# If a function is smooth over an Interval, does that mean that the function is differentiable over that interval?

Wolfram Mathworld gives a clear definition:

Smooth Function

A smooth function is a function that has continuous derivatives up to some desired order over some domain.

But I don't see it any different than defining differentiablitty for the function, after all if a function is differentiable over an Interval then its derivative must be continuous over that interval. I have tried this on multiple functions and it proves to be true, In this case I recognise this as a better definition for differentiablitty over an interval than the one liner 'Differentiable at every point though the interval'.

• The derivative of a differentiable function need not be continuous. A popular counter-example is $f(x)=x^2\sin\frac{1}{x}$ for $x\neq0$ and $f(0)=0$. May 2, 2019 at 20:30
• I have mostly used "smooth" to mean "differentiable indefinitely" (often incorrectly called "infinitely many times", including by me), which in particular means that any derivative must be continuous as it is differentiable. May 2, 2019 at 21:10

Smoothness, in the Wolfram definition, is more properly referred to as a function being $$C_n$$, where $$n$$ is the order of derivative that is continuous. For example, the function mentioned by Thorgott, which has its first derivative discontinuous, would be $$C_0$$, since it does not have a continuous first derivative, while a function with a discontinuous second derivative but a continuous first derivative would be $$C_1$$.
It's a term used for convenience. One big place it's used is when you are doing a proof that relies on some property of the derivative. For instance, in checking whether an extreme point is a min or a max using the second derivative test, you require the function to have a second derivative that is continuous at the point you are interested in. We can say that it is $$C_n$$, "smooth to second order", or "twice differentiable". All are equivalent.
One particularly useful class of smooth functions is $$C_\infty$$, the functions with infinite continuous derivatives. To continue my example of an application, these functions can be analyzed up to any order derivative you want and you'll still have differentiability. So for example, if you are working on a proof that requires high order derivatives and you aren't sure if it works at all, you might as well assume your function is smooth in the first place and see if it does. It takes away the need to define special cases when the derivatives aren't continuous.
• Thanks Micheal, for your precise explanation. I along with Wolfram Alpha and Pre-Calculus actually didn't quite agree with Thorgott's example, this function isn't defined for f(0) by definition, I know that plugging zero gives zero but so does plugging infinity in $Tanh(x)-1$ you still don't include infinity in the domain, I think saying that smoothness indicates differentiablitty is valid atleast for all elementary compositions, and it does make checking for differentiablitty much easier, sometimes its needed in proves but isn't a prerequisite anyway atleast that's the case for Arc length. May 3, 2019 at 7:17