We know that there exist some functions $f(x)$ such that their derivative $f'(x)$ is strictly greater than the function itself. for example the function $5^x$ has a derivative $5^x\ln(5)$ which is greater than $5^x$. Exponential functions in general are known to be proportional to their derivatives. The question I have is whether it is possible for a function to grow "even faster" than this. To be more precise let's take the ratio $f'(x)/f(x)$ for exponential functions this ratio is a constant. For most elementary functions we care about, this ratio usually tends to 0. But are there functions for which this ratio grows arbitrarily large? If so, is there an upper limit for how large the ratio $f'(x)/f(x)$ can grow? I also ask a similar question for integrals.

  • "Arbitrarily large" where? Are you taking a limit as $x\to\infty$, or some finite interval, as in Ross Millikan's answer? – mr_e_man Jun 13 at 4:52
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    By the chain rule, $f'(x)/f(x) = \bigl(\log f(x)\bigr)'$. Thus your question translates to "are there functions such that the derivative of the logarithm grows arbitrarily large?" – jochen Jun 13 at 14:37
  • @jochen which is of course saying that there are functions whose derivative grows arbitrarily fast, by letting $\log(f) \to f$ – Brevan Ellefsen Jun 13 at 22:29
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    Note the ratio you're talking about may not be a good way of measuring if "the derivative of a function grows arbitrarily faster than the function itself". Consider $f(x)\equiv x$. The derivative doesn't grow at all, the function itself grows in a constant rate. Still the ratio can be as large as you want if only you get close enough to $0$ from the positive side – not because the numerator gets large but because the denominator gets tiny. – Kamil Maciorowski Jun 13 at 22:33
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    Take the derivative of a function that grows slower than $e^x$, and you get a function that grows even slower. Take a function that grows faster than $e^x$ and you get a function that grows even faster. Functions that grow like $e^x$ are the fixed point. – Nick Alger Jun 14 at 4:41
up vote 102 down vote accepted

Consider the differential equation $$ \frac{f'}{f} = g $$ where $g$ is the fast-growing function you want. For instance, for $g(x) = e^x$ (and say the initial condition $f(0) =1$) you get $$f(x) = e^{e^x-1} $$ The ratio $f'/f$ grows arbitrarily large.

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    If $g$ is a continuous function defined on all of $\mathbb{R}$, will this differential equation also define a total function on $\mathbb{R}$? This reminds me of other differential equations which fail to define total functions, such as $y' = y^2$. – Tanner Swett Jun 13 at 15:14
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    @TannerSwett $f(x) = Ae^{\int g(x)}$ for some constant $A$. So, yes, it is defined on all of $\Bbb R$ in that case. – Paul Sinclair Jun 13 at 16:16
  • Isn't $f$ the function you want, rather than $g$? – Ovi Jun 14 at 3:31
  • @Ovi g is the fast-growing function you want (of your choice) for the ratio. – Clement C. Jun 14 at 3:56

You can always try functions of the form $f(x) = e^{g(x)}$, where $g(x)$ is an antiderivative of a function with large growth. For example, if $f(x) = e^{e^x}$, then $$\frac{f'(x)}{f(x)} = e^x,$$ which is, of course, exponential.

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    +1, but would be nice and probably instructive if you could also show how to derive $f(x) = e^{g(x)}$ (how to start and separate the variables etc.), since right now it looks like you kind of pulled that rabbit out of a hat. – Mehrdad Jun 13 at 22:24

Even $f(x)=\frac 1x$ has derivative $\frac {-1}{x^2}$ and second derivative $\frac 2{x^3}$ which grow arbitrarily faster than $f(x)$ as $x \to 0$. The ratio $\frac {f'(x)}{f(x)}=-\frac 1x$ which is not bounded as $x \to 0$. The important message is that derivatives accentuate short range changes, so if you have a function that changes quickly in a short distance the derivative is large.

I don't understand what "a similar question for integrals" means. Integrals are smoothing functions, so the integral of a function can't grow faster than the function times the length of the integral.

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    I think OP is talking about the antiderivative rather than a definite integral. – MackTuesday Jun 13 at 22:09

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