Differentiability

In the last learning module we saw how to compute the tangent plane to a surface $z=f(x,y)$ at $(a,b,f(a,b))$ by using partial derivatives. Of course, that assumed that the partial derivatives existed, and that the surface was smooth enough for the tangent plane to make sense! In this module we'll examine what that means.

We begin with functions of one variable. We say a function $f(x)$ is differentiable at $x=a$ if the limit $$\lim_{x\to a} \frac{f(x)-f(a)}{x-a}$$ exists. However, the equation $$\lim_{x\to a} \frac{f(x)-f(a)}{x-a}=c$$ means exactly the same thing as $$\lim_{x \to a} \frac{f(x)-f(a)-c(x-a)}{x-a}=0.$$ In other words, a function $f(x)$ is differentiable at $a$ if and only if there is a linear function $$L(x) = f(a) + c(x-a)$$ such that $$\lim_{x \to a} \frac{f(x)-L(x)}{x-a}=0.$$ That last expression generalizes naturally to functions of two (or more) variables.

A function $f(x,y)$ is differentiable at $(a,b)$ if there is a linear function $$L(x,y)=f(a,b) + c_1 (x-a) + c_2 (y-b)$$ such that $$\lim_{(x,y) \to (a,b)} \frac{|f(x,y)-L(x,y)|}{\|\langle x-a, y-b \rangle \|}=0.$$

You may wonder why we didn't just say `a function is differentiable at $(a,b)$ if its partial derivatives exist at $(a,b)$?' Because it isn't true! If a function is differentiable, then its partial derivatives exist (and equal $c_1$ and $c_2$), but a function can have partial derivatives without being differentiable!

Example 1: Compute the partial derivatives of $$f(x,y) = \begin{cases} \frac{xy}{x^2+y^2} & (x,y) \ne (0,0); \cr 0 & (x,y)=(0,0). \end{cases}$$ at $(x,y)=(0,0)$. Solution: Since $f(h,0)=0$ for all $h$,

$$f_x(0,0) = \lim_{h \to 0} \frac{f(h,0)-f(0,0)}{h} = \lim_{h \to 0}\frac{0}{h}=0.$$ Similarly, $f(0,h)=0$ for all $h$, so $$f_y(0,0) = \lim_{h \to 0} \frac{f(0,h)-f(0,0)}{h} = \lim_{h \to 0}\frac{0}{h}=0.$$

Example 2: Is $$f(x,y) = \begin{cases} \frac{xy}{x^2+y^2} & (x,y) \ne (0,0); \cr 0 & (x,y)=(0,0). \end{cases}$$ differentiable at $(x,y)=(0,0)$?

Solution: No. $f(x,y)$ isn't even continuous at $(x,y)=(0,0)$, since $f(x,y)=\frac{1}{2}$ whenever $x=y\ne 0$ and $f(x,y)=\frac{-1}{2}$ whenever $x = -y \ne 0$. The function is well-behaved on the $x$-axis and the $y$-axis, so it has partial derivatives, but it behaves badly in every other direction.

So for a function to be differentiable, we need more than just the existence of partial derivatives. We need continuity of partial derivatives:

Theorem: Suppose that $f(x,y)$ is continuous on a region $D$ containing $(a,b)$, and that $f_x$ and $f_y$ exist and are continuous on $D$. Then $f(x,y)$ is differentiable at $(a,b)$.

The composition of differentiable functions is differentiable. In particular, if $x(t)$ and $y(t)$ are differentiable functions of a parameter $t$, and if $f(x,y)$ is a differentiable function of $x$ and $y$, then $f(x(t),y(t))$ is a differentiable function of $t$. In the next slide, we'll compute its derivative.