9.4 Partial Derivatives
INTRODUCTION
In this section we consider functions of two or more variables and how to find the instantaneous rate of change—that is, the derivative—of such functions with respect to each variable.
Functions of Two Variables
Recall from calculus that a function of two variables is a rule of correspondence that assigns to each ordered pair of real numbers (x, y) of a subset of the xy-plane one and only one number z in the set R of real numbers. The set of ordered pairs (x, y) is called the domain of the function and the set of corresponding values of z is called the range. A function of two variables is usually written z = f(x, y). The variables x and y are called the independent variables of the function, and z is called the dependent variable. The graph of a function z = f(x, y) is a surface in 3-space. See FIGURE 9.4.1.
Level Curves
For a function z = f(x, y), the curves defined by f(x, y) = c, for suitable c, are called the level curves of f. The word level arises from the fact that we can interpret the equation f(x, y) = c as the projection onto the xy-plane of the curve of intersection, or trace, of z = f(x, y) and the (horizontal or level) plane z = c. See FIGURE 9.4.2.
EXAMPLE 1 Level Curves
The level curves of the function f(x, y) = y2 – x2 are defined by y2 – x2 = c. As shown in FIGURE 9.4.3, when c > 0 or c < 0, a member of this family of curves is a hyperbola. For c = 0, we obtain the lines y = x and y = –x. ≡
Functions of Three or More Variables
Functions of three or more variables are defined analogously to functions of two variables. For example, a function of three variables is a rule of correspondence that assigns to each ordered triple of real numbers (x, y, z) of a subset of 3-space one and only one number w in the set R of real numbers. We write w = F(x, y, z).
Level Surfaces
Although we cannot draw a graph of a function of three variables w = F(x, y, z), we can draw the surfaces defined by F(x, y, z) = c for suitable values of c. These surfaces are called level surfaces. This is an unfortunate, though standard, choice of words, since level surfaces are usually not level.
EXAMPLE 2 Level Surfaces
Describe the level surfaces of the function F(x, y, z) = (x2 + y2)/z.
SOLUTION
For c ≠ 0 the level surfaces are given by
= c or x2 + y2 = cz.
A few members of this family of paraboloids are shown in FIGURE 9.4.4. ≡
Partial Derivatives
The derivative of a function of one variable y = f(x) is given by
In exactly the same manner, we can define a derivative of a function of two variables with respect to each variable. If z = f(x, y), then the partial derivative with respect to x is
(1)
and the partial derivative with respect to y is
(2)
provided each limit exists.
In (1) the variable y does not change in the limiting process; that is, y is held fixed. Similarly, in (2) the variable x is held fixed. The two partial derivatives (1) and (2) then represent the rates of change of f with respect to x and y, respectively. On a practical level:
To compute ∂z/∂x, use the laws of ordinary differentiation while treating y as a constant.
To compute ∂z/∂y, use the laws of ordinary differentiation while treating x as a constant.
EXAMPLE 3 Partial Derivatives
If z = 4x3y2 – 4x2 + y6 + 1, find ∂z/ ∂x and ∂z/ ∂y.
SOLUTION
We hold y fixed and treat constants in the usual manner. Thus,
= 12x2y2 – 8x.
By treating x as a constant, we obtain
= 8x3y + 6y5. ≡
Alternative Symbols
The partial derivatives ∂z/∂x and ∂z/∂y are often represented by alternative symbols. If z = f(x, y), then
.
Higher-Order and Mixed Derivatives
For a function of two variables z = f(x, y), the partial derivatives ∂z/∂x and ∂z/∂y are themselves functions of x and y. Consequently, we can compute second and higher partial derivatives. Indeed, we can find the partial derivative of ∂z/∂x with respect to y, and the partial derivative of ∂z/∂y with respect to x. The latter types of partial derivatives are called mixed partial derivatives. In summary, for z = f(x, y):
Second-order partial derivatives:
Third-order partial derivatives:
Mixed second-order partial derivatives:
Alternative Symbols
The second- and third-order partial derivatives are denoted by fxx, fyy, fxxx, and so on. The subscript notation for mixed second partial derivatives is fxy or fyx. Note that
Although we shall not prove it, if a function f has continuous second partial derivatives, then the order in which a mixed second partial derivative is done is irrelevant; that is,
fxy = fyx. (3)
Functions of Three or More Variables
The rates of change of a function of three variables w = F(x, y, z) in the x-, y-, and z-directions are ∂w/∂x, ∂w/∂y, and ∂w/∂z, respectively. To compute, say, ∂w/∂x, we differentiate with respect to x in the usual manner while holding both y and z constant. In this manner we extend the process of partial differentiation to functions of any number of variables.
EXAMPLE 4 Partial Derivatives
If F(x, y, t) = e–3πt cos 4x sin 6y, then the partial derivatives with respect to x, y, and t are, in turn,
Fx(x, y, t) = –4e−3πt sin 4x sin 6y,
Fy(x, y, t) = 6e−3πt cos 4x cos 6y,
Ft(x, y, t) = –3πe−3πt cos 4x sin 6y. ≡
Chain Rule
The Chain Rule for functions of one variable states that if y = f(u) is a differentiable function of u, and u = g(x) is a differentiable function of x, then the derivative of the composite function is
(4)
For a composite function of two variables z = f(u, v), where u = g(x, y) and v = h(x, y), we would naturally expect two formulas analogous to (4), since we can compute both ∂z/∂x and ∂z/∂y. The Chain Rule for functions of two variables is summarized as follows:
THEOREM 9.4.1 Chain Rule
If z = f(u, v) is differentiable and u = g(x, y) and v = h(x, y) have continuous first partial derivatives, then
(5)
EXAMPLE 5 Chain Rule
If z = u2 – v3 and u = e2x−3y, v = sin(x2 – y2), find ∂z/∂x and ∂z/∂y.
SOLUTION
Since ∂z/∂u = 2u and ∂z/∂v = –3v2, it follows from (5) that
(6)
(7) ≡
Of course, in Example 5 we could substitute the expressions for u and v in the original function and then find the partial derivatives directly. In the same manner, the answers (6) and (7) can be expressed in terms of x and y.
Special Case
If z = f(u, v) is differentiable and u = g(t) and v = h(t) are differentiable functions of a single variable t, then Theorem 9.4.1 implies that the ordinary derivative dz/dt is
(8)
Generalizations
The results given in (5) and (8) immediately generalize to any number of variables. If z = f(u1, u2, ..., un) and each of the variables u1, u2, u3, ..., un are functions of x1, x2, ..., xk, then under the same assumptions as in Theorem 9.4.1 we have
(9)
where i = 1, 2, ..., k. Similarly, if the ui, i = 1, ..., n, are differentiable functions of a single variable t, then
(10)
Tree Diagrams
The results in (5) can be memorized in terms of a tree diagram. The dots in the first diagram in FIGURE 9.4.5(a) indicate the fact that z depends on u and v; u and v depend, in turn, on x and y. To compute ∂z/∂x for example, we read the diagram vertically downward starting from z and following the two blue polygonal paths leading to x, multiply the partial derivatives on each path, and then add the products. The result given in (8) is represented by the second tree diagram in Figure 9.4.5(b).
We shall use tree diagrams in the next two examples to illustrate special cases of (9) and (10).
EXAMPLE 6 Using Tree Diagrams
If r = x2 + y5z3 and x = uve2s, y = u2 – v2s, z = sin(uvs2), find ∂r/∂s.
SOLUTION
From the blue and green paths in the tree diagram in Figure 9.4.5(c) we obtain
≡
EXAMPLE 7 Using Tree Diagrams
If z = u2v3w4 and u = t2, v = 5t – 8, w = t3 + t, find dz/dt.
SOLUTION
In this case the tree diagram in Figure 9.4.5(d) indicates that
ALTERNATIVE SOLUTION
Differentiate z = t4(5t – 8)3(t3 + t)4 by the Product Rule. ≡
REMARKS
If w = F(x, y, z) has continuous partial derivatives of any order, then analogous to (3), the mixed partial derivatives are equal:
Fxyz = Fyzx = Fzyx, Fxxy = Fyxx = Fxyx,
and so on.
9.4 Exercises Answers to selected odd-numbered problems begin on page ANS-23.
In Problems 1–6, sketch some of the level curves associated with the given function.
- f(x, y) = x + 2y
- f(x, y) = y2 – x
- f(x, y) =
- f(x, y) =
- f(x, y) =
- f(x, y) = tan−1(y – x)
In Problems 7–10, describe the level surfaces but do not graph.
- F(x, y, z) =
- F(x, y, z) = x2 + y2 + z2
- F(x, y, z) = x2 + 3y2 + 6z2
- F(x, y, z) = 4y – 2z + 1
- Graph some of the level surfaces associated with F(x, y, z) = x2 + y2 – z2 for c = 0, c > 0, and c < 0.
- Given that
F(x, y, z) = ,
find the x-, y-, and z-intercepts of the level surface that passes through (–4, 2, –3).
In Problems 13–32, find the first partial derivatives of the given function.
- z = x2 – xy2 + 4y5
- z = –x3 + 6x2y3 + 5y2
- z = 5x4y3 – x2y6 + 6x5 – 4y
- z = tan(x3y2)
- z =
- z = 4x3 – 5x2 + 8x
- z = (x3 – y2)−1
- z = (–x4 + 7y2 + 3y)6
- z = cos2 5x + sin2 5y
- z =
- f(θ, ϕ) = ϕ2 sin
- f(x, y) =
- g(u, v) = ln(4u2 + 5v3)
- h(r, s) =
- w = xy ln(xz)
- F(u, v, x, t) = u2w2 – uv3 + vw cos(ut2) + (2x2t)4
- G(p, q, r, s) =
In Problems 33 and 34, verify that the given function satisfies Laplace’s equation:
.
- z = ln(x2 + y2)
- z =
In Problems 35 and 36 verify that the given function satisfies the wave equation:
.
- u = cos at sin x
- u = cos(x + at) + sin(x – at)
- The molecular concentration C(x, t) of a liquid is given by C(x, t) = t−1/2 e−x2/kt. Verify that this function satisfies the diffusion equation:
.
- The pressure P exerted by an enclosed ideal gas is given by P = k(T/V), where k is a constant, T is temperature, and V is volume. Find:
(a) the rate of change of P with respect to V,
(b) the rate of change of V with respect to T, and
(c) the rate of change of T with respect to P.
In Problems 39–48, use the Chain Rule to find the indicated partial derivatives.
- z = ; u = x3, v = x – y2;
- z = u2 cos 4v; u = x2y3, v = x3 + y3;
- z = 4x – 5y2; x = u4 – 8v3, y = (2u – v)2;
- w = (u2 + v2)3/2; u = e−t sin θ, v = e−t cos θ;
- R = rs2t4; r = , s = , t = ;
- Q = ln(pqr); p = t2 sin−1 x, q = , r = tan−1 ;
In Problems 49–52, use (8) to find the indicated derivative.
- z = ln(u2 + v2); u = t2, v = t−2;
- z = u3v – uv4; u = e−5t, v = sec 5t;
- w = cos(3u + 4v); u = 2t +
- If u = f(x, y) and x = r cos θ, y = r sin θ, show that Laplace’s equation ∂2u/∂x2 + ∂2u/∂y2 = 0 becomes
.
- Van der Waals’ equation of state for the real gas CO2 is
.
If dT/dt and dV/dt are rates at which the temperature and volume change, respectively, use the Chain Rule to find dP/dt.
- The equation of state for a thermodynamic system is F(P, V, T) = 0, where P, V, and T are pressure, volume, and temperature, respectively. If the equation defines V as a function of P and T, and also defines T as a function of V and P, show that
- The voltage across a conductor is increasing at a rate of 2 volts/min and the resistance is decreasing at a rate of 1 ohm/min. Use I = E/R and the Chain Rule to find the rate at which the current passing through the conductor is changing when R = 50 ohms and E = 60 volts.
- The length of the side labeled x of the triangle in FIGURE 9.4.6 increases at a rate of 0.3 cm/s, the side labeled y increases at a rate of 0.5 cm/s, and the included angle θ increases at a rate of 0.1 rad/s. Use the Chain Rule to find the rate at which the area of the triangle is changing at the instant x = 10 cm, y = 8 cm, and θ = π/6.
- A particle moves in 3-space so that its coordinates at any time are x = 4 cos t, y = 4 sin t, z = 5t, t ≥ 0. Use the Chain Rule to find the rate at which its distance
from the origin is changing at t = 5π/2 seconds.