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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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    You're assuming that $f$ has a derivative, so of course $f$ is differentiable. All differentiable functions are continuous.2012-05-13
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    Is your desired conclusion perhaps that $f$ is differentiable *everywhere*? If so, the answer is that it is differentiable exactly where $h(x)\neq 0$.2012-05-13
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    Ok. So If you use the basic-standard-differentation rules for a function and you get a meaningfull result (not dividing with zero), then you can deduce that the function is a differentiable continuous function?2012-05-13
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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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