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 $f'(x)$)


  • 6
    $\begingroup$ $f$ is not differentiable at $1$, $f'$ does not exist at $1$ $\endgroup$
    – Robert Z
    Sep 23 at 7:34
  • 1
    $\begingroup$ @RobertZ So can I say that if $f'$ is not conitnuous at c, $f$ will not be differentiable at c? $\endgroup$ Sep 23 at 7:36
  • $\begingroup$ No, there are functions that are differentiable but have a discontinuous derivative $\endgroup$ Sep 23 at 7:57
  • $\begingroup$ Why dont you use the definition of derivative at 1? $\endgroup$ Sep 23 at 7:57
  • $\begingroup$ 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 $c$). $\endgroup$ Sep 23 at 8:06

1 Answer 1


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.


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