Page 147 - Foundations Of Differential Calculus

130 7. On the Differentiation of Functions of Two or More Variables
If we let dV = Pdx + Qdy, since dy = tdx + xdt,wehave

dV = Pdx + Qt dx + Qx dt.
Since the two forms should be equal, we have

P + Qt = nTx n−1 = nV
x
n
because V = Tx . Since t = y/x we have
Px + Qy = nV,

and this equation thus defines the relation between P and Q. Hence if one
n
is known, the other is easily discovered. Since Qx = x Θ, it follows that
Qx, as well as Qy and Px, is an n-dimensional function of x and y.
223. Hence, if in the differential of any homogeneous function in x and y
we substitute x and y for dx and dy, respectively, the result will be equal to
the original function whose differential is given multiplied by the dimension.
2
2

I. If V = x + y , then n = 1, and since
xdx + ydy
,
dV = 2 2
x + y
we have
2
x + y 2
2
2
= V = x + y .
2 2
x + y
3
y + x 3
II. If V = , then n = 2 and
y − x
3 2 2 3 3 3
2y dy − 3y xdy +3yx dx − 2x dx + y dx − x dy
dV = 2 .
(y − x)
When we substitute x for dx and y for dy,wehave
4 3 3 4 3 3
2y − 2y x +2yx − 2x 2y +2x
2 = =2V.
(y − x) y − x
1
III. If V = 2 , then n = −4 and
2
2
(y + y )
4ydy +4xdx
dV = − 3 .
2
2
(y + x )
When we substitute x for dx and y for dy,wehave
2
4y +4x 2
− 3 = −4V.
2
2
(x + y )