Page 146 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 6] PARTIAL DERIVATIVES 137

3 1

@ðF; GÞ F x F v

@u G x G v 1 4v 1 12v
@ðx; vÞ
¼ ¼ ¼
@x @ðF; GÞ F u 2u 1 1 8uv
¼
F v
1 4v
@ðu; vÞ G u G v
provided 1 8uv 6¼ 0. Similarly, the other partial derivatives are obtained.
6.33. If Fðu; v; w; x; yÞ¼ 0, Gðu; v; w; x; yÞ¼ 0, Hðu; v; w; x; yÞ¼ 0, find

@v @x @w
; ; :
@y x @v w @u y
ðaÞ ðbÞ ðcÞ
From 3 equations in 5 variables, we can (theoretically at least) determine 3 variables in terms of the
remaining 2. Thus, 3 variables are dependent and 2 are independent. If we were asked to determine @v=@y,
we would know that v is a dependent variable and y is an independent variable, but would not know the

@v
remaining independent variable. However, the particular notation serves to indicate that we are to
@y x
obtain @v=@y keeping x constant, i.e., x is the other independent variable.
(a)Differentiating the given equations with respect to y,keeping x constant, gives
F u u y þ F v v y þ F w w y þ F y ¼ 0 G u u y þ G v v y þ G w w y þ G y ¼ 0
ð1Þ ð2Þ
H u u y þ H v v y þ H w w y þ H y ¼ 0
ð3Þ
Solving simultaneously for v y ,we have

F u F y F w

G u G y @ðF; G; HÞ
G w

@v H u H y H w
@ðu; y; wÞ

@y
v y ¼ ¼ ¼
x F u F v F w @ðF; G; HÞ
G u G v

@ðu; v; wÞ
G w

H u H v H w
Equations (1), (2), and (3)can also be obtained by using differentials as in Problem 6.31.
The Jacobian method is very suggestive for writing results immediately, as seen in this problem and

@v
Problem 6.31. Thus, observe that in calculating the result is the negative of the quotient of two
@y x
Jacobians, the numerator containing the independent variable y,the denominator containing the
dependent variable v in the same relative positions. Using this scheme, we have
@ðF; G; HÞ @ðF; G; HÞ

@x @w
@ðv; y; uÞ @ðu; x; vÞ

ðbÞ ¼ ðcÞ ¼
@v w @ðF; G; HÞ @u y @ðF; G; HÞ
@ðx; y; uÞ @ðw; x; vÞ
2
2
@ z 3z þ x
3
6.34. If z xz y ¼ 0, prove that ¼ .
@x @y 2 3
ð3z xÞ
Differentiating with respect to x,keeping y constant and remembering that z is the dependent variable
depending on the independent variables x and y,we find
2 @z @z @z z
3z x z ¼ 0 and ð1Þ ¼
2
@x @x @x 3z x
Differentiating with respect to y,keeping x constant, we find
2 @z @z @z 1
3z x 1 ¼ 0 and ð2Þ ¼
2
@y @y @y 3z x