Tutorials › Calculus III › The Difference Quotient for a Partial Derivative

Home  /  Tutorials  /  Computing Partial Derivatives

Computing Partial Derivatives · Tutorial 301 of 999

The Difference Quotient for a Partial Derivative

Learn how fixing one input turns a partial derivative into an ordinary one-variable difference quotient, and how to calculate that limit directly.

Beginner 10 min read

What You'll Learn

  • Define the partial derivative with respect to one input using a two-sided limit.
  • Distinguish changing one coordinate from changing several coordinates at once.
  • Compute first partial derivatives directly from their defining limits.
  • Check the limit by simplifying the difference quotient before substituting zero.
  • Recognize when a partial derivative does not exist.

Why Hold One Input Fixed?

A function of two variables assigns an output to each input pair. To measure how the output changes as one input varies, we must keep the other input unchanged. This isolates the effect of a single coordinate and turns the calculation into a familiar one-variable difference quotient.

For example, when studying how \(f(x,y)\) changes with \(x\) near \((a,b)\), we compare \(f(a+h,b)\) with \(f(a,b)\). The first input has changed by \(h\); the second remains \(b\). Dividing the output change by \(h\) gives an average rate of change along that fixed-\(y\) slice. Taking the limit as \(h\) approaches zero gives the rate at the point, if that limit exists.

Definition by a Difference Quotient

Let \(f\) be a function defined on an open set \(D\) in \(\mathbb{R}^2\), and let \((a,b)\) be a point in \(D\). Because \(D\) is open, sufficiently small positive and negative changes in either coordinate remain in the domain. The partial derivative with respect to \(x\) is defined by changing \(x\) while holding \(y=b\) fixed.

Definition: If the following two-sided limit exists as a finite real number, it is the partial derivative of \(f\) with respect to \(x\) at \((a,b)\):
$$ f_x(a,b) = \lim_{h\to 0} \frac{f(a+h,b)-f(a,b)}{h}. $$
The partial derivative with respect to \(y\) is defined, when its finite limit exists, by
$$ f_y(a,b) = \lim_{h\to 0} \frac{f(a,b+h)-f(a,b)}{h}. $$

In the first quotient, only the \(x\)-coordinate changes. In the second, only the \(y\)-coordinate changes. The same increment symbol \(h\) is used in both definitions, but each limit is calculated separately. A partial derivative exists only if its own quotient has a finite two-sided limit; the existence of one partial derivative does not by itself guarantee the existence of the other.

The open-domain condition ensures that both signs of a sufficiently small increment are allowed. At a boundary point, a one-sided limit might be available, but that is not the two-sided partial derivative defined here unless the function is defined on an open neighborhood of the point.

Why This Is an Ordinary Derivative on a Slice

The definition is a direct application of the ordinary one-variable derivative. To see why, fix \(y=b\) and define a one-variable function \(g(t)=f(t,b)\) for values of \(t\) near \(a\). Its derivative at \(a\), by the ordinary difference-quotient definition, is

$$ g'(a) = \lim_{h\to 0}\frac{g(a+h)-g(a)}{h}. $$

Substituting \(g(a+h)=f(a+h,b)\) and \(g(a)=f(a,b)\) gives exactly the definition of \(f_x(a,b)\). Thus, whenever the limit exists, \(f_x(a,b)=g'(a)\). For the \(y\)-partial derivative, fix \(x=a\), define \(q(t)=f(a,t)\), and apply the same one-variable definition at \(t=b\). This explains why the method works: a partial derivative is an ordinary derivative of the function restricted to a coordinate-parallel slice.

Each quotient first measures a change over a nonzero increment. The limit then asks whether those average rates approach one finite value as the increment shrinks to zero. It is important to simplify the quotient before setting \(h\) to zero: the quotient itself is undefined at \(h=0\), even when its limit exists.

Computing Both Partials Directly

Worked Example: A Polynomial at a Point

Find both first partial derivatives of \(f(x,y)=x^2+xy+y^2\) at \((1,2)\), using the defining limits.

First calculate the function value:

$$ f(1,2)=1^2+(1)(2)+2^2=1+2+4=7. $$

For the \(x\)-partial, hold \(y=2\) fixed:

$$ \begin{aligned} f_x(1,2) &=\lim_{h\to0}\frac{f(1+h,2)-f(1,2)}{h}\\ &=\lim_{h\to0}\frac{(1+h)^2+(1+h)(2)+2^2-7}{h}\\ &=\lim_{h\to0}\frac{1+2h+h^2+2+2h+4-7}{h}\\ &=\lim_{h\to0}\frac{4h+h^2}{h}\\ &=\lim_{h\to0}(4+h)=4. \end{aligned} $$

For the \(y\)-partial, hold \(x=1\) fixed:

$$ \begin{aligned} f_y(1,2) &=\lim_{h\to0}\frac{f(1,2+h)-f(1,2)}{h}\\ &=\lim_{h\to0}\frac{1^2+(1)(2+h)+(2+h)^2-7}{h}\\ &=\lim_{h\to0}\frac{1+2+h+4+4h+h^2-7}{h}\\ &=\lim_{h\to0}\frac{5h+h^2}{h}\\ &=\lim_{h\to0}(5+h)=5. \end{aligned} $$

As a check, the simplified \(x\)-quotient \(4+h\) approaches \(4\), and the simplified \(y\)-quotient \(5+h\) approaches \(5\). The two rates differ because they describe changes along different coordinate directions.

Keeping the Other Input Fixed

The fixed input may appear as a constant in the calculation, but it still affects the output change. In the next example, the \(x\)-quotient is calculated with \(y=-1\) throughout, while the \(y\)-quotient uses \(x=2\) throughout.

Worked Example: A Product with a Linear Term

For \(f(x,y)=x^2y+3y\), find \(f_x(2,-1)\) and \(f_y(2,-1)\) directly from the limits.

The value at the point is

$$ f(2,-1)=2^2(-1)+3(-1)=-4-3=-7. $$

To find the \(x\)-partial, keep \(y=-1\):

$$ \begin{aligned} f_x(2,-1) &=\lim_{h\to0}\frac{f(2+h,-1)-f(2,-1)}{h}\\ &=\lim_{h\to0}\frac{(2+h)^2(-1)+3(-1)-(-7)}{h}\\ &=\lim_{h\to0}\frac{-(4+4h+h^2)-3+7}{h}\\ &=\lim_{h\to0}\frac{-4h-h^2}{h}\\ &=\lim_{h\to0}(-4-h)=-4. \end{aligned} $$

To find the \(y\)-partial, keep \(x=2\):

$$ \begin{aligned} f_y(2,-1) &=\lim_{h\to0}\frac{f(2,-1+h)-f(2,-1)}{h}\\ &=\lim_{h\to0}\frac{4(-1+h)+3(-1+h)-(-7)}{h}\\ &=\lim_{h\to0}\frac{-4+4h-3+3h+7}{h}\\ &=\lim_{h\to0}\frac{7h}{h}=7. \end{aligned} $$

Check the simplified quotients: \(-4-h\) approaches \(-4\), and \(7\) remains \(7\) for every nonzero \(h\). Thus the two partial derivatives at the point are \(-4\) and \(7\), respectively.

When the Limit Does Not Exist

A difference quotient need not approach a single finite number. The next example shows why a two-sided limit matters: the quotient approaches different values from the two sides, so the partial derivative in that coordinate does not exist.

Worked Example: A Partial Derivative That Fails to Exist

Let \(f(x,y)=|x|+2y\). Examine both partial derivatives at \((0,3)\). First,

$$ f(0,3)=|0|+2(3)=6. $$

For the \(x\)-partial, keep \(y=3\):

$$ \begin{aligned} \frac{f(0+h,3)-f(0,3)}{h} &=\frac{|h|+6-6}{h}\\ &=\frac{|h|}{h}. \end{aligned} $$

When \(h>0\), \(|h|/h=1\), so the right-hand limit is \(1\). When \(h<0\), \(|h|/h=-1\), so the left-hand limit is \(-1\). Since the one-sided limits differ, the two-sided limit does not exist. Therefore \(f_x(0,3)\) does not exist.

For the \(y\)-partial, keep \(x=0\):

$$ \begin{aligned} f_y(0,3) &=\lim_{h\to0}\frac{f(0,3+h)-f(0,3)}{h}\\ &=\lim_{h\to0}\frac{2(3+h)-6}{h}\\ &=\lim_{h\to0}\frac{2h}{h}=2. \end{aligned} $$

The check is explicit: the \(y\)-quotient equals \(2\) for every nonzero \(h\), while the \(x\)-quotient has unequal one-sided limits. One partial exists and the other does not.

Common Mistake: Changing Both Inputs

A partial derivative measures a change in one coordinate, not a change along an arbitrary path. If both inputs are changed in the quotient, the result generally measures a different rate and cannot be substituted for either partial derivative.

Common mistake: For \(f(x,y)=xy\) at \((1,2)\), changing both inputs gives
$$ \frac{f(1+h,2+h)-f(1,2)}{h} = \frac{(1+h)(2+h)-2}{h} = \frac{3h+h^2}{h} = 3+h, $$
which approaches \(3\). This is not the \(x\)-partial derivative, because the \(y\)-input changed too. To calculate the \(x\)-partial, keep \(y=2\):
$$ f_x(1,2) = \lim_{h\to0}\frac{f(1+h,2)-f(1,2)}{h} = \lim_{h\to0}\frac{2(1+h)-2}{h} = \lim_{h\to0}2=2. $$
The fixed-coordinate requirement is part of the definition, not an optional convenience.

A Reliable Procedure

When evaluating a partial derivative directly from its definition, write the fixed coordinate into the function arguments at every step. Then simplify the numerator, factor or cancel the increment where possible, and only then take the limit.

1
Choose the coordinate.
For an \(x\)-partial, change \(a\) to \(a+h\) and leave \(b\) unchanged; reverse these roles for a \(y\)-partial.
2
Write the full quotient.
Subtract the original function value from the perturbed value, then divide by \(h\).
3
Simplify before taking the limit.
Use algebra to remove any factor of \(h\), while remembering that \(h\ne0\) inside the quotient.
4
Check the limit from both sides.
A partial derivative exists only when both sides approach the same finite value.

This direct method is useful even when later techniques make some derivative calculations shorter. The limit definition clarifies exactly what is being measured, which input is fixed, and what must be true for the derivative to exist.

Check Your Understanding

For each question, begin from the difference-quotient definition rather than a differentiation shortcut.

  1. For \(f(x,y)=x^2+y\), write the difference quotient whose limit defines \(f_x(1,2)\). Which input is held fixed?
  2. For \(g(x,y)=xy\), compute \(g_y(2,3)\) directly from the limit.
  3. Why must the quotient be simplified before setting \(h=0\)?
  4. For \(p(x,y)=|y|\), does \(p_y(1,0)\) exist? Compare the limits for positive and negative \(h\).
  5. In the quotient for \(f_x(a,b)\), what changes if the second input is changed from \(b\) to \(b+h\) as well?

Previous

Next

Back to Tutorials