If x = 5 and y= 10 , what will the following expression return : a) x* y < = y b) (x+ y) ! = y % 2
Answers
Answer:
Section 14.4 (3/23/08)
Chain Rules with two variables
Overview: In this section we discuss procedures for differentiating composite functions with two variables. Then we consider second-order and higher-order derivatives of such functions.
Topics:
• Using the Chain Rule for one variable
• The general Chain Rule with two variables
• Higher order partial derivatives
Using the Chain Rule for one variable
Partial derivatives of composite functions of the forms z = F (g(x, y)) can be found directly with the
Chain Rule for one variable, as is illustrated in the following three examples.
Example 1 Find the x-and y-derivatives of z = (x
2
y
3 + sin x)
10
.
Solution To find the x-derivative, we consider y to be constant and apply the one-variable Chain
Rule formula d
dx(f
10) = 10 f
9 df
dx from Section 2.8. We obtain
∂
∂x[(x
2
y
3 + sin x)
10] = 10(x
2
y
3 + sin x)
9 ∂
∂x(x
2
y
3 + sin x)
= 10(x
2
y
3 + sin x)
9
(2xy
3 + cos x).
Similarly, we find the y-derivative by treating x as a constant and using the same
one-variable Chain Rule formula with y as variable:
∂
∂y [(x
2
y
3 + sin x)
10] = 10(x
2
y
3 + sin x)
9 ∂
∂y (x
2
y
3 + sin x)
= 10(x
2
y
3 + sin x)
9
(3x
2
y
2
).
Example 2 The radius (meters) of a spherical balloon is given as a function r = r(P, T) of the
atmospheric pressure P (atmospheres) and the temperature T (degrees Celsius). At
one moment the radius is ten meters, the rate of change of the radius with respect to
atmospheric pressure is −0.01 meters per atmosphere, and the rate of change of the
radius with respect to the temperature is 0.002 meter per degree. What are the rates of
change of the volume V =
4
3
πr3
of the balloon with respect to P and T at that time?
Solution We first take the P-derivative with T constant and then take the T-derivative with
P constant, using the Chain Rule for one variable in each case to differentiate r
3
. We
obtain
∂V
∂P =
∂
∂P
4
3
πr
3
=
1
3
πr
2 ∂r
∂P
∂V
∂P =
∂
∂T
4
3
πr
3
=
1
3
πr
2 ∂r
∂T .
Setting r = 10, ∂r/∂P = −0.01, and ∂r/∂T = 0.002 then gives
∂V
∂P =
1
3
π(102
)(−0.01) = −
1
3
π
.= −1.05 cubic meters
atmosphere
∂V
∂T =
1
3
π(102
)(0.002) = 1
15π
.= 0.21 cubic meters
degree .
3
p. 318 (3/23/08) Section 14.4, Chain Rules with two variables
Example 3 What are the x- and y-derivatives of z = F(g(x, y)) at x = 5, y = 6 if
g(5, 6) = 10, F0
(10) = −7, gx(5, 6) = 3, and gy(5, 6) = 11?
Solution By the Chain Rule formula d
dt[F (u(t))] = F
0
(u(t)) u
0
(t) for one variable with first x
and then y in place of t, we obtain
h
∂
∂x{F(g(x, y))}
i
x=5,y=6
= F
0
(g(5, 6)) gx(5, 6)
= F
0
(10) gx(5, 6) = (−7)(3) = −21
h
∂
∂y {F(g(x, y))}
i
x=5,y=6
= F
0
(g(5, 6)) gy(5, 6)
= F
0
(10) gy(5, 6) = (−7)(11) = −77.
Partial derivatives of composite functions of the forms F(t) = f (x(t), y(t)) and F(s, t)
= f (x(s, t), y(s, t)) can be found directly with the Chain Rule for one variable if the “outside” function
z = f(x, y) is given in terms of power functions, exponential functions, logarithms, trigonometric
functions, and inverse trigonometric functions rather than just by a letter name. This is illustrated
in the following example.
Example 4 Find the t-derivative of z = f (x(t), y(t)), where f(x, y) = x
5
y
6
, x(t) = e
t
, and
y(t) = √
t.
Solution Because f(x, y) is a product of powers of x and y, the composite function f (x(t), y(t))
can be rewritten as a function of t. We obtain
f (x(t), y(t)) = [x(t)]5
[y(t)]6 = (e
t
)
5
(t
1/2
)
6 = e
5t
t
3
.
Then the Product and Chain Rules for one variable give
d
dt[f (x(t), y(t))] = d
dt (e
5t
t
3
) = e
5t d
dt (t
3
) + t
3 d
dt (e
5t
)
= 3t
2
e
5t + t
3
e
5t d
dt (5t) = 3t
2
e
5t + 5t
3
e
5t
.
The general Chain Rule with two variables
We the following general Chain Rule is needed to find derivatives of composite functions in the form
z = f(x(t), y(t)) or z = f (x(s, t), y(s, t)) in cases where the outer function f has only a letter name. We
begin with functions of the first type.
Theorem 1 (The Chain Rule) The t-derivative of the composite function z = f (x(t), y(t)) is
d
dt[f (x(t), y(t))] = fx (x(t), y(t)) x
0
(t) + fy (x(t), y(t)) y
0
(t). (1)
We assume in this theorem and its applications that x = x(t) and y = y(t) have first derivatives
at t and that z = f(x, y) has continuous first-order derivatives in an open circle centered at (x(t), y(t)).
Learn equation (1) as the following statement: the t-derivative of the composite function equals
the x-derivative of the outer function z = f(x, y) at the point (x(t), y(t)) multiplied by the t-derivative
of the inner function x = x(t), plus the y-derivative of the outer function at (x(t), y(t)) multipled by the
t-derivative of the inner function y = y(
Explanation: