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Is the following sentence true assuming that $f$ and $g$ are differentiable and their derivatives are continous? I'd say yes, but don't know how to show it.

$$g(x) \in \Theta(f(x)) \iff \frac{d}{dx} g(x) \in \Theta\left(\frac{d}{dx}f(x)\right)$$

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Hi, Rob, welcome to Math.SE, thanks for your question. – gt6989b Apr 23 '13 at 15:24
Related:… – gt6989b Apr 23 '13 at 15:32

At least one direction fails. Take the function $f(x)=x$, and modify it by adding a bump function that, for every positive integer $n$, is $0$ between $n$ and $n+e^{-n}$, climbs to $1$ at $n+e^{-2n}$, and falls back to $0$ at $n+e^{-3n}$. All derivatives exist, but the first derivative is very wild for large $x$.

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No, Big-Theta control over a function does not imply control over its derivatives. Imagine that $g(x)$ is "oscillating around" the function $f(x)$.

Here's an example.

Suppose that $f(x)$ is the constant function $f(x) \equiv 10$, say. If $g(x)= \sin e^x$, then it is a bounded function, taking values in $[0,2]$. However, its derivative $g'(x) = e^x \cos e^x $ is certainly not $\Theta(f'(x))$.

The same idea works when $f(x)$ is in a far more general class of functions. Suppose that $$g(x)=f(x) + \sin e^x$$ Then (for many functions $f(x)$) $f(x)$ and $g(x)$ are "the same size", but $g(x)$ is increasingly rapidly oscillating around $f(x)$, so we have $$ g'(x) \notin \Theta(f'(x)). $$

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Don't think this is true, here is a construction similar to a related question:

$$\sin x = \Theta \left(\sin\left(x^2\right)\right),$$

but taking derivatives,

$$\cos x \neq \Theta \left(2x \cos \left(x^2\right)\right).$$

This now raises an interesting question: for which classes of functions is this actually true? For one, polynomials are clearly an example...

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What if it is asymptotically positive function, thus not permitting sin and cos ? – Rob Apr 23 '13 at 15:36

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