The chain rule

5.4 USEFUL THEOREMS OF PARTIAL DIFFERENTIATION
5.4 Useful theorems of partial differentiation
So far our discussion has centred on a function f(x, y) dependent on two variables,
x and y. Equally, however, we could have expressed x as a function of f and y,
or y as a function of f and x. To emphasise the point that all the variables are
of equal standing, we now replace f by z. This does not imply that x, y and z
are coordinate positions (though they might be). Since x is a function of y and z,
it follows that
∂x
∂x
dy +
dz
(5.11)
dx =
∂y z
∂z y
and similarly, since y = y(x, z),
dy =
∂y
∂x
dx +
z
∂y
∂z
dz.
(5.12)
x
We may now substitute (5.12) into (5.11) to obtain
∂x
∂x
∂x
∂y
∂y
dx +
+
dx =
dz.
∂y z ∂x z
∂y z ∂z x
∂z y
(5.13)
Now if we hold z constant, so that dz = 0, we obtain the reciprocity relation
−1
∂y
∂x
=
,
∂y z
∂x z
which holds provided both partial derivatives exist and neither is equal to zero.
Note, further, that this relationship only holds when the variable being kept
constant, in this case z, is the same on both sides of the equation.
Alternatively we can put dx = 0 in (5.13). Then the contents of the square
brackets also equal zero, and we obtain the cyclic relation
∂z
∂x
∂y
= −1,
∂z x ∂x y ∂y z
which holds unless any of the derivatives vanish. In deriving this result we have
used the reciprocity relation to replace (∂x/∂z)−1
y by (∂z/∂x)y .
5.5 The chain rule
So far we have discussed the differentiation of a function f(x, y) with respect to
its variables x and y. We now consider the case where x and y are themselves
functions of another variable, say u. If we wish to find the derivative df/du,
we could simply substitute in f(x, y) the expressions for x(u) and y(u) and then
differentiate the resulting function of u. Such substitution will quickly give the
desired answer in simple cases, but in more complicated examples it is easier to
make use of the total differentials described in the previous section.
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