It is possible that both `f` and `g` are differentiable e.g. `f(x)=xe^x` and `g(x)=e^(-x).` But it's **not necessary**. One counter example would be

`f(x)=sgn(x)={(1, if x>0),(0, if x=0),(-1, if x<0):}`

`g(x)=|x|={(x, if x>=0),(-x, if x<0):}`

It is easy to see that `f(x)g(x)=x`, but neither `f` nor `g` are differentiable at 0. More so `f` isn't even continuous at 0.

Graph shows graphs of functions `f` (red) and `g` (blue).

It is true that `f(0)=g(0)=0` but if we change definition of `g` like this

`g(x)={(1, if x>=0),(-1, if x<0):}`

then we would still have `f(x)g(x)=x` but `g(0)=1`

**To sum it up if** `f(x)g(x)=x` **it's possible but not necessary for either** `f` **or** `g` **to be differentiable** or** that** `f(0)=g(0)=0.`

This is false. If `f(x)g(x)=x,` and `f` and `g` are differentiable, then taking the derivative of both sides using the product rule results in

`f'(x)g(x)+f(x)g'(x)=1` for all `x.`

But if `f(0)=g(0)=0,` then the equality doesn't hold for `x=0,` which contradicts what was just said.

The last two sentences in the question seem a little mixed up. An example isn't enough to show that the statement is false (an example where it is possible obviously proves the statement is true, however). Only a proof can show that the statement is false.

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