# Clarifications on differentiability and continuity

If there is a function $$f(x)$$ such that f'(x) has a jump discontinuity:

$$f(x)=\begin{cases} 2x & 0 \leq x \leq 1 \\ 3x-1 & 1 < x \leq 2 \\ \end{cases}$$

then what is the derivative at $$x=1$$?

The graph looks like this: (orange is $$f(x)$$ and purple is $$f'(x)$$) • $f$ is not differentiable at $1$, $f'$ does not exist at $1$ Sep 23 at 7:34
• @RobertZ So can I say that if $f'$ is not conitnuous at c, $f$ will not be differentiable at c? Sep 23 at 7:36
• No, there are functions that are differentiable but have a discontinuous derivative Sep 23 at 7:57
• Why dont you use the definition of derivative at 1? Sep 23 at 7:57
• See the “basic example” in Mark McClure’s answer to this question: math.stackexchange.com/questions/292275/… It is the classical example of a function $f$ such that $f$ is differentiable (at $c$) but $f’$ is discontinuous (at $c$). Sep 23 at 8:06

We know that by definition,

$$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$

so if the limit fails to exist, the derivative must also fail to exist. (as in, $$f$$ is not differentiable at $$x.$$)

Recall also the concept of one-sided limits: $$\lim_{x \to a} g(x)$$ exists if and only if the one-sided limits $$\lim_{x \to a^-} g(x)$$ and $$\lim_{x \to a^+} g(x)$$ exist and agree.

So, let's see what happens when we apply this to our limit definition of $$f'(1),$$ taking limits from both sides.

Approaching from the left, $$h < 0$$ implies $$1 + h < 1,$$ so:

$$\lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^-} \frac{2(1 + h) - 2}{h} = \lim_{h \to 0^-} \frac{2h}{h} = 2$$

Similarly, approaching from the right, $$h > 0$$ implies $$1 + h > 1,$$ so:

$$\lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^-} \frac{(3(1 + h) - 1)- 2}{h} = \lim_{h \to 0^-} \frac{3h}{h} = 3$$

So, because $$2 \neq 3,$$ the two one-sided limits don't agree, so the limit fails to exist and $$f$$ is not differentiable at $$1.$$

That said, this doesn't mean that all differentiable functions have continuous derivatives: see here for a counterexample.