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Is the following statement true?

If the derivative of a function $f(x)$ is a product $g(x)\cdot \dfrac{1}{h(x)}$ where $g$ and $h$ are continuous functions, then the function $f(x)$ is a differentiable continuous function.

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    @bemyguest: The "basic-standard-differentiation rules" (by which I assume you mean *formulas* and *theorems*) either include prerequisites of differentiability or else *yield* conclusions of differentiability. You seem to be either using rules without understanding them, or trying to put the cart in front of the horse. Perhaps you can be more specific about your problem, instead of trying to be too abstract?2012-05-13

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As already mentioned in the comments above, the statement/question doesn't really make sense. When you talk about the derivative of $f$, then you are assuming that the function is actually differentiable.

Example: Now, one might think of the example of the sign function $ \text{sgn}(x) = \begin{cases}-1 &\text{ if } x < 0 \\ 0 &\text{ if } x = 0 \\ 1 &\text{ if } x >0 \end{cases} $ (Note that one might not say that the sign function is defined at $0$).

Here the derivative of the function exists for all real numbers except zero. And the derivative is zero, and so you could say that the derivative is equal to the product of two continuous functions (ex the zero function and $\frac{0}{1}$), but again, one of those functions (for example the $g$ in your case), would not be defined at zero. So the sign function isn't really a counter example to your statement. Remember that included in the definition of a function is a domain.