Gilbert Strang, Calculus, 3rd edition (Wellesley-Cambridge Press; MIT OpenCourseWare RES.18-001) §13.2 Partial Derivatives, #43 · Multivariable Calculus · Partial derivatives

The figure for this exercise is in the book and isn't reproduced…

The proof of $f_{x y} = f_{y x}$ studies $f (x, y)$ in a small rectangle. The top-bottom difference is $g (x) = f (x, B) - f (x, A)$ . The difference at the corners 1, 2, 3, 4 is:
$Q = [f_{4} - f_{3}] - [f_{2} - f_{1}]$
$= g (b) - g (a) (definition of g)$
$= (b - a) g_{x} (c) (Mean Value Theorem)$
$= (b - a) [f_{x} (c, B) - f_{x} (c, A)] (compute g_{x})$
$= (b - a) (B - A) f_{x y} (c, C) (MVT again)$

(a) The right-left difference is $h (y) = f (b, y) - f (a, y)$ . The same $Q$ is $h (B) - h (A)$ . Change the steps to reach $Q = (B - A) (b - a) f_{y x} (c^{*}, C^{*})$ .

(b) The two forms of $Q$ make $f_{x y}$ at $(c, C)$ equal to $f_{y x}$ at $(c^{*}, C^{*})$ . Shrink the rectangle toward $(a, A)$ . What assumption yields $f_{x y} = f_{y x}$ at that typical point?

The figure for this exercise is in the book and isn't reproduced here.

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Gilbert Strang, Calculus, 3rd edition (Wellesley-Cambridge Press; MIT OpenCourseWare RES.18-001) · §13.2 Partial Derivatives, #43

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