justification of the rule, while differentiating \(f=qg\) is primarily a way to remember its form.