10.2: First-Order Partial Derivatives

Motivating Questions

• How are the first-order partial derivatives of a function \(f\) of the independent variables \(x\) and \(y\) defined?
• Given a function \(f\) of the independent variables \(x\) and \(y\text{,}\) what do the first-order partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) tell us about \(f\text{?}\)

The derivative plays a central role in first semester calculus because it provides important information about a function. Thinking graphically, for instance, the derivative at a point tells us the slope of the tangent line to the graph at that point. In addition, the derivative at a point also provides the instantaneous rate of change of the function with respect to changes in the independent variable.

Now that we are investigating functions of two or more variables, we can still ask how fast the function is changing, though we have to be careful about what we mean. Thinking graphically again, we can try to measure how steep the graph of the function is in a particular direction. Alternatively, we may want to know how fast a function's output changes in response to a change in one of the inputs. Over the next few sections, we will develop tools for addressing issues such as these. Preview Activity \(\PageIndex{1}\) explores some issues with what we will come to call partial derivatives.

Preview Activity \(\PageIndex{1}\)

Suppose we take out an $18,000 car loan at interest rate \(r\) and we agree to pay off the loan in \(t\) years. The monthly payment, in dollars, is

\[ M(r,t) = \frac{1500r}{1-\left(1+\frac{r}{12}\right)^{-12t}}. \nonumber \]

1. What is the monthly payment if the interest rate is \(3\%\) so that \(r = 0.03\text{,}\) and we pay the loan off in \(t=4\) years?
2. Suppose the interest rate is fixed at \(3\%\text{.}\) Express \(M\) as a function \(f\) of \(t\) alone using \(r=0.03\text{.}\) That is, let \(f(t) = M(0.03, t)\text{.}\) Sketch the graph of \(f\) on the left of Figure \(\PageIndex{1}\). Explain the meaning of the function \(f\text{.}\)

Figure \(\PageIndex{1}\). Left: Graphs \(f(t)= M(0.03, t)\text{.}\) Right: Graph \(g(r) = M(r,4)\text{.}\)

3. Find the instantaneous rate of change \(f'(4)\) and state the units on this quantity. What information does \(f'(4)\) tell us about our car loan? What information does \(f'(4)\) tell us about the graph you sketched in (b)?
4. Express \(M\) as a function of \(r\) alone, using a fixed time of \(t=4\text{.}\) That is, let \(g(r) = M(r, 4)\text{.}\) Sketch the graph of \(g\) on the right of Figure \(\PageIndex{1}\). Explain the meaning of the function \(g\text{.}\)
5. Find the instantaneous rate of change \(g'(0.03)\) and state the units on this quantity. What information does \(g'(0.03)\) tell us about our car loan? What information does \(g'(0.03)\) tell us about the graph you sketched in (d)?

First-Order Partial Derivatives

In Section 9.1, we studied the behavior of a function of two or more variables by considering the traces of the function. Recall that in one example, we considered the function \(f\) defined by

\[ f(x,y) = \frac{x^2 \sin(2 y)}{32}, \nonumber \]

which measures the range, or horizontal distance, in feet, traveled by a projectile launched with an initial speed of \(x\) feet per second at an angle \(y\) radians to the horizontal. The graph of this function is given again on the left in Figure \(\PageIndex{2}\). Moreover, if we fix the angle \(y = 0.6\text{,}\) we may view the trace \(f(x,0.6)\) as a function of \(x\) alone, as seen at right in Figure \(\PageIndex{2}\).

Figure \(\PageIndex{2}\). Left: The trace of \(z=\frac{x^2 \sin(2 y)}{32}\) with \(y = 0.6\text{.}\)

Since the trace is a one-variable function, we may consider its derivative just as we did in the first semester of calculus. With \(y=0.6\text{,}\) we have

\[ f(x,0.6) = \frac{\sin(1.2)}{32}x^2, \nonumber \]

and therefore

\[ \frac{d}{dx}[f(x,0.6)] = \frac{\sin(1.2)}{16}x. \nonumber \]

When \(x=150\text{,}\) this gives

\[ \frac{d}{dx}[f(x,0.6)]|_{x=150} = \frac{\sin(1.2)}{16}150 \approx 8.74~\mbox{feet per feet per second} , \nonumber \]

which gives the slope of the tangent line shown on the right of Figure \(\PageIndex{2}\). Thinking of this derivative as an instantaneous rate of change implies that if we increase the initial speed of the projectile by one foot per second, we expect the horizontal distance traveled to increase by approximately 8.74 feet if we hold the launch angle constant at \(0.6\) radians.

By holding \(y\) fixed and differentiating with respect to \(x\text{,}\) we obtain the first-order partial derivative of \(f\) with respect to \(x\). Denoting this partial derivative as \(f_x\text{,}\) we have seen that

\[ f_x(150, 0.6) = \frac{d}{dx}f(x,0.6)|_{x=150} = \lim_{h\to 0}\frac{f(150+h, 0.6) - f(150, 0.6)}{h}. \nonumber \]

More generally, we have

\[ f_x(a,b) = \lim_{h\to0} \frac{f(a+h, b)-f(a,b)}{h}, \nonumber \]

provided this limit exists.

In the same way, we may obtain a trace by setting, say, \(x=150\) as shown in Figure \(\PageIndex{3}\).

Figure \(\PageIndex{3}\). The trace of \(z=\frac{x^2 \sin(2 y)}{32}\) with \(x = 150\text{.}\)

This gives

\[ f(150, y) = \frac{150^2}{32}\sin(2y), \nonumber \]

and therefore

\[ \frac{d}{dy}[f(150,y)] = \frac{150^2}{16}\cos(2y). \nonumber \]

If we evaluate this quantity at \(y=0.6\text{,}\) we have

\[ \frac{d}{dy}[f(150,y)]|_{y=0.6} = \frac{150^2}{16}\cos(1.2) \approx 509.5 ~\mbox{feet per radian} . \nonumber \]

Once again, the derivative gives the slope of the tangent line shown on the right in Figure \(\PageIndex{3}\). Thinking of the derivative as an instantaneous rate of change, we expect that the range of the projectile increases by 509.5 feet for every radian we increase the launch angle \(y\) if we keep the initial speed of the projectile constant at 150 feet per second.

By holding \(x\) fixed and differentiating with respect to \(y\text{,}\) we obtain the first-order partial derivative of \(f\) with respect to \(y\). As before, we denote this partial derivative as \(f_y\) and write

\[ f_y(150, 0.6) = \frac{d}{dy}f(150,y)|_{y=0.6} = \lim_{h\to 0}\frac{f(150, 0.6+h) - f(150, 0.6)}{h}. \nonumber \]

As with the partial derivative with respect to \(x\text{,}\) we may express this quantity more generally at an arbitrary point \((a,b)\text{.}\) To recap, we have now arrived at the formal definition of the first-order partial derivatives of a function of two variables.

Definition \(\PageIndex{1}\)

The first-order partial derivatives of \(f\) with respect to \(x\) and \(y\) at a point \((a,b)\) are, respectively,

\[\begin{align*} f_x(a,b) & = \lim_{h \to 0} \frac{f(a+h,b)-f(a,b)}{h}, \ \mbox{and}\\[4pt] f_y(a,b) & = \lim_{h \to 0} \frac{f(a,b+h)-f(a,b)}{h}, \end{align*}\]

provided the limits exist.

Activity \(\PageIndex{2}\)

Consider the function \(f\) defined by

\[ f(x,y) = \frac{xy^2}{x+1} \nonumber \]

at the point \((1,2)\text{.}\)

1. Write the trace \(f(x,2)\) at the fixed value \(y=2\text{.}\) On the left side of Figure \(\PageIndex{4}\), draw the graph of the trace with \(y=2\) around the point where \(x=1\text{,}\) indicating the scale and labels on the axes. Also, sketch the tangent line at the point \(x=1\text{.}\)

Figure \(\PageIndex{4}\). Traces of \(f(x,y) = \frac{xy^2}{x+1}\text{.}\)

2. Find the partial derivative \(f_x(1,2)\) and relate its value to the sketch you just made.
3. Write the trace \(f(1,y)\) at the fixed value \(x=1\text{.}\) On the right side of Figure \(\PageIndex{5}\), draw the graph of the trace with \(x=1\) indicating the scale and labels on the axes. Also, sketch the tangent line at the point \(y=2\text{.}\)
4. Find the partial derivative \(f_y(1,2)\) and relate its value to the sketch you just made.

As these examples show, each partial derivative at a point arises as the derivative of a one-variable function defined by fixing one of the coordinates. In addition, we may consider each partial derivative as defining a new function of the point \((x,y)\text{,}\) just as the derivative \(f'(x)\) defines a new function of \(x\) in single-variable calculus. Due to the connection between one-variable derivatives and partial derivatives, we will often use Leibniz-style notation to denote partial derivatives by writing

\[ \frac{\partial f}{\partial x}(a, b) = f_x(a,b), \ \mbox{and} \ \frac{\partial f}{\partial y}(a, b) = f_y(a,b). \nonumber \]

To calculate the partial derivative \(f_x\text{,}\) we hold \(y\) fixed and thus we treat \(y\) as a constant. In Leibniz notation, observe that

\[ \frac{\partial }{\partial x} (x) = 1 \ \mbox{and} \ \frac{\partial }{\partial x}(y) = 0. \nonumber \]

To see the contrast between how we calculate single variable derivatives and partial derivatives, and the difference between the notations \(\frac{d}{dx}[ \ ]\) and \(\frac{\partial}{\partial x}[ \ ]\text{,}\) observe that

\[\begin{align*} & \frac{d}{dx}[3x^2 - 2x + 3] = 3\frac{d}{dx}[x^2] - 2\frac{d}{dx}[x] + \frac{d}{dx}[3] = 3\cdot 2x - 2,\\[4pt] \mbox{and} \ & \frac{\partial}{\partial x}[x^2y - xy + 2y] = y\frac{\partial}{\partial x}[x^2] - y\frac{\partial}{\partial x}[x] + \frac{\partial}{\partial x}[2y] = y\cdot 2x - y \end{align*}\]

Thus, computing partial derivatives is straightforward: we use the standard rules of single variable calculus, but do so while holding one (or more) of the variables constant.

Activity \(\PageIndex{3}\)

1. If \(f(x,y) = 3x^3 - 2x^2y^5\text{,}\) find the partial derivatives \(f_x\) and \(f_y\text{.}\)
2. If \(f(x,y) = \displaystyle\frac{xy^2}{x+1}\text{,}\) find the partial derivatives \(f_x\) and \(f_y\text{.}\)
3. If \(g(r,s) = rs\cos(r)\text{,}\) find the partial derivatives \(g_r\) and \(g_s\text{.}\)
4. Assuming \(f(w,x,y) = (6w+1)\cos(3x^2+4xy^3+y)\text{,}\) find the partial derivatives \(f_w\text{,}\) \(f_x\text{,}\) and \(f_y\text{.}\)
5. Find all possible first-order partial derivatives of \(q(x,t,z) = \displaystyle \frac{x2^tz^3}{1+x^2}.\)

Interpretations of First-Order Partial Derivatives

Recall that the derivative of a single variable function has a geometric interpretation as the slope of the line tangent to the graph at a given point. Similarly, we have seen that the partial derivatives measure the slope of a line tangent to a trace of a function of two variables as shown in Figure \(\PageIndex{5}\).

Figure \(\PageIndex{5}\). Tangent lines to two traces of the distance function.

Now we consider the first-order partial derivatives in context. Recall that the difference quotient \(\frac{f(a+h)-f(a)}{h}\) for a function \(f\) of a single variable \(x\) at a point where \(x=a\) tells us the average rate of change of \(f\) over the interval \([a,a+h]\text{,}\) while the derivative \(f'(a)\) tells us the instantaneous rate of change of \(f\) at \(x=a\text{.}\) We can use these same concepts to explain the meanings of the partial derivatives in context.

Activity \(\PageIndex{4}\)

The speed of sound \(C\) traveling through ocean water is a function of temperature, salinity and depth. It may be modeled by the function

\[ C=1449.2+4.6T-0.055T^2+0.00029T^3+(1.34-0.01T)(S-35)+0.016D. \nonumber \]

Here \(C\) is the speed of sound in meters/second, \(T\) is the temperature in degrees Celsius, \(S\) is the salinity in grams/liter of water, and \(D\) is the depth below the ocean surface in meters.

1. State the units in which each of the partial derivatives, \(C_T\text{,}\) \(C_S\) and \(C_D\text{,}\) are expressed and explain the physical meaning of each.
2. Find the partial derivatives \(C_T\text{,}\) \(C_S\) and \(C_D\text{.}\)
3. Evaluate each of the three partial derivatives at the point where \(T=10\text{,}\) \(S=35\) and \(D=100\text{.}\) What does the sign of each partial derivatives tell us about the behavior of the function \(C\) at the point \((10,35, 100)\text{?}\)

Using tables and contours to estimate partial derivatives

Remember that functions of two variables are often represented as either a table of data or a contour plot. In single variable calculus, we saw how we can use the difference quotient to approximate derivatives if, instead of an algebraic formula, we only know the value of the function at a few points. The same idea applies to partial derivatives.

Activity \(\PageIndex{5}\)

The wind chill, as frequently reported, is a measure of how cold it feels outside when the wind is blowing. In Table \(\PageIndex{1}\), the wind chill \(w\text{,}\) measured in degrees Fahrenheit, is a function of the wind speed \(v\text{,}\) measured in miles per hour, and the ambient air temperature \(T\text{,}\) also measured in degrees Fahrenheit. We thus view \(w\) as being of the form \(w = w(v, T)\text{.}\)

Table \(\PageIndex{1}\). Wind chill as a function of wind speed and temperature.

\(v \backslash T\)	\(-30\)	\(-25\)	\(-20\)	\(-15\)	\(-10\)	\(-5\)	\(0\)	\(5\)	\(10\)	\(15\)	\(20\)
\(5\)	\(-46\)	\(-40\)	\(-34\)	\(-28\)	\(-22\)	\(-16\)	\(-11\)	\(-5\)	\(1\)	\(7\)	\(13\)
\(10\)	\(-53\)	\(-47\)	\(-41\)	\(-35\)	\(-28\)	\(-22\)	\(-16\)	\(-10\)	\(-4\)	\(3\)	\(9\)
\(15\)	\(-58\)	\(-51\)	\(-45\)	\(-39\)	\(-32\)	\(-26\)	\(-19\)	\(-13\)	\(-7\)	\(0\)	\(6\)
\(20\)	\(-61\)	\(-55\)	\(-48\)	\(-42\)	\(-35\)	\(-29\)	\(-22\)	\(-15\)	\(-9\)	\(-2\)	\(4\)
\(25\)	\(-64\)	\(-58\)	\(-51\)	\(-44\)	\(-37\)	\(-31\)	\(-24\)	\(-17\)	\(-11\)	\(-4\)	\(3\)
\(30\)	\(-67\)	\(-60\)	\(-53\)	\(-46\)	\(-39\)	\(-33\)	\(-26\)	\(-19\)	\(-12\)	\(-5\)	\(1\)
\(35\)	\(-69\)	\(-62\)	\(-55\)	\(-48\)	\(-41\)	\(-34\)	\(-27\)	\(-21\)	\(-14\)	\(-7\)	\(0\)
\(40\)	\(-71\)	\(-64\)	\(-57\)	\(-50\)	\(-43\)	\(-36\)	\(-29\)	\(-22\)	\(-15\)	\(-8\)	\(-1\)

1. Estimate the partial derivative \(w_v(20,-10)\text{.}\) What are the units on this quantity and what does it mean? (Recall that we can estimate a partial derivative of a single variable function \(f\) using the symmetric difference quotient \(\frac{f(x+h)-f(x-h)}{2h}\) for small values of \(h\text{.}\) A partial derivative is a derivative of an appropriate trace.)
2. Estimate the partial derivative \(w_T(20,-10)\text{.}\) What are the units on this quantity and what does it mean?
3. Use your results to estimate the wind chill \(w(18, -10)\text{.}\) (Recall from single variable calculus that for a function \(f\) of \(x\text{,}\) \(f(x+h) \approx f(x) + hf'(x)\text{.}\))
4. Use your results to estimate the wind chill \(w(20, -12)\text{.}\)
5. Consider how you might combine your previous results to estimate the wind chill \(w(18, -12)\text{.}\) Explain your process.

Activity \(\PageIndex{6}\)

Shown below in Figure \(\PageIndex{6}\) is a contour plot of a function \(f\text{.}\) The values of the function on a few of the contours are indicated to the left of the figure.

Figure \(\PageIndex{6}\). A contour plot of \(f\text{.}\)

1. Estimate the partial derivative \(f_x(-2,-1)\text{.}\) (Hint: How can you find values of \(f\) that are of the form \(f(-2+h)\) and \(f(-2-h)\) so that you can use a symmetric difference quotient?)
2. Estimate the partial derivative \(f_y(-2,-1)\text{.}\)
3. Estimate the partial derivatives \(f_x(-1,2)\) and \(f_y(-1,2)\text{.}\)
4. Locate, if possible, one point \((x,y)\) where \(f_x(x,y)= 0\text{.}\)
5. Locate, if possible, one point \((x,y)\) where \(f_x(x,y)\lt 0\text{.}\)
6. Locate, if possible, one point \((x,y)\) where \(f_y(x,y)>0\text{.}\)
7. Suppose you have a different function \(g\text{,}\) and you know that \(g(2,2) = 4\text{,}\) \(g_x(2,2) > 0\text{,}\) and \(g_y(2,2) > 0\text{.}\) Using this information, sketch a possibility for the contour \(g(x,y)=4\) passing through \((2,2)\) on the left side of Figure \(\PageIndex{7}\). Then include possible contours \(g(x,y) = 3\) and \(g(x,y) = 5\text{.}\)

Figure \(\PageIndex{7}\). Plots for contours of \(g\) and \(h\text{.}\)

8. Suppose you have yet another function \(h\text{,}\) and you know that \(h(2,2) = 4\text{,}\) \(h_x(2,2) \lt 0\text{,}\) and \(h_y(2,2) > 0\text{.}\) Using this information, sketch a possible contour \(h(x,y)=4\) passing through \((2,2)\) on the right side of Figure \(\PageIndex{9}\). Then include possible contours \(h(x,y) = 3\) and \(h(x,y) = 5\text{.}\)

Summary

• If \(f=f(x,y)\) is a function of two variables, there are two first order partial derivatives of \(f\text{:}\) the partial derivative of \(f\) with respect to \(x\text{,}\)
  \[ \frac{\partial f}{\partial x}(x,y) = f_x(x,y) = \lim_{h \to 0} \frac{f(x+h,y) - f(x,y)}{h}, \nonumber \]

  and the partial derivative of \(f\) with respect to \(y\text{,}\)

  \[ \frac{\partial f}{\partial y}(x,y) = f_y(x,y) = \lim_{h \to 0} \frac{f(x,y+h) - f(x,y)}{h}, \nonumber \]

  where each partial derivative exists only at those points \((x,y)\) for which the limit exists.

• The partial derivative \(f_x(a,b)\) tells us the instantaneous rate of change of \(f\) with respect to \(x\) at \((x,y) = (a,b)\) when \(y\) is fixed at \(b\text{.}\) Geometrically, the partial derivative \(f_x(a,b)\) tells us the slope of the line tangent to the \(y=b\) trace of the function \(f\) at the point \((a,b,f(a,b))\text{.}\)
• The partial derivative \(f_y(a,b)\) tells us the instantaneous rate of change of \(f\) with respect to \(y\) at \((x,y) = (a,b)\) when \(x\) is fixed at \(a\text{.}\) Geometrically, the partial derivative \(f_y(a,b)\) tells us the slope of the line tangent to the \(x=a\) trace of the function \(f\) at the point \((a,b,f(a,b))\text{.}\)