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Are there cases when this relation holds? $$f(x) \gt g(x) \implies f'(x) \gt g'(x)$$

I.e. what are the conditions on $f(x)$ and $g(x)$ for that to be true? Is it even possible to determine them? In case it is always valid, how can it be proved?

rubik
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    There are cases where this holds, but off the top of my head, I can't think of any conditions more interesting than "I computed the values of the functions and their derivatives. Turned out it held!" – user2357112 May 23 '14 at 13:12
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    a slightly different situation: you can prove that $f(x)>g(x)$ for $\textit{some} \ x$ using the fact that $f'(x) -g'(x)>0$ – Alex May 23 '14 at 13:14
  • @Alex I don't think so. What about a horizontal line $y=0$ and a strictly negative function with strictly positive derivative which, say, approaches $-\infty$ as $x\to -\infty$ and approaches $y=0$ from below as $x\to \infty$. There must be examples of such functions. – Seth May 23 '14 at 13:23
  • Seth, I never said you can do that $\forall \ f,g$. But $\exists \ f,g$ for which this is provable. Take e.g. $f=e^{-x}, \ g = 1-x, \ x_0 =0$ and $x \in [0, \infty)$ – Alex May 23 '14 at 13:27
  • the converse is more likely to be true, just depending on the initial conditions. – Denis May 23 '14 at 13:28
  • Let us look for the class of differentiable functions for which $f(x)\geq 0$ implies $f'(x))\geq 0$. Constants work but no other polynomials because all degree one polynomials are lines and so 'go negative'. Certainly exponential polynomials with positive coefficients fit the bill. – JP McCarthy May 23 '14 at 13:34
  • How about looking at solutions of odes of the form $y'=F(y)$ with $F$ a suitably nice, positive real valued function? – JP McCarthy May 23 '14 at 13:38
  • @becko arctan x and y=π/2? Why do people keep saying this? – Seth May 23 '14 at 13:52
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    @becko: $(\arctan x)' > (\pi/2)'$ for all $x$, but it is never true that $\arctan x > \pi / 2$. – user2357112 May 23 '14 at 21:58
  • @Seth deleting comments – a06e May 23 '14 at 22:18

4 Answers4

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Take any two functions $f,g$ that are bounded on the interval $[a,b]$. By adding a constant to one of them, you can make one always greater than the other, or always less than the other, and the derivatives are unchanged. So the answer is that the derivatives have nothing to do with which function is greater.

vadim123
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  • That's indeed right. After some thinking, it's clear that it cannot be always true. But my question was more like "can we find some conditions under which that my hold or is it never true"? – rubik May 23 '14 at 16:00
  • Well I can think of one example when the relation holds: $f(x) = e^x$ and $g(x) = 0$. But since you and others have made clear that it does not hold in general, I can conclude that it holds for a few specific cases. – rubik May 23 '14 at 16:08
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Let $h(x)=f(x)-g(x)$, What you claim is $$h(x)\gt 0\implies h'(x)\gt 0 $$ which clearly is not true. For example, if $h(x)=x^2+1 $, then $h(x)$ is always positive while $$h'(x)=2x$$ is not.

Fermat
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It is clearly not true in general. Take $f\colon x\in\mathbb{R}\mapsto 3$ and $g\colon x\in\mathbb{R}\mapsto \sin x$.

Clement C.
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I think,when $f(x)$ and $g(x)$ are power functions ,on the interval $(1,+\infty)$,if the power of $f(x)$ is larger $g(x)$,we may conclude $f'(x) \gt g'(x)$

user152441
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