Some questions about differentiability

[athrun200]

We have a corollary that

"If the partial derivatives $f_{x}$ and $f_{y}$ of a function f(x,y) are continuous throughtout an open region R, then f is differentiable at every point of R."(Thomas F. 1994. Calculus. U.S.A. Wesley p824)

But I wonder can we prove a function is not differentiable by showing that $f_{x}$ or $f_{y}$ are not continuous?
i.e. is the converse of this statement true?

By the way, are there any books have a proof on this corollary?
Most of the Calculus book state the corollary of theorem only without prove.

 

[Erland]

We have a corollary that

But I wonder can we prove a function is not differentiable by showing that $f_{x}$ or $f_{y}$ are not continuous?
i.e. is the converse of this statement true?

By the way, are there any books have a proof on this corollary?
Most of the Calculus book state the corollary of theorem only without prove.

The converse is not true. A counterexample is the function defined by f(x,y) = x^2 sin(1/x) for x=/=0 and f(0,y)=0 for all y. For this function, we have fx(x,y) = 2x sin(1/x) - cos(1/x) for x=/=0 and fx(0,y)=0, and fy identically zero. Clearly, fx is not continuous at the y-axis (x=0). But since (x^2 sin (1/x))/ sqrt(x^2 +y^2) tends to 0 as (x,y) tends to (0,0), the function f is differentiable at the origin (and likewise anywhere on the y-axis).

For the proof you wanted, look up any textbook in advanced calculus or real analysis.

 

[Bacle]

Just to comment that the proof is not that hard; the linear approximation is f_xΔx+f_yΔy, i.e.. this is the equation of the plane that is tangent to the surface. Try showing that you can approximate the surface to any degree you want , i.e., in an ε-δ sense, with the tangent plane as described above.

 

[athrun200]

Thx everyone!