First of all, I think that before stating the general question it would be okay to make some concrete example of what do I have in mind.
Let us take the function $f(x)=|x|$.
We could write this function as $f(x)= \begin{cases}x&{x>0}\\-x&x<0 \\ 0& {x=0} \end{cases}$.
This function is differentiable everywhere except at the point $x=0$.
Now let us multiply this function with, say, function $g(x)=x$.
Now we have $f(x)g(x)=\begin{cases}{x^2}&{x>0}\\{-x^2}&{x<0}\\ 0&{x=0} \end{cases}$.
Clearly $f(x)g(x)$ is differentiable everywhere.
What did we do here?
We multiplied function which was not differentiable at one point with some other function which is of class $C^{\infty}$ and we obtained function which is everywhere differentiable.
Now I would like to ask the question which deals with the general case:
Suppose that we have some real function of a real variable $f$ that is continuous on some set $(a,b)$ and that is not differentiable on some subset of $(a,b)$ (the subset could be only one point as in the above described example or it could be the whole set $(a,b)$ so that we have everywhere continuous but nowhere differentiable function). Could it be that there always exists some function $g$ (which could depend on $f$) of class $C^{\infty}$ (function that is infinitely times differentiable) which is not the zero function (so $g$ is not the function $g(x)=0$) and which is such that we have that the function $fg$ (the product of the functions $f$ and $g$) is differentiable on the set $(a,b)$?