Proof of the illegality of the equality
𝜕𝑓
𝜕𝑥
∗
𝜕𝑥
𝜕𝑢
=
𝜕𝑓
𝜕𝑢
, where
f and x are functions of x & y and u & v, respectively
Settings
Let f be a function of two variables x and y, where x and y are functions of
two other variables u and v; that is, 𝑓 = 𝑓 (𝑥, 𝑦), 𝑤ℎ𝑒𝑟𝑒 𝑥 = 𝑥(𝑢, 𝑣 ), 𝑦 =
𝑦(𝑢, 𝑣 ).
The change in y due the change in x can be represented by 𝜕𝑓 = 𝑓𝑥 ∗ 𝜕𝑥,
where 𝑓𝑥 indicates the rate of change of f due to the change in x, and 𝜕𝑥
represents a small change in x; then we also have
𝜕𝑓
𝜕𝑥
= 𝑓𝑥 by dividing 𝜕𝑥 on
both side of the equation. Apply the same method to function x where 𝒙 =
𝒙(𝒖, 𝒗), we also get 𝜕𝑥 = 𝑥𝑢 ∗ 𝜕𝑢 and
𝜕𝑥
𝜕𝑢
= 𝑥𝑢 to represent the change in x
due to the change in u. Since f is dependent to x and x is dependent to u,
thus we can use the relationship between f and x, x and u to indicate the
rate of change in f due to the change in u.
Proof
The equality
𝜕𝑓
𝜕𝑥
∗
𝜕𝑥
𝜕𝑢
=
𝜕𝑓
𝜕𝑢
holds and only holds true if the 𝜕𝑥 in both
components are equal. Recall that 𝒙 = 𝒙(𝒖, 𝒗) and 𝒚 = 𝒚(𝒖, 𝒗), where the
value of x and y are determined by u and v. The partial derivative
𝜕𝑥
𝜕𝑢
represents the change in x due to a small change in u at a fixed point of v,
the partial derivative
𝜕𝑓
𝜕𝑥
represents the change in f due to a small change in x
at a fixed point of y; note that the choice of points of v and y is completely
arbitrary. At different points of v,
𝜕𝑥
𝜕𝑢
= 𝑥𝑢 changes its value; since 𝜕𝑢 is
always defined as a infinitely small change in u, thus only a change in 𝜕𝑥 will
result in the change in
𝜕𝑥
𝜕𝑢
= 𝑥𝑢 . Thus, the value of 𝜕𝑥, which is determined
by an arbitrary value of v, is in an indeterminate form; that is, we do not
know the exactly value of it because it have many viable values. The same
idea can be applied to the 𝜕𝑥 in partial derivative
𝜕𝑓
𝜕𝑥
to get the same
conclusion for 𝜕𝑥 in this partial derivative.
There is one and surly not only one alternative proof of it. Since 𝑥 = (𝑢, 𝑣),
thus we have 𝜕𝑥 =
𝜕𝑥
𝜕𝑢
∗ 𝑑𝑢 +
𝜕𝑥
𝜕𝑣
∗ 𝑑𝑣 to represent the total derivative of
function x. The two partial derivatives in the equation,
𝜕𝑥
𝜕𝑢
and
𝜕𝑥
𝜕𝑣
, as
mentioned above, are determined at some particular fixed point of v and u,
respectively. Since the choices of those values are arbitrary, thus the values
of two partial derivatives varies with different values of v or u. Since 𝜕𝑥 =
𝜕𝑥
𝜕𝑢
∗ 𝑑𝑢 +
𝜕𝑥
𝜕𝑣
∗ 𝑑𝑣 and 𝜕𝑥 is related to the two partial derivatives, thus the
value of 𝜕𝑥 also varies with different values of
𝜕𝑥
𝜕𝑢
and
𝜕𝑥
𝜕𝑣
, whose values
depends on the value of v and u. The same idea can be applied to both 𝜕𝑥 in
𝜕𝑓
𝜕𝑥
and
𝜕𝑥
𝜕𝑢
, thus the two 𝜕𝑥 may have different values and does not equal to
each other. Therefore, the equality
𝜕𝑓
𝜕𝑥
∗
𝜕𝑥
𝜕𝑢
=
𝜕𝑓
𝜕𝑢
does not hold true.
In an abstract point of view to look at this equality, f is a function of two
variables x and y, and
𝜕𝑓
𝜕𝑥
∗
𝜕𝑥
𝜕𝑢
only represents the change in f due to the
change in u with v being some fixed points for relationship of u and v in the
function 𝒙 = 𝒙(𝒖, 𝒗); thus, it only take in account the effects of x on f but
not that of y on f. Therefore, it’s necessary to add the change in f due to the
change in u for relationship in function 𝑦 = 𝑦(𝒖, 𝒗) as well.
In short, to conclude the basic idea of the proof: the two 𝜕𝑥’s value all
depends on the arbitrary value of their fixed point of v, thus the two 𝜕𝑥
2
does not necessarily equal to each other, and thus they cannot cancel each
other, the the equality
3
𝜕𝑓
𝜕𝑥
∗
𝜕𝑥
𝜕𝑢
=
𝜕𝑓
𝜕𝑢
does not hold true.