Introduction
A function of two variables can describe an output that depends on two changing inputs. For example, temperature may depend on location, and production cost may depend on both output quantity and labor time.
A first partial derivative answers a focused question: how does the output change as one input varies, when the other input is held fixed? It is a rate in one coordinate direction, not automatically a description of what happens when both inputs change.
This interpretation gives partial derivatives practical meaning. Their values indicate the direction and rate of change, while their units specify how much output changes per unit of the chosen input.
Definition: Change Along One Coordinate
Let \(f(x,y)\) be a real-valued function, and let \((a,b)\) be an interior point of its domain. To study change in the \(x\)-direction, keep \(y=b\) and vary \(x\). To study change in the \(y\)-direction, keep \(x=a\) and vary \(y\).
These limits are ordinary one-variable derivatives of coordinate slices. For example, define \(g(x)=f(x,b)\). Then \(g'(a)\), when it exists, is exactly \(f_x(a,b)\), because \[ \frac{g(a+h)-g(a)}{h} = \frac{f(a+h,b)-f(a,b)}{h}. \] Likewise, the function \(k(y)=f(a,y)\) has derivative \(k'(b)=f_y(a,b)\), when that derivative exists. Thus each partial derivative is a familiar one-variable rate, taken along a chosen coordinate line.
Rates, Signs, and Units
If \(f\) measures an output and \(x\) measures an input, then \(f_x\) measures output change per unit of \(x\), with \(y\) held fixed. In the same way, \(f_y\) measures output change per unit of \(y\), with \(x\) held fixed. The units follow by dividing the output units by the corresponding input units.
A partial derivative is a rate at a point, not generally a promise that a finite change in the input will produce exactly that rate times the change. That equality does hold for an affine model along a coordinate line, as in the temperature and cost calculations below. For a nonlinear model, the derivative describes the limiting rate for increasingly small changes.
Worked Example: Temperature Change Across a Map
Worked Example: Temperature Change Across a Map
Suppose a local temperature model is \[ T(x,y)=20+0.6x-0.3y, \] where \(x\) is eastward distance in kilometers, \(y\) is northward distance in kilometers, and \(T\) is measured in degrees Celsius. Find the two coordinatewise rates at \((5,2)\), and interpret them.
First evaluate the temperature at the point: \[ T(5,2)=20+0.6(5)-0.3(2)=20+3-0.6=22.4. \]
For the \(x\)-rate, keep \(y=2\): \[ T(5+h,2)=20+0.6(5+h)-0.3(2)=22.4+0.6h. \] Therefore, \[ \frac{T(5+h,2)-T(5,2)}{h} =\frac{22.4+0.6h-22.4}{h} =\frac{0.6h}{h}=0.6 \] for \(h\ne 0\), and hence \(T_x(5,2)=0.6\).
For the \(y\)-rate, keep \(x=5\): \[ T(5,2+h)=20+0.6(5)-0.3(2+h)=22.4-0.3h. \] Thus, \[ \frac{T(5,2+h)-T(5,2)}{h} =\frac{22.4-0.3h-22.4}{h} =\frac{-0.3h}{h}=-0.3, \] so \(T_y(5,2)=-0.3\).
Both rates have units of degrees Celsius per kilometer. At this location, with northward position fixed, moving east increases the model temperature at \(0.6\) degrees Celsius per kilometer. With eastward position fixed, moving north decreases it at \(0.3\) degrees Celsius per kilometer.
Check: The model is affine, so these coordinatewise rates are constant. Increasing \(x\) by \(1\) km while keeping \(y=2\) changes the formula by \(0.6\) degrees; increasing \(y\) by \(1\) km while keeping \(x=5\) changes it by \(-0.3\) degrees. These agree with the computed partial derivatives.
Worked Example: Cost per Item and per Hour
Worked Example: Cost per Item and per Hour
Consider the cost model \[ C(q,h)=120+8q+3h+0.5qh, \] where \(q\) is the number of items produced, \(h\) is labor time in hours, and \(C\) is cost in dollars. Find the coordinatewise rates at \(q=10\), \(h=4\).
The cost at this operating point is \[ C(10,4)=120+8(10)+3(4)+0.5(10)(4) =120+80+12+20=232 \] dollars.
To find the rate with respect to \(q\), hold \(h=4\) fixed. For a change \(s\) in quantity, \[ C(10+s,4)=120+8(10+s)+3(4)+0.5(10+s)(4). \] Expanding each term gives \[ C(10+s,4)=120+80+8s+12+20+2s=232+10s. \] Consequently, \[ \frac{C(10+s,4)-C(10,4)}{s} =\frac{232+10s-232}{s}=10 \] for \(s\ne0\), so \(C_q(10,4)=10\) dollars per item.
Now hold \(q=10\) fixed and vary \(h\): \[ C(10,4+s)=120+8(10)+3(4+s)+0.5(10)(4+s). \] Expanding, \[ C(10,4+s)=120+80+12+3s+20+5s=232+8s. \] Thus, \[ \frac{C(10,4+s)-C(10,4)}{s} =\frac{232+8s-232}{s}=8 \] for \(s\ne0\), and \(C_h(10,4)=8\) dollars per hour.
Interpretation and check: At this operating point, the cost rate is \(10\) dollars per additional item when labor time stays at \(4\) hours, and \(8\) dollars per additional labor hour when production stays at \(10\) items. The fixed-variable substitutions show that each stated rate matches the change in the model along its corresponding coordinate line.
Why the Coordinatewise Interpretation Works
The reason a partial derivative measures change in one direction is built directly into its definition: its difference quotient compares two function values whose coordinates differ in only one place. The other coordinate is identical in the numerator’s two evaluations.
For \(f_x(a,b)\), compare points \((a+h,b)\) and \((a,b)\). For \(f_y(a,b)\), compare \((a,b+h)\) and \((a,b)\).
Divide the change in the output by the change \(h\) in the selected input. The other input remains fixed.
Let \(h\) approach zero. If the limit exists, it is the instantaneous rate along that coordinate line.
This also gives a short justification of the slice interpretation. For fixed \(b\), define \(g(x)=f(x,b)\). Substituting this definition into the one-variable difference quotient yields \[ g'(a)=\lim_{h\to0}\frac{g(a+h)-g(a)}{h} =\lim_{h\to0}\frac{f(a+h,b)-f(a,b)}{h} =f_x(a,b), \] whenever the limit exists. The same argument with \(k(y)=f(a,y)\) gives \(k'(b)=f_y(a,b)\). No input other than the chosen coordinate varies in either calculation.
Worked Example: Slope of a Height Model
Worked Example: Slope of a Height Model
Suppose a surface’s height above a reference level is modeled by \[ H(x,y)=50-0.02x^2-0.01y^2, \] where \(x\) and \(y\) are horizontal distances in meters and \(H\) is height in meters. Find the coordinatewise rates at \((10,20)\).
First evaluate the height: \[ H(10,20)=50-0.02(10^2)-0.01(20^2) =50-2-4=44 \] meters.
To compute the \(x\)-rate from the definition, hold \(y=20\) fixed: \[ H(10+s,20)-H(10,20) =-0.02\big((10+s)^2-10^2\big). \] Since \((10+s)^2=100+20s+s^2\), this becomes \[ H(10+s,20)-H(10,20)=-0.4s-0.02s^2. \] For \(s\ne0\), the difference quotient is \[ \frac{H(10+s,20)-H(10,20)}{s}=-0.4-0.02s. \] Taking the limit as \(s\to0\) gives \(H_x(10,20)=-0.4\) meters of height per meter of horizontal distance.
For the \(y\)-rate, hold \(x=10\) fixed: \[ H(10,20+s)-H(10,20) =-0.01\big((20+s)^2-20^2\big). \] Because \((20+s)^2=400+40s+s^2\), \[ H(10,20+s)-H(10,20)=-0.4s-0.01s^2. \] Therefore, \[ \frac{H(10,20+s)-H(10,20)}{s}=-0.4-0.01s \] for \(s\ne0\), and the limit as \(s\to0\) is \(H_y(10,20)=-0.4\) meters per meter.
Check: Differentiating the one-variable coordinate slices gives \(H(x,20)=46-0.02x^2\), whose derivative at \(x=10\) is \(-0.04(10)=-0.4\). The other slice is \(H(10,y)=48-0.01y^2\), whose derivative at \(y=20\) is \(-0.02(20)=-0.4\). Both checks agree with the limits.
A Pitfall: A Partial Rate Is Not a Combined Change
A partial derivative describes change when one coordinate varies and the other stays fixed. If both inputs change, neither partial derivative alone gives the resulting change. The two-variable temperature model makes this distinction concrete.
At \((5,2)\), the temperature is \(22.4\) degrees Celsius, and \(T_x(5,2)=0.6\) degrees Celsius per kilometer. Suppose someone claims that increasing \(x\) by \(1\) km while also increasing \(y\) by \(1\) km must raise the temperature by \(0.6\) degrees. That incorrectly applies the \(x\)-rate while ignoring that \(y\) also changed.
Evaluate the model at the actual new location: \[ T(6,3)=20+0.6(6)-0.3(3)=20+3.6-0.9=22.7. \] The change is \[ T(6,3)-T(5,2)=22.7-22.4=0.3 \] degrees Celsius, not \(0.6\). The \(x\)-change contributes \(0.6\) degrees in this affine model, while the \(y\)-change contributes \(-0.3\) degrees. Their combined effect is \(0.6-0.3=0.3\) degrees. The partial derivatives remain correct; the mistake is using a coordinatewise rate as if it described a change in multiple inputs.
Check Your Understanding
Use the definition and coordinatewise interpretation to answer the following questions.
- For a function \(f(x,y)\), which coordinate is held fixed when forming the difference quotient for \(f_y(a,b)\)?
- If \(P(x,y)\) is measured in dollars and \(x\) in kilograms, what are the units of \(P_x\)?
- For \(f(x,y)=7+2x-5y\), what do the signs of \(f_x\) and \(f_y\) say about the output along each coordinate direction?
- For \(H(x,y)=50-0.02x^2-0.01y^2\), why must \(y\) remain fixed when calculating \(H_x(10,20)\)?
- If both inputs change, why is it not generally valid to use just one partial derivative to state the output’s change?