5.4 USEFUL THEOREMS OF PARTIAL DIFFERENTIATION
5.4 Useful theorems of partial differentiation
So far our discussion has centred on a functionf(x, y) dependent on two variables,
xandy. Equally, however, we could have expressedxas a function offandy,
oryas a function offandx. To emphasise the point that all the variables are
of equal standing, we now replacefbyz. This does not imply thatx,yandz
are coordinate positions (though they might be). Sincexis a function ofyandz,
it follows that
dx=
(
∂x
∂y
)
z
dy+
(
∂x
∂z
)
y
dz (5.11)
and similarly, sincey=y(x, z),
dy=
(
∂y
∂x
)
z
dx+
(
∂y
∂z
)
x
dz. (5.12)
We may now substitute (5.12) into (5.11) to obtain
dx=
(
∂x
∂y
)
z
(
∂y
∂x
)
z
dx+
[(
∂x
∂y
)
z
(
∂y
∂z
)
x
+
(
∂x
∂z
)
y
]
dz. (5.13)
Now if we holdzconstant, so thatdz= 0, we obtain thereciprocity relation
(
∂x
∂y
)
z
=
(
∂y
∂x
)− 1
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 casez, is the same on both sides of the equation.
Alternatively we can putdx= 0 in (5.13). Then the contents of the square
brackets also equal zero, and we obtain thecyclic relation
(
∂y
∂z
)
x
(
∂z
∂x
)
y
(
∂x
∂y
)
z
=− 1 ,
which holds unless any of the derivatives vanish. In deriving this result we have
used the reciprocity relation to replace (∂x/∂z)−y^1 by (∂z/∂x)y.
5.5 The chain rule
So far we have discussed the differentiation of a functionf(x, y) with respect to
its variablesxandy. We now consider the case wherexandyare themselves
functions of another variable, sayu. If we wish to find the derivativedf/du,
we could simply substitute inf(x, y) the expressions forx(u)andy(u)andthen
differentiate the resulting function ofu. 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.