# Relationship between continuous and differentiable functions

Firstly, I am wondering about a special case of the non differentiability of a discontinuous function along some interval. If the function is simply undefined at some point it is not a function and thus not differentiable at that point. But suppose it is defined and there is a jump discontinuity by a constant $C$. Then the derivative at that point and everywhere around that point is continuous and defined. So is this a discontinuous function with a continuous derivative?

Secondly, I have found some instances where a function is differentiable, continuous but its derivative is not continuous. For example, the node of a cusp on the y axis will have a unique derivative but the derivative function is not continuous. Or another example, $y = x, a < x < c$ $y=2x, c \leq x < b$. The derivative is jump discontinuous because it shifts up at c, but it is not undefined at c. Are these valid examples? I'm guessing it is possible to integrate jump discontinuous functions.

• Why do you say that if there is a discontinuity, then the derivative is defined there? Apr 21, 2017 at 7:57
• " But suppose it is defined and there is a jump discontinuity by a constant CC. Then the derivative at that point and everywhere around that point is continuous and defined." Apr 21, 2017 at 7:59
• Both your examples are invalid. You have to understand the meaning of continuity and differentiability properly by looking at definitions. Once you get that you will know that a function which is differentiable is automatically continuous. So that if the function has a jump discontinuity at a point the function can't be differentiable at that point. Apr 21, 2017 at 8:00
• oh yes, sorry. well at the point of discontinuity, the derivative exists because we can only add positive $dx$. It doesn't matter how small $dx$ gets does it? Apr 21, 2017 at 8:00
• @gebra see my answer below as to why the derivative does not exist at the point of discontinuity Apr 21, 2017 at 8:58

If a function is 'jumps up' at a point $x$, its derivative can't be defined there. Suppose we have a function which is equal to $a$ at $x$, and $b$ just before $x$, with $a\neq b$.
Now, use the definition of derivative: $$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.$$ If we work out $$\lim_{h\to 0^-}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0^-}\frac{b-a}{h}=\infty$$ we see that the derivative is not defined there.