More on differentiability

The definition of a function being differentiable is the following lovely statement:

A function f : Rⁿ $\rightarrow$ R^m is differentiable at the point x = a if there exists a linear function L : Rⁿ $\rightarrow$ R^m so that

$\displaystyle \lim_{{\mathbf{x} \rightarrow \mathbf{a}}}^{}$ $\displaystyle {\frac{{ \Vert \mathbf{f}(\mathbf{x}) - \mathbf{f}(\mathbf{a}) -\mathbf{L}(\mathbf{x}-\mathbf{a})\Vert}}{{\Vert\mathbf{x}-\mathbf{a}\Vert}}}$ = 0.

The matrix associated with the linear function L is the partial derivative matrix Df(a).

In case you don't find that above definition intuitively obvious, you can read on to gain some more intuition about what differentiability means.

Recall that for a limit in one-dimension to exist, the limit from the right and the left have to be the same.

However, the above limit as x $\to$ a is a limit in Rⁿ. Since it is too hard to think in n-dimensions, let's just think of two dimensions, R². If you can understand the differentiability condition in two dimensions, that's good enough for now. As you know from the previous reading, in two dimensions, differentiability at a point means the existence of a tangent plane. (I realize that it is not so obvious why the above limit definition of differentiability implies the existence of a tangent plane. But, just take it by faith and move on.)

For the limit in two dimensions to exist, the limit must be the same as x approaches a along any path, such as the paths sketched below. Since there are an infinite number of paths (rather than just two paths for the one-dimensional case), you can see that the two-dimensional case is more complicated.

The important consequence of this fact is that the existence of a derivative is much stronger than the existence of partial derivatives. Partial derivatives involve the limit of f only along directions parallel to the coordinate axes (in the above picture, such limits would be arrows coming straight in from the left, right, top, and bottom).

If a function varies smoothly along the paths from coming in to a from positive and negative x direction, the partial derivative with respect to x at the point a will exist. If a function varies smoothly along the paths coming in from the positive and negative y direction, the partial derivative with respect to y at the point a will exist.

But just because the function behaves "nicely" along those four directions, it doesn't mean it behaves nicely along every path coming into a.

For example, look at the following surface. At the origin (i.e., a = (0, 0)), the partial derivatives exist and are zero. (If one moves in the positive or negative x or y direction, the function is constant.)

However, if you approach the origin from a path coming from any other direction, the slope of the path will be nonzero. You'll be climbing uphill as you reach the origin.

If the function were differentiable at the origin, it would have a tangent plane at the origin. (The nice limit definition says that the slope of all paths would approach that from the tangent plane.)

If a tangent plane existed, the slopes would have to match the partial derivatives (as discussed earlier). In this case, since the partial derivatives are zero, the only option for the tangent plane is the horizontal plane, as shown below. Clearly this plane is not a tangent plane, however, as it is not tangent to paths approach the origin from all directions. This surface is a graph of a function that has partial derivatives at the origin but is not differentiable at the origin.

For the above figures, if you drag the blue dot on the sliders, you can look at other surfaces that are not differentiable. The partial derivatives at the origin do not change, so the candidate for the tangent plane is still the horizontal. However, you can see that the horizontal plane is not a tangent plane.

Note that for these functions, you could write down the matrix of partial derivatives. It would simply be [0 0]. But that matrix would not correspond to the derivative.

The following is another example function that has partial derivatives at the origin but is not differentiable. For this example, we have an equation for the function.

f (x, y) = $\begin{displaymath}\begin{cases} \displaystyle \frac{x^2y}{x^2+y^2} & \text{if } (x,y) \ne (0,0) 0 & \text{if } (x,y) = (0.0) \end{cases}\end{displaymath}$

We can show partial derivatives exist at (0,0) but that function is not differentiable at (0,0). We have to use the limit definition,:

$\displaystyle {\frac{{\partial f}}{{\partial x}}}$ (0, 0)

= $\displaystyle \lim_{{h \rightarrow 0}}^{}$ $\displaystyle {\frac{{f(0+h,0)-f(0,0)}}{{h}}}$

$\displaystyle {\frac{{\partial f}}{{\partial x}}}$ (0, 0)	= $\displaystyle \lim_{{h \rightarrow 0}}^{}$ $\displaystyle {\frac{{f(0+h,0)-f(0,0)}}{{h}}}$
	= $\displaystyle \lim_{{h \rightarrow 0}}^{}$ $\displaystyle {\frac{{0- 0}}{{h}}}$ = $\displaystyle \lim_{{h \rightarrow 0}}^{}$ 0 = 0

Similarly, you can show that $\displaystyle {\frac{{\partial f}}{{\partial y}}}$ (0, 0) = 0.

If the function had a tangent plane at (0,0), then it would have to be the flat plane with equation z = f (0, 0) = 0.

As you may see from graph, if approached the origin from a diagonal, the tangent to the path would not be in the plane z = 0.