再0

(1)

Prove that if is a function of two variables that is differentiable at , then is continuous at .

Hint: Show that

36. (a) The function

was graphed in Figure 4. Show that and both exist but is not differentiable at . [Hint: Use the result of Exercise 35.]

(b) Explain why fxand fyare not continuous at 0, 0.

0, 0

f fy0, 0

fx0, 0

fx, y

⁰^x²^xy^y² ^if^if x, y 0, 0

x, y 0, 0

x, y l 0, 0lim fa x, b y f a, b

a, b

a, b f 35. f

32. Suppose you need to know an equation of the tangent plane to a surface at the point . You don’t have an equation for but you know that the curves

both lie on . Find an equation of the tangent plane at . 33–34 ^■ Show that the function is differentiable by ﬁnding values of and that satisfy Deﬁnition 7.

34.

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

fx, y xy 5y² fx, y x² y² 33.

²

¹

P S

r2u 1 u², 2u³ 1, 2u 1

r1t 2 3t, 1 t², 3 4t t² S

P2, 1, 3

S

THE CHAIN RULE

Recall that the Chain Rule for functions of a single variable gives the rule for differentiating a composite function: If and , where and are differentiable functions, then is indirectly a differentiable function of and

For functions of more than one variable, the Chain Rule has several versions, each of them giving a rule for differentiating a composite function. The ﬁrst version (The- orem 2) deals with the case where and each of the variables and is, in turn, a function of a variable . This means that is indirectly a function of , , and the Chain Rule gives a formula for differentiating as a function of . We assume that is differentiable (Deﬁnition 11.4.7). Recall that this is the case when and are continuous (Theorem 11.4.8).

THE CHAIN RULE (CASE 1) Suppose that is a differentiable function of and , where and are both differentiable functions of . Then is a differentiable function of and

PROOF A change of in produces changes of in and in . These, in turn, produce a change of in , and from Deﬁnition 11.4.7 we have

where and as . [If the functions and are not

deﬁned at ¹l 00, 0, we can deﬁne them to be 0 there.] Dividing both sides of this ²l 0 x, y l 0, 0 ¹ ²

z &f

&x x &f

&yy ¹x ²y z

zt x x y y

t

dz dt &f

&x dx dt &f

&y dy dt z t

t

y ht

x tt

y x

z f x, y

2

fy

fx

f t

z

z f tt, ht t z t

y z f x, y x

dy dt dy

dx dx

1 dt

t y

t f x tt

y f x

11.5

(2)

equation by , we have

If we now let , then because is differentiable

and therefore continuous. Similarly, . This, in turn, means that and , so

■

Since we often write in place of , we can rewrite the Chain Rule in the form

EXAMPLE 1 If , where and , ﬁnd when

.

SOLUTION The Chain Rule gives

It’s not necessary to substitute the expressions for and in terms of . We simply observe that when we have x sin 0 0 and y cos 0 1. Therefore,

■

The derivative in Example 1 can be interpreted as the rate of change of with respect to as the point moves along the curve with parametric equations , . (See Figure 1.) In particular, when , the point is and is the rate of increase as we move along the curve through

. If, for instance, represents the temperature at the

point , then the composite function represents the temperature at points on and the derivative represents the rate at which the temperature changes along .

EXAMPLE 2 The pressure (in kilopascals), volume (in liters), and temperature (in kelvins) of a mole of an ideal gas are related by the equation T PV 8.31T.

V P

V

C

dzdt C

z Tsin 2t, cos t

x, y z Tx, y x²y 3xy⁴

0, 1 dzdt 6 C

0, 1 y cos t t 0 x, y

x sin 2tt x, y C

z dz

dt

^t⁰ 0 32 cos 0 0 0sin 0 6

t 0 x y t

2xy 3y⁴2 cos 2t x² 12xy³sin t

dz dt &z

&x dx

dt &z

&y dy dt

t 0 z x²y 3xy⁴ x sin 2t y cos t dzdt dz

dt &z

&x dx

dt &z

&y dy dt

&f&x

&z&x

&f

&x dx dt &f

&y dy dt

&f

&x dx dt &f

&y dy

dt 0 dx

dt 0 dy dt

&f

&x lim

t l 0

x

t &f

&y lim

t l 0

y

t lim

t l 0¹ lim

t l 0

x

t lim

t l 0² lim

t l 0

y

t dz

dt lim

t l 0

z

t

²l 0 t l 0 x tt t tt l 0y l 0 t ¹l 0

z

t &f

&x

x

t &f

&y

y

t ¹ x

t ²y

t

■ Notice the similarity to the deﬁnition of the differential:

dz &z

&xdx &z

&ydy

FIGURE 1

The curve x=sin 2t, y=cos t x (0, 1)

y

(3)

Find the rate at which the pressure is changing when the temperature is and increasing at a rate of and the volume is 100 L and increasing at a rate of

.

SOLUTION If represents the time elapsed in seconds, then at the given instant we

have , , , . Since

the Chain Rule gives

The pressure is decreasing at a rate of about kPas. ■

We now consider the situation where but each of and is a function of two variables and : , . Then is indirectly a function of and and we wish to ﬁnd and . Recall that in computing we hold ﬁxed and compute the ordinary derivative of with respect to . Therefore, we can apply Theorem 2 to obtain

A similar argument holds for and so we have proved the following version of the Chain Rule.

THE CHAIN RULE (CASE 2) Suppose that is a differentiable function of and , where and are differentiable func- tions of s and t. Then

EXAMPLE 3 If , where and , ﬁnd and . SOLUTION Applying Case 2 of the Chain Rule, we get

2ste^st²sins²t s²e^st²coss²t ■

&z

&t &z

&x

&t &z

&y

&t e^xsin y2st e^xcos ys²

t²e^st²sins²t 2ste^st²coss²t

&z

&s &z

&x

&s &z

&y

&s e^xsin yt² e^xcos y2st

&z&t

&z&s y s²t

x st² z e^xsin y

&z

&t &z

&x

&t &z

&y

&t

&z

&s &z

&x

&s &z

&y

&s

y hs, t

x ts, t

y x

z f x, y

3

&z&s

&z

&t &z

&x

&t &z

&y

&t

z t &z&t s

&z&t

&z&s t

z s y hs, t

x ts, t

t s

y z f x, y x

0.042

8.31

100 0.1 8.31300

100² 0.2 0.04155 dP

dt &P

&T dT

dt &P

&V dV

dt 8.31 V

dT

dt 8.31T V²

dV dt P 8.31T

V dVdt 0.2 V 100

dTdt 0.1 T 300t

0.2 Ls 0.1 Ks 300 K

(4)

Case 2 of the Chain Rule contains three types of variables: and are indepen- dent variables, and are called intermediate variables, and is the dependent vari- able. Notice that Theorem 3 has one term for each intermediate variable and each of these terms resembles the one-dimensional Chain Rule in Equation 1.

To remember the Chain Rule it’s helpful to draw the tree diagram in Figure 2. We draw branches from the dependent variable to the intermediate variables and to indicate that is a function of and . Then we draw branches from and to the independent variables and . On each branch we write the corresponding partial derivative. To ﬁnd we ﬁnd the product of the partial derivatives along each path from to and then add these products:

Similarly, we ﬁnd by using the paths from to .

Now we consider the general situation in which a dependent variable is a function of intermediate variables , , each of which is, in turn, a function of independent variables , . Notice that there are terms, one for each intermediate variable. The proof is similar to that of Case 1.

THE CHAIN RULE (GENERAL VERSION) Suppose that is a differentiable function of the variables , , and each is a differentiable function of the variables , , . Then is a function of , , and

for each , , .

EXAMPLE 4 Write out the Chain Rule for the case where and

, , , and .

SOLUTION We apply Theorem 4 with and . Figure 3 shows the tree diagram. Although we haven’t written the derivatives on the branches, it’s under- stood that if a branch leads from to , then the partial derivative for that branch is

. With the aid of the tree diagram we can now write the required expressions:

■

EXAMPLE 5 If , where , , and ,

ﬁnd the value of when , , .

SOLUTION With the help of the tree diagram in Figure 4, we have

4x³yre^t x⁴ 2yz³2rse^t 3y²z²r²sin t

&u

&s &u

&x

&s &u

&y

&s &u

&z

&s t 0 s 1 r 2

&u&su x⁴y y²z³ x rse^t y rs²e^t z r²s sin t

V

&^w

&^v &^w

&x

&^v &^w

&y

&^v &^w

&z

&^v &^w

&t

&^v

&^w

&u &^w

&x

&u &^w

&y

&u &^w

&z

&u &^w

&t

&u

&y&u y u

m 2 n 4

t tu,^v z zu,^v

y yu,^v x xu,^v

w f x, y, z, t

V

m . . . , 2 i 1

&u

&tⁱ &u

&x¹

&tⁱ &u

&x²

&tⁱ &u

&xⁿ

&tⁱ

tm

. . . , t2

t1

u tm

. . . , t2

t1

m

xj

xn

. . . , x2

x1

n

u

4

n tm

. . . , t1

m xn

. . . , x1

n

u z t

&z&t

&z

&s &z

&x

&s &z

&y

&s z s &z&ss t

y x y

z x z x y

z y

x

t s

FIGURE 2 z

y x

s t s t

⳵x

⳵s ⳵x

⳵t ⳵y

⳵t

⳵y

⳵s

⳵z

⳵x ⳵z

⳵y

FIGURE 3 w

y t

x

u v u v u v

z

u v

FIGURE 4 u

y z

x

s

r t r s t r s t

(5)

When , , and , we have , , and , so

■

EXAMPLE 6 If and is differentiable, show that satisﬁes the equation

SOLUTION Let and . Then and the Chain

Rule gives

Therefore

■

EXAMPLE 7 If has continuous second-order partial derivatives and

and , ﬁnd (a) and (b) .

SOLUTION

(a) The Chain Rule gives

(b) Applying the Product Rule to the expression in part (a), we get

But, using the Chain Rule again (see Figure 5), we have

Putting these expressions into Equation 5 and using the equality of the mixed second- order derivatives, we obtain

2 &z ■

&x 4r² &²z

&x² 8rs &²z

&x &y 4s² &²z

&y²

&²z

&r² 2 &z

&x 2r

^2r ^&x^&²^z² ^2s &y &x^&²^z

^2s

^2r &x &y^&²^z 2s &²z

&y²

&²z

&x &y 2r &²z

&y² 2s

&

&r

^&z^&y

^&x^&

^&z^&y

^&x^&r ^&y^&

^&y^&z

^&y^&r

&²z

&x² 2r &²z

&y &x 2s

&

&r

^&z^&x

^&x^&

^&z^&x

^&x^&r ^&y^&

^&x^&z

^&y^&r

2 &z

&x 2r &

&r

^&z^&x

^2s ^&r^&

^&z^&y

&²z

&r² &

&r

^2r ^&z^&x ^2s ^&z^&y

5

&z

&r &z

&x

&r &z

&y

&r &z

&x 2r &z

&y 2s

&²z&r²

&z&r y 2rs

x r² s² z f x, y

t &t

&s s &t

&t

^2st ^&x^&f ^2st ^&y^&f

⁰

&t

&t &f

&x

&t &f

&y

&t &f

&x 2t &f

&y 2t

&t

&s &f

&x

&s &f

&y

&s &f

&x 2s &f

&y 2s

ts, t f x, y

y t² s² x s² t²

t &t

&s s &t

&t 0

t ts, t f s² t², t² s² f

&u

&s 642 164 00 192 z 0 y 2

x 2 t 0

s 1 r 2

FIGURE 5

⳵z

⳵x y x

r s r s

(6)

IMPLICIT DIFFERENTIATION

The Chain Rule can be used to give a more complete description of the process of implicit differentiation that was introduced in Sections 2.6 and 11.3. We suppose that an equation of the form deﬁnes implicitly as a differentiable function of , that is, , where for all in the domain of . If is differentiable, we can apply Case 1 of the Chain Rule to differentiate both sides of the equation with respect to . Since both and are functions of , we obtain

But , so if we solve for and obtain

To derive this equation we assumed that deﬁnes implicitly as a func- tion of . The Implicit Function Theorem, proved in advanced calculus, gives con- ditions under which this assumption is valid. It states that if is deﬁned on a disk

containing where , , and and are continuous on

the disk, then the equation deﬁnes as a function of near the point and the derivative of this function is given by Equation 6.

EXAMPLE 8 Find if .

SOLUTION The given equation can be written as

so Equation 6 gives

■

Now we suppose that is given implicitly as a function by an equation

of the form . This means that for all in the

domain of . If and are differentiable, then we can use the Chain Rule to differentiate the equation as follows:

But &

&x y 0

& and

&x x 1

&F

&x

&x &F

&y

&x &F

&z

&x 0 Fx, y, z 0f

F f

x, y

Fx, y, f x, y 0

Fx, y, z 0z z f x, y

dy

dx Fx

Fy 3x² 6y

3y² 6x x² 2y y² 2x Fx, y x³ y³ 6xy 0 x³ y³ 6xy

y

a, b

x y

Fx, y 0

Fy

Fx

Fya, b 0 Fa, b 0

a, b,

F x

y Fx, y 0

dy dx

&F

&x

&F

&y

Fx

Fy

6

dydx

&F&y 0 dxdx 1

&F

&x dx dx &F

&y dy dx 0

x y

x x

Fx, y 0

F f x

Fx, f x 0 y f x

x

y Fx, y 0

■ The solution to Example 8 should be compared to the one in Example 2 in Section 2.6.

(7)

so this equation becomes

If , we solve for and obtain the ﬁrst formula in Equations 7. The formula for is obtained in a similar manner.

Again, a version of the Implicit Function Theorem gives conditions under which our assumption is valid. If is deﬁned within a sphere containing , where

, , and , , and are continuous inside the sphere, then the equation deﬁnes as a function of and near the point

and this function is differentiable, with partial derivatives given by (7).

EXAMPLE 9 Find and if .

SOLUTION Let . Then, from Equations 7, we

have

&z ■

&y Fy

F_z 3y² 6xz

3z² 6xy y² 2xz z² 2xy

&z

&x Fx

Fz 3x² 6yz

3z² 6xy x² 2yz z² 2xy Fx, y, z x³ y³ z³ 6xyz 1

x³ y³ z³ 6xyz 1

&z

&y

&z

&x

a, b, c F_za, b, c 0Fx, y, z 0 Fx Fyz F_z x y

Fa, b, c 0 F a, b, c

&z

&y

&F

&y

&F

&z

&x

&F

&x

&F

&z

7

&z&y &z&x

&F&z 0

&F

&x &F

&z

&x 0

■ The solution to Example 9 should be compared to the one in Example 4 in Section 11.3.

9. If , where is differentiable, , ,

, , , , ,

and , ﬁnd when .

10. Let , where are

differentiable, , , ,

, , , , and

. Find and .

11. Suppose is a differentiable function of and , and . Use the table of values to calculate t^u0, 0 and t^v0, 0.

tu,^v f e^u sin^v, e^u cos^v

y x f

Wt1, 0

Ws1, 0

Fv2, 3 10

Fu2, 3 1 vt1, 0 4

vs1, 0 5 v1, 0 3

ut1, 0 6 us1, 0 2

u1, 0 2

F, u, and v Ws, t Fus, t,^vs, t

t 3 dzdt

fy2, 7 8

fx2, 7 6 h3 4

h3 7 t3 5

t3 2

y ht

x tt

z f x, y f 1– 4 ^■ Use the Chain Rule to ﬁnd or .

1. , ,

2. , ,

, , ,

4. , , ,

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

5– 8 ^■ Use the Chain Rule to ﬁnd and .

5. , ,

6. , ,

, ,

8. , ,

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

 s t

 3s t z sin tan

 ss² t² r st

z e^r cos 7.

y 1 se^t x se^t

z xy

y st x s t

z x² xy y²

&z&t

&z&s

z e^t cos t y e^t sin t

x e^t w xy yz²

z 1 2t y 1 t

x t² w xe^y^z

3.

y cos t x sin t

z x lnx 2y

y st xt

z sin x cos y

dwdt dzdt

EXERCISES

11.5

3 6 4 8

6 3 2 5

1, 2

0, 0

fy

fx

t f

(8)

estimate that the average temperature is rising at a rate of 0.15°Cyear and rainfall is decreasing at a rate of 0.1 cmyear. They also estimate that, at current production

levels, and .

(a) What is the signiﬁcance of the signs of these partial derivatives?

(b) Estimate the current rate of change of wheat production, .

31. The speed of sound traveling through ocean water with salinity 35 parts per thousand has been modeled by the equation

where is the speed of sound (in meters per second), is the temperature (in degrees Celsius), and is the depth below the ocean surface (in meters). A scuba diver began a leisurely dive into the ocean water; the diver’s depth and surrounding water temperature over time are recorded in the following graphs. Estimate the rate of change (with respect to time) of the speed of sound through the ocean water experienced by the diver 20 minutes into the dive. What are the units?

32. The radius of a right circular cone is increasing at a rate of ins while its height is decreasing at a rate of ins.

At what rate is the volume of the cone changing when the radius is 120 in. and the height is 140 in.?

The length 艎, width , and height of a box change with time. At a certain instant the dimensions are and m, and 艎 and are increasing at a rate of 2 ms while is decreasing at a rate of 3 ms. At that instant ﬁnd the rates at which the following quantities are changing.

(a) The volume (b) The surface area (c) The length of a diagonal

34. The voltage in a simple electrical circuit is slowly decreasing as the battery wears out. The resistance is slowly increasing as the resistor heats up. Use Ohm’s Law,

, to ﬁnd how the current is changing at the moment

when , A, Vs, and

.

35. The pressure of 1 mole of an ideal gas is increasing at a rate of kPas and the temperature is increasing at a rate of

Ks. Use the equation in Example 2 to ﬁnd the rate of change of the volume when the pressure is 20 kPa and the temperature is K.

36. If a sound with frequency is produced by a source traveling along a line with speed vsand an observer is traveling

fs

320 0.15

0.05

dRdt 0.03 s

dVdt 0.01 I 0.08

R 400

I V IR

R V

h

w

w h 2 艎 1 m

h 33. w

2.5 1.8

t (min) T

10 12

10 20 30 40

14 16

8 t (min) D

5 10

10 20 30 40

15 20

D

T C

C 1449.2 4.6T 0.055T² 0.00029T³ 0.016D dWdt

&W&R 8

&W&T 2 12. Suppose is a differentiable function of and , and

Use the table of values in Exercise 11 to calculate and

13–16 ^■ Use a tree diagram to write out the Chain Rule for the given case. Assume all functions are differentiable.

, where ,

14. , where , ,

15. ,

where , ,

16. , where ,

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

17–21 ^■ Use the Chain Rule to ﬁnd the indicated partial derivatives.

17. , , ;

, , when , ,

18. , , ;

, , when , ,

19. ,

, , ;

, when

20. , , , ;

, when

21. , , , ;

, , when

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

22–24 ^■ Use Equation 6 to ﬁnd .

22. 23.

24.

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

25–28 ^■ Use Equations 7 to ﬁnd and . 25.

27. 28.

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

The temperature at a point is , measured in degrees Celsius. A bug crawls so that its position after

seconds is given by , where

and are measured in centimeters. The temperature func-

tion satisﬁes and . How fast is the

temperature rising on the bug’s path after 3 seconds?

30. Wheat production in a given year, , depends on the average temperature and the annual rainfall . ScientistsT R

W Ty2, 3 3 Tx2, 3 4

y

x x s1 t , y 2 ¹³t

t

Tx, y

x, y

29.

yz lnx z

x z arctanyz

x yz cosx y z

x² y² z² 3xyz 26.

&z&y

&z&x sin x cos y sin x cos y

sxy 1 x²y y⁵ x²y³ 1 ye^x²

dydx p 2, r 3, 0

&u

&

&u

&r

&u

&p

z p r y pr sin

x pr cos u x² yz

u 3,^v 1

&M

&^v

&M

&u

z u ^v y u ^v

x 2u^v M xe^y^z²

x y 1

&R

&y

&R

&x

w 2xy v 2x y

u x 2y

R lnu²^v²^w²

t 0 y 2 x 1

&u

&t

&u

&y

&u

&x

s x y sin t r y x cos t

u sr² s²

w 0 v 1 u 2

&z

&^w

&z

&^v

&z

&u

y u ^ve^w x u^v²^w³

z x² xy³

t t^w, x, y,z

s s^w, x, y,z

u f s, t

r rx, y, z

q qx, y, z

p px, y, z

v f p, q, r

z zt, u

y yt, u

x xt, u

w f x, y, z

y yr, s, t

x xr, s, t

u f x, y

13.

t^s1, 2.

t^r1, 2

tr, s f 2r s, s² 4r.

y x f

(9)

is a solution of the wave equation

[Hint: Let , .]

42. If , where and , show

that

43. If , where , , ﬁnd .

(Compare with Example 7.)

44. If , where , , ﬁnd

(a) , (b) , and (c) .

45. If , where , , show that

46. Suppose , where and .

(a) Show that

(b) Find a similar formula for .

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

47. Suppose that the equation implicitly deﬁnes each of the three variables , , and as functions of the

other two: , , . If is dif-

ferentiable and , , and are all nonzero, show that

&z

&x &x

&y &y

&z 1 F_z Fy

Fx

F x hy, z

y tx, z

z f x, y x y z Fx, y, z 0

&²z&s &t &z

&x

&²x

&t² &z

&y

&²y

&t²

&²z

&t² &²z

&x²

^&x^&t

²²&x &y^&²^z

&x

&t

&y

&t &²z

&y²

^&y^&t

²

y hs, t

x ts, t

z f x, y

&²z

&x² &²z

&y² &²z

&r² 1 r²

&²z

&² 1 r

&z

&r y r sin x r cos

z f x, y

&²z&r &

&z&

&z&r

y r sin x r cos

z f x, y

&²z&r &s y 2rs

x r² s² z f x, y

&²u

&x² &²u

&y² e^2s

^&^&s²^u² ^&^&t²^u²

y e^ssin t x e^scos t

u f x, y

v x at u x at

&²z

&t² a² &²z

&x² with speed along the same line from the opposite direc-

tion toward the source, then the frequency of the sound heard by the observer is

where is the speed of sound, about . (This is the Doppler effect.) Suppose that, at a particular moment, you are in a train traveling at and accelerating at

. A train is approaching you from the opposite direction on the other track at , accelerating at

, and sounds its whistle, which has a frequency of 460 Hz. At that instant, what is the perceived frequency that you hear and how fast is it changing?

37– 40 ^■ Assume that all the given functions are differentiable.

If , where and , (a) ﬁnd

and and (b) show that

38. If , where and , show

that

If , show that .

40. If , where and , show that

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

41– 46 ^■ Assume that all the given functions have continuous second-order partial derivatives.

41. Show that any function of the form z f x at tx at

^&z^&x

²^x^{s t}

^&z^&y

² ^&z^&s^y^{s t}^&z^&t

z f x, y

&z

&x &z

&y 0 z f x y

39.

^&u^&x

²

^&u^&y

²^e^2s

^&u^&s

²

^&u^&t

²

y e^ssin t x e^scos t

u f x, y

^&x^&z&z&

²

^&z^&y

²

^&z^&r

² ^r¹²

^&^&z

²

&z&r

y r sin x r cos

z f x, y

37.

1.4 ms²

40 ms 1.2 ms²

34 ms

332 ms c

fo

^c^c^v^v^o^s

^f^s

vo

DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR

Recall that if , then the partial derivatives and are deﬁned as

and represent the rates of change of in the - and -directions, that is, in the directions of the unit vectors and .i j

y z x

fyx⁰, y0 lim

hl 0

fx⁰, y0 h f x⁰, y0 h

fxx⁰, y0 lim

hl 0

fx⁰ h, y⁰ f x⁰, y0

1 h

fy

fx

z f x, y

再0



11.5





























11.5



































11.6