Is 'a' differentiable in f when f is a product of a differentiable and non-differentiable function? Recently, I was studying differentiable and non-differentiable functions and I wondered whether this "conjecture" of mine is true:
1)
"If $f(x)$ is a function that is the product of $g(x)$ and $h(x)$ where $g(x)$ is differentiable at point "a" and $h(x)$ is not, then, by the product rule, $f(x)$ is not differentiable at point $x=a$. "
However, I then endeavoured to apply this theorem to rational functions, sum functions and other functions and quickly I saw that the same "conjecture" holds true for my applied cases. So, I then made a new conjecture:
2) "If $f(x)$ is a function combining the functions $g(x)$ and $h(x)$ by some operation, and $g(x)$ is differentiable at point $x=a$ and $h(x)$ is not, then $f(x)$ is not differentiable at point $x=a$".
My first endeavour to prove this "new conjecture" was to simply check the validity of it in all the different possible operations, that is, addition, subtraction Et Cetera, where, to speed the process, I could prove that if the conjecture holds for one operation then it must also hold for the inverse of that operation. But, I simply do not have the time to devote myself to such things, though it may be not at all time-consuming a task, and so I tried to find a contradiction of this conjecture but I was unable to. So, my question is, can anyone hint me at a contradiction or at a method of proof as time-consuming as a proof by contradiction, but before all that tell me whether this conjecture is even valid, been spoken of, or simply downright idiotic. 
(Note: it would appreciated if my lack of mathematical terminology could be fixed by editing when necessary)
(Note: It would be helpful to avoid the trivial case where $x=0$)
 A: What you can say is that $f$ may not be differentiable. For example, take $g(x)=0$, and $h(x)=|x|$. Then clearly $f(x)=g(x)h(x)=0$ is differentiable.
What you can say, though, is the following:

If $g$ is differentiable in $x$ and $g'(x)\ne 0$, then $g(x)h(x)$ is differentiable if and only if $h(x)$ is.

If you want something more general than the product, suppose that:
$$
f(x) = F(g(x),h(x))
$$
for a differentiable (2-variable) function $F$. Then if $g,h$ are differentiable, $f$ is differentiable, and:
$$
f'(x) = \frac{\partial F}{\partial g}\frac{dg}{dx} + \frac{\partial F}{\partial h}\frac{dh}{dx},
$$
which implies that:

if $F$ above is differentiable, and $h$ is not (at a point x), then $f'$ is differentiable if and only if:
  $$
\frac{\partial F}{\partial h} = 0.
$$

A: You consider $f(x)=F(g(x),h(x))$ where $F$ may be one of several "standard operations" and wonder if it is possible to have $f,g$ differentiable at $a$ and $h$ not.
For $F(u,v)=u+v$ your conjecture is ture because we obtain that $h(x)=f(x)-g(x)$ is the difference of differentiable functions. Similarly for $F(u,v)=u-v$.
For $F(u,v)=uv$ we have the trivial counterexample $g(x)=0$ for all $x$. But also $g(a)=0$ with $h$ continuous (except perhaps at $a$) is a counterexample.
However, if additionally we have $g(a)\ne 0$, then $h(x)=\frac{f(x)}{g(x)}$ is differentiable by the quotient rule.
The argument for $F(u,v)=\frac uv$ (with $h(a)\ne0$) is similar.
For $F(u,v)=\frac vu$ we find immediately that $h(x)=f(x)g(x)$ is differentiable.
Another interstting question would be about $f(x)=g(h(x))$ or $f(x)=h(g(x))$. For these $h$ can be quite wild unless $g$ is injective near $a$.
