Math 13 Fall 2004
Calculus of Vector-valued Functions
Example of a function that has
both partial derivatives at (0,0), but is not differentiable
October 4, 2004
Define a scalar-valued function of two variables
| > | f := (x, y) -> x^3 / (x^2 + y^2); |
Have a look at its graph
| > | plot3d(f(x, y), x = -1..1, y = -1..1); |
![[Plot]](Math13-images/NoDerivative_2.gif)
f seems to be continuous everywhere and indeed it is.
But f is not differentiable at (0, 0) (see the handout for details).
Let's plot its partial derivatives
| > | plot3d(diff(f(x, y), x), x = -1..1, y = -1..1);
plot3d(diff(f(x, y), y), x = -1..1, y = -1..1); |
![[Plot]](Math13-images/NoDerivative_3.gif)
![[Plot]](Math13-images/NoDerivative_4.gif)
They are obviously discontinuous!!!
| > | factor(diff(f(x, y), x));
factor(diff(f(x, y), y)); |
| > |