Page 152 - Foundations Of Differential Calculus

7. On the Differentiation of Functions of Two or More Variables 135
Therefore, the functions P and Q are related in such a way that if both of
them are differentiated as we have done, the quantities q and r are equal to
each other. For the sake of brevity, at least in this chapter, the quantities r
and q will conveniently be symbolized by ∂P/∂y, in that P is differentiated
with only y variable, which is indicated by ∂y in the denominator. In this
way we obtain the finite quantity r. In like manner ∂Q/∂x will symbolize
the finite quantity q, since by this is indicated that the function Q is dif-
ferentiated with only x variable, so we ought to divide the differential by
∂x.

232. We will use this method of symbolizing, although there is some
danger of ambiguity therein. We avoid this with the use of the partial
symbol, with the result that the complications in describing the conditions
of differentiation are avoided. Thus we can briefly express the relationship
between P and Q by stating

∂P ∂Q
= .
∂y ∂x
In fractions of this kind, beyond the usual significance is which the denom-
inator indicates the divisor, here the differential of the numerator is to be
taken with only that quantity variable which is indicted by the differential
in the denominator. In this way by the division of differentials these frac-
tions ∂P/∂y and ∂Q/∂x exist from calculus and indicate finite quantities,
which in this case are equal to each other. Once this method is agreed upon,
the quantities p and s can be denoted by p = ∂P/∂x and s = ∂Q/∂y if, as
we have noted, the differentiation of the numerator by the denominator is
properly restricted.
233. There is a wonderful agreement between this property and that prop-
erty of homogeneous functions which we previously have shown. Let V be a
homogeneous function in x and y of dimension n.Welet dV = Pdx+Qdy,
and we have shown that nV = Px + Qy. Hence
nV Px
Q = − .
y y

We let dP = pdx + rdy, so that
∂P
= r,
∂y

which is thus shown to be equal to ∂Q/∂x. Let Q be differentiated with
only x variable, and under the present hypothesis we have

nP dx Pdx xp dx
dQ = − − ,
y y y