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Lecture 29 : Mixed Derivative Theorem, MVT and Extended MVT

If f : R2 R, then fx is a function from R2 to R (if it exists). So one can analyze the existence of

2f f 2f f
fxx = (fx )x = 2
= ( ) and fxy = (fx )y = = ( )
x x x yx y x
which are partial derivatives of fx with respect x or y and, similarly the existence of fyy and fyx .
These are called second order partial derivatives of f .

The following example shows that, in general, fxy need not be equal to fyx .
2 2
Example : Let f (x, y) = xy xx2 y
+y 2
if (x, y) 6= (0, 0) and f (0, 0) = 0. Note that

f (h, k) f (h, 0) h(h2 k 2 )

fy (h, 0) = lim = lim =h
k0 k k0 h2 + k 2

and
fy (h, 0) fy (0, 0)
fyx (0, 0) = lim = 1.
h0 h
Similarly, fxy (0, 0) = 1.

Theorem 29.1(Mixed derivative theorem) : If f (x, y) and its partial derivatives fx , fy , fxy
and fyx are defined in a neighborhood of (x0 , y0 ) and all are continuous at (x0 , y0 ) then fxy (x0 , y0 ) =
fyx (x0 , y0 ).

We will not present the proof of this result here. The proof is given in the text book.

Mean Value Theorem : We will present the MVT for functions of several variables which is a
consequence of MVT for functions of one variable.

Theorem 29.2: Suppose f : R2 R is differentiable. Let X0 = (x0 , y0 ) and X = (x0 + h, y0 + k).

Then there exists C which lies on the line joining X0 and X such that

f (X) = f (X0 ) + f 0 (C)(X X0 )

i.e, there exists c (0, 1) such that

f (x0 + h, y0 + k) = f (x0 , y0 ) + hfx (C) + kfy (C) where C = (x0 + ch, y0 + ck).

Proof : Define : [0, 1] R by

(t) = f (x0 + th, y0 + tk), t [0, 1].

Note that by the Chain Rule is differentiable and

dx dy
0 = fx + fy = hfx + kfy .
dt dt
By the MVT, there exist c (0, 1) such that

(1) (0) = 0 (c).

This proves the result.

Remark : In the previous result if we fix X0 and X then it is enough to assume that the function
f is differentiable on the line segment joining X and X0 .
 2

Problem : If f (x, y) is constant if and only if fx = 0 and fy = 0.

We will now take up the extended mean value theorem which we need.

Theorem 29.3(EMVT): Let f, X, X0 be as in the previous theorem. Suppose fx and fy are

continuous and they have continuous partial derivatives. Then, there exists C which lies on the line
joining X0 and X such that
1
f (X) = f (X0 ) + f 0 (X0 )(X X0 ) + (X X0 )f 00 (C)(X X0 )
2

fxx fxy
where f 00 = . That is, there exists c (0, 1) such that
fyx fyy
1
f (x0 + h, y0 + k) = f (x0 , y0 ) + (hfx + kfy )(X0 ) + (h2 fxx + 2hkfxy + k 2 fyy )(C)
2
where C = (x0 + ch, y0 + ck).

Proof (*): Consider the function (t) defined in the proof of the previous result. Since fx and
fy are continuous f is differentiable. Therefore, as given in the proof of the previous theorem, is
differentiable and
dx dy
0 = fx + fy = hfx + kfy .
dt dt
Since fx and fy have continuous partial derivatives, they are differentiable. Denote

0 (t) = hfx (x0 + th, y0 + tk) + kfy (x0 + th, y0 + tk) = F (x0 + th, y0 + tk), t [0, 1].

Again by the Chain Rule,

2f 2f 2f 2f
00 = hFx + kFy = h (hfx + kfy ) + k (hfx + kfy ) = h(h 2 + k ) + k(h + k 2 ).
x y x yx xy y
By the mixed derivative theorem,

00 = h2 fxx + 2hkfxy + k 2 fyy .

By the EMVT for , there exists c (0, 1) such that

00 (c)
(1) = (0) + 0 (0) + .
2
By substituting , 0 and 00 in the above equation we get the result.

Remarks : 1. We will consider f 00 , given in the statement of the previous theorem, as a notation.
We do not say that the function f is twice differentiable.

2. We will recall the EMVT when we will deal with the second derivative test for local maxima
and minima of f (x, y) in the next lecture.

3. Whatever we discussed above can be generalized to the functions of three variables.

4. The matrix given in the statement of the previous theorem is called Hessian matrix. We should be
able to guess what should be the corresponding Hessian matrix for the functions of three variables.

5. Note that we applied the MVT and the EMVT for the function to get the MVT and the
EMVT for f (x, y). Similarly by assuming that f (x, y) has continuous partial derivatives of order
n and applying Taylors theorem for the function , we can obtain Taylors Theorem for f (x, y).