Subsection 2.2.3 Where is the Derivative Undefined?
According to Definition 2.2.1, the derivative \(f'(a)\) exists precisely when the limit \(\lim\limits_{x\rightarrow a} \frac{f(x)-f(a)}{x-a}\) exists. That limit is also the slope of the tangent line to the curve \(y=f(x)\) at \(x=a\text{.}\) That limit does not exist when the curve \(y=f(x)\) does not have a tangent line at \(x=a\) or when the curve does have a tangent line, but the tangent line has infinite slope. We have already seen some examples of this.
- In Example 2.2.7, we considered the function \(f(x)=\frac{1}{x}\text{.}\) This function “blows up” (i.e. becomes infinite) at \(x=0\text{.}\) It does not have a tangent line at \(x=0\) and its derivative does not exist at \(x=0\text{.}\)
- In Example 2.2.10, we considered the function \(f(x)=|x|\text{.}\) This function does not have a tangent line at \(x=0\text{,}\) because there is a sharp corner in the graph of \(y=|x|\) at \(x=0\text{.}\) (Look at the graph in Example 2.2.10.) So the derivative of \(f(x)=|x|\) does not exist at \(x=0\text{.}\)
Here are a few more examples.
Visually, the function
\(H(x) = \begin{cases} 0 & \text{if }x \le 0 \\ 1 & \text{if }x \gt 0 \end{cases}\)
does not have a tangent line at \((0,0)\text{.}\) Not surprisingly, when \(a=0\) and \(h\) tends to \(0\) with \(h \gt 0\text{,}\)
\begin{gather*}
\frac{H(a+h)-H(a)}{h}
=\frac{H(h)-H(0)}{h}
=\frac{1}{h}
\end{gather*}
blows up. The same sort of computation shows that \(f'(a)\) cannot possibly exist whenever the function \(f\) is not continuous at \(a\text{.}\) We will formalize, and prove, this statement in Theorem 2.2.14, below.
Example 2.2.12 \(\diff{}{x}x^{1/3}\)
Visually, it looks like the function \(f(x) = x^{1/3}\text{,}\) sketched below, (this might be a good point to recall that cube roots of negative numbers are negative — for example, since \((-1)^3=-1\text{,}\) the cube root of \(-1\) is \(-1\)),
has the \(y\)–axis as its tangent line at \((0,0)\text{.}\) So we would expect that \(f'(0)\) does not exist. Let's check. With \(a=0\text{,}\)
\begin{align*}
f'(a)\amp= \lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}
=\lim_{h\rightarrow 0}\frac{f(h)-f(0)}{h}
=\lim_{h\rightarrow 0}\frac{h^{1/3}}{h}\\
\amp=\lim_{h\rightarrow 0}\frac{1}{h^{2/3}}
=DNE
\end{align*}
as expected.
Example 2.2.13 \(\diff{}{x}\sqrt{|x|}\)
We have already considered the derivative of the function \(\sqrt{x}\) in Example 2.2.9. We'll now look at the function \(f(x) = \sqrt{|x|}\text{.}\) Recall, from Example 2.2.10, the definition of \(|x|\text{.}\)
When \(x \gt 0\text{,}\) we have \(|x|=x\) and \(f(x)\) is identical to \(\sqrt{x}\text{.}\) When \(x \lt 0\text{,}\) we have \(|x|=-x\) and \(f(x)=\sqrt{-x}\text{.}\) So to graph \(y=\sqrt{|x|}\) when \(x \lt 0\text{,}\) you just have to graph \(y=\sqrt{x}\) for \(x \gt 0\) and then send \(x\rightarrow -x\) — i.e. reflect the graph in the \(y\)–axis. Here is the graph.
The pointy thing at the origin is called a cusp. The graph of \(y=f(x)\) does not have a tangent line at \((0,0)\) and, correspondingly, \(f'(0)\) does not exist because
\begin{gather*}
\lim_{h\rightarrow 0^+}\frac{f(h)-f(0)}{h}
=\lim_{h\rightarrow 0^+}\frac{\sqrt{|h|}}{h}
=\lim_{h\rightarrow 0^+}\frac{1}{\sqrt{h}}
=DNE
\end{gather*}
Theorem 2.2.14
If the function \(f(x)\) is differentiable at \(x=a\text{,}\) then \(f(x)\) is also continuous at \(x=a\text{.}\)
Proof
The function \(f(x)\) is continuous at \(x=a\) if and only if the limit of
\begin{gather*}
f(a+h) - f(a) = \frac{f(a+h)-f(a)}{h}\ h
\end{gather*}
as \(h\rightarrow 0\) exists and is zero. But if \(f(x)\) is differentiable at \(x=a\text{,}\) then, as \(h\rightarrow 0\text{,}\) the first factor, \(\frac{f(a+h)-f(a)}{h}\) converges to \(f'(a)\) and the second factor, \(h\text{,}\) converges to zero. So the product provision of our arithmetic of limits Theorem 1.4.3 implies that the product \(\frac{f(a+h)-f(a)}{h}\ h\) converges to \(f'(a)\cdot 0=0\) too.
Notice that while this theorem is useful as stated, it is (arguably) more often applied in its contrapositive 7 If you have forgotten what the contrapositive is, then quickly reread Footnote 1.3.5 in Section 1.3. form:
Theorem 2.2.15 The contrapositive of Theorem 2.2.14
If \(f(x)\) is not continuous at \(x=a\) then it is not differentiable at \(x=a\text{.}\)
As the above examples illustrate, this statement does not tell us what happens if \(f\) is continuous at \(x=a\) — we have to think!