Originally Posted by

**Runty** This question is right out of Taylor & Mann Advanced Calculus, Third Edition, and is a REALLY tricky one. Due to how context-heavy this is, I can't write it word-for-word.

We're to prove the following theorem:

Let $\displaystyle f(x,y)$ and its first partial derivatives $\displaystyle f_1, f_2$ be defined in a neighborhood of the point $\displaystyle (a,b)$, and suppose that $\displaystyle f_1$ and $\displaystyle f_2$ are differentiable at that point. Then $\displaystyle f_{12}(a,b)=f_{21}(a,b)$.

This theorem uses part of another theorem's proof to start things off, before getting to the parts that I can't figure out.

Let $\displaystyle h$ be a number different from zero such that the point $\displaystyle (a+h,b+h)$ is inside a square having its center at $\displaystyle (a,b)$. We then consider the following expression:

$\displaystyle D=f(a+h,b+h)-f(a+h,b)-f(a,b+h)+f(a,b)$

If we introduce the function

$\displaystyle \phi (x)=f(x,b+h)-f(x,b)$,

we can express $\displaystyle D$ in the form

$\displaystyle D=\phi (a+h)-\phi (a)$ (*)

Now $\displaystyle \phi$ has the derivative

$\displaystyle \phi '(x)=f_1(x,b+h)-f_1(x,b)$

Hence $\displaystyle \phi$ is continuous, and we may apply the mean-value theorem for derivatives to (*), obtaining the following:

$\displaystyle D=h\phi '(a+\theta_1 h)=h(f_1(a+\theta_1 h,b+h)-f_1(a+\theta_1 h,b))$, where $\displaystyle 0<\theta_1 <1$

That ends the part of the separate proof; now to parts which I'm in the dark about (these are from the actual question).

From the fact that $\displaystyle f_1$ is differentiable at $\displaystyle (a,b)$, one can write

$\displaystyle f_1(a+\theta_1 h,b+h)=f_1(a,b)+f_{11}(a,b)\theta_1 h+f{12}(a,b)h+\epsilon_1 |h|$, where $\displaystyle \epsilon_1\rightarrow 0$ as $\displaystyle h\rightarrow 0$.

Explain why this is so.