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Since, by hypothesis, fx and fy are continuous, it follows that fx x 1 x; y y fx x; y 1 ; fy x; y 2 y fy x; y 2
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where 1 ! 0, 2 ! 0 as x ! 0 and y ! 0. Thus, f fx x fy y 1 x 2 y as required. De ning x dx; y dy, we have f fx dx fy dy 1 dx 2 dy: We call df fx dx fy dy the di erential of f (or z) or the principal part of f (or z).
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6.15. If z f x; y x2 y 3y, nd (a) z; b dz: c Determine z and dz if x 4, y 3, x 0:01, y 0:02. (d) How might you determine f 5:12; 6:85 without direct computation
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PARTIAL DERIVATIVES
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Solution: a z f x x; y y f x; y f x x 2 y y 3 y y g fx2 y 3yg 2xy x x2 3 y x 2 y 2x x y x 2 y | {z } | {z }
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The sum (A) is the principal part of z and is the di erential of z, i.e., dz. b dz 2xy x x2 3 y 2xy dx x2 3 dy Another method: c dz @z @z dx dy 2xy dx x2 3 dy @x @y
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z f x x; y y f x; y f 4 0:01; 3 0:02 f 4; 3 f 3:99 2 3:02 3 3:02 g f 4 2 3 3 3 g 0:018702 dz 2xy dx x2 3 dy 2 4 3 0:01 43 3 0:02 0:02 Note that in this case z and dz are approximately equal, because x dx and y dy are su ciently small.
(d) We must nd f x x; y y when x x 5:12 and y y 6:85. We can accomplish this by choosing x 5, x 0:12, y 7, y 0:15. Since x and y are small, we use the fact that f x x; y y f x; y z is approximately equal to f x; y dz, i.e., z dz. Now z f x; y f 5; 7 5 2 7 3 7 154 dz 2xy dx x2 3 dy 2 5 7 0:12 52 3 0:15 5:1: Then the required value is 154 5:1 159:1 approximately. The value obtained by direct computation is 159.01864.
6.16. (a) Let U x2 ey=x . Find dU. (b) Show that 3x2 y 2y2 dx x3 4xy 6y2 dy can be written as an exact di erential of a function  x; y and nd this function.
(a) Method 1:   @U y x2 ey=x 2 2xe y=x ; @x x Then Method 2: dU x2 d ey=x ey=x d x2 x2 e y=x d y=x 2xe y=x dx   x dy y dx x2 ey=x 2xe y=x dx 2xe y=x ye y=x dx xe y=x dy x2 (b) Method 1: Suppose that Then 3x2 y 2y2 dx x3 4xy 6y2 dy d (1) @ 3x2 y 2y2 ; @x (2) @ @ dx dy: @x @y dU   @U 1 x2 ey=x @y x
@U @U dx dy 2xe y=x ye y=x dx xe y=x dy @x @y
@ x3 4xy 6y2 @y
From (1), integrating with respect to x keeping y constant, we have  x3 y 2xy2 F y
PARTIAL DERIVATIVES
[CHAP. 6
where F y is the constant of integration. x 4xy F y x 4xy 6y
3 3 0 2
Substituting this into (2) yields
from which F 0 y 6y2 ; i.e., F y 2y3 c
Hence, the required function is  x3 y 2xy2 2y3 c, where c is an arbitrary constant. Note that by Theorem 3, Page 122, the existence of such a function is guaranteed, since if If @P=@y 6 P 3x2 y 2y2 and Q x3 4xy 6y2 , then @P=@y 3x2 4y @Q=@x identically. @Q=@x this function would not exist and the given expression would not be an exact di erential. Method 2: 3x2 y 2y2 dx x3 4xy 6y2 dy 3x2 y dx x3 dy 2y2 dx 4xy dy 6y2 dy d x3 y d 2xy2 d 2y3 d x3 y 2xy2 2y3 d x3 y 2xy2 2y3 c Then the required function is x3 y 2xy2 2y3 c. This method, called the grouping method, is based on one s ability to recognize exact di erential combinations and is less than Method 1. Naturally, before attempting to apply any method, one should determine whether the given expression is an exact di erential by using Theorem 3, Page 122. See Theorem 4, Page 122.
DIFFERENTIATION OF COMPOSITE FUNCTIONS 6.17. Let z f x; y and x  t , y t where f ; ; are assumed di erentiable. Prove
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