Differentiability for Functions of Two Variables Local Linearity
Recall that when we zoom in on a “sufficiently nice” function of two variables, we see a plane.
What is meant by “sufficiently nice”? Suppose we zoom in on the function z=f(x,y) centering our zoom on the point (a,b) and we see a plane. What can we say about the plane? • The partial derivatives for the plane at the point must be the same as the partial derivatives for the function. • Therefore, the equation for the tangent plane is
In particular. . .The Partial Derivatives Must Exist If the partial derivatives don’t exist at the point (a,b), the function f cannot be locally planar at (a,b). Example: (as given in text) A cone with vertex at the origin cannot be locally planar there, as it is clear that the x and y cross sections are not differentiable there.
Not enough: A Puny Condition Whoa! The existence of the partial derivatives doesn’t even guarantee continuity at the point! Suppose we have afunction • Notice several things: • Both partial derivatives exist at x=0. • The function is not locally planar at x=0. • The function is not continuous at x=0.
Directional Derivatives? It’s not even good enough for all of the directional derivatives to exist! Just take a function that is a bunch of straight lines through the origin with random slopes. (One for each direction in the plane.)
Directional Derivatives? It’s not even good enough for all of the directional derivatives to exist! Locally Planar at the origin? What do you think?
Directional Derivatives? If you don’t believe this is a function, just look at it from “above”. There’s one output (z value) for each input (point (x,y)).
Differentiability The function z = f(x,y) is differentiable (locally planar) at the point (a,b) if and only if the partial derivatives of f exist and are continuous in a small disk centered at (a,b).
Differentiability: A precise definition A function f(x,y) is said to be differentiable at the point (a,b) provided that there exist real numbers m and n and a function E(x,y) such that for all x and y and
E(x) for One-Variable Functions E(x) measures the vertical distance between f (x) and Lp(x) But E(x)→0 is not enough, even for functions of one variable! What happens to E(x) as x approaches p?