Partial Derivatives

Theorem The Mix Derivative

Theorem (Symmetry of Second Partial Derivative): If the function $f : R^2 \rightarrow R$ has continuous derivatives $f_{xy}$ and $f_{yx}$ at some point $(a, b)$, then

Proof:

W.L.O.G, for $h > 0, k > 0$, define

Define $F(x) = f(x, b + k) - f(x, b)$, which is continuous and differentiable (since $f_x$ exists on $R$). By mean value theorem, $\exists a_1 \in [a, a + h]$ for which

Apply again mean value theorem, $\exists b_1 \in [b, b + k]$, s.t.,

We can also define $G(y) = f(a + h, y) - f(a, y)$ and rewrite

Similarly, by applying mean value theorem twice, $\exists b_2 \in [b, b + k], a_2 \in [a, a + h]$, s.t,

Putting together

Because both $f_{xy}$ and $f_{yx}$ are continuous, as $h \rightarrow 0$ and $k \rightarrow 0$, $f_{xy}(a_1, b_1) \rightarrow f_{xy}(a, b)$ and $f_{yx}(a_2, b_2) \rightarrow f_{yx}(a, b)$, as desired.

$\blacksquare$.

Theorem The Increment

If function $f : R^2 \rightarrow R$ has continuous derivative $f_x$ and $f_y$ at some point $(a, b)$, then

where $\epsilon_1, \epsilon_2 \rightarrow 0$ as $h,k \rightarrow 0$.

Proof:

By mean value theorem, $\exists a_1$ between $a$ and $a + h$, $\exists b_1$ between $b$ and $b + k$, s.t.,

Define $\epsilon_1 = f_x(a_1, b) - f_x(a, b)$ and $\epsilon_2 = f_y(a + h, b_1) - f_y(a, b)$, we infer that

The claim $\epsilon_1, \epsilon_2 \rightarrow 0$ as $h,k \rightarrow 0$ results from the continuity of $f_x$ and $f_y$.

$\blacksquare$.

Reference

1. George B. Thomas, Jr., Maurice D. Weir, Joel Hass, THOMAS’ CALCULUS