# Functions that are always less than their derivatives

I was wondering if there are functions for which $$f'(x) > f(x)$$ for all $x$. Only examples I could think of were $e^x - c$ and simply $- c$ in which $c > 0$. Also, is there any significance in a function that is always less than its derivative?

Edit: Thank you very much for all the replies. It seems almost all functions that apply are exponential by nature... Are there more examples like - 1/x?

Again are there any applications/physical manifestations of these functions? [for example an object with a velocity that is always greater than its position/acceleration is always greater than its velocity]

• Off the top of my head, any bounded, monotonically increasing function in the bottom half-plane. – BallpointBen Jul 28 '17 at 21:43
• Ixion's answer gives the full, most general solution (though some particular families of solutions might be writable in nicer forms), and should be accepted. – Robin Saunders Jul 29 '17 at 9:56
• +1! But please fix the title, changing "its" to "their". The way the title is written, for a moment it looked like you were considering derivatives of all orders. And now I'm curious about this side question, haha! – Pedro A Jul 30 '17 at 23:23
• Jean Marie.Thanks for the link. – Peter Szilas Jan 31 '19 at 10:24

## 14 Answers

If $y'(x)>y(x)\quad\forall x\in\mathbb{R}$, we can define $f(x)=y'(x)-y(x)$ which is positive forall $x$. Suppose that $y'(x)$ is continuous function so that $f(x)$ is continuous too. Now with this element we can build the differential equation $$y'(x)=y(x)+f(x)$$ and its solutions are given by: $$y(x)=e^{x}\left(c+\int_{x_0}^{x}e^{-s}f(s)ds\right)$$

Again are there any applications/physical manifestations of these functions? [for example an object with a velocity that is always greater than its position/acceleration is always greater than its velocity]

I don't know if there's application of this interesting property, but I'm sure that you can't compare velocity with the position because they are not homogeneous quantities.

Assuming $f(x)>0$, $f:\mathbb{R}\mapsto\mathbb{R}$

$f'(x) > f(x) \iff \frac{d}{dx}\ln(f(x))>1$

So you can turn any function $g$ where $g'(x)>1$ into this type of function by taking the exponential of it:

$\frac{d}{dx}g(x)>1 \implies \frac{d}{dx}\ln(e^{g(x)})>1 \implies \frac{d}{dx} e^{g(x)}>e^{g(x)}$

• You assume $f(x)>0$ in the beginning – Hagen von Eitzen Jul 28 '17 at 12:20
• @HagenvonEitzen : Then he could just use $\hat{f}(x) \equiv e^{f(x)}$ as his starting point for any given $f$. That way one always has $\hat{f}(x)>0$. – MPW Jul 28 '17 at 17:43
• Ixion's answer gives the full generalization by allowing $\frac{df}{dx} - f(x)$ to be any function which is everywhere-positive. – Robin Saunders Jul 29 '17 at 10:05
• @RobinSaunders No, he assumes continuity of $f'(x)$. – Adayah Jul 29 '17 at 10:13
• I'm pretty sure that condition isn't actually needed. – Robin Saunders Jul 29 '17 at 10:19

A simple example is $f(x)=-x^2-3$

A more interesting problem is to find a function $f:\mathbb{R}\rightarrow\mathbb{R}$, whose image is $\mathbb{R}$ and satisfies $f'(x)>f(x)$ for all $x\in\mathbb{R}$. One of those functions is

$$\sinh(x),$$

because

$$\frac{d}{dx}\sinh(x)=\cosh(x)>\sinh(x)$$ for all $x\in \mathbb{R}$.

$$(e^{-x}f(x))'=e^{-x} (f'(x)-f(x)) >0$$ so $$e^{-x}f(x)$$ is an increasing function. Since it is $$0$$ at $$0$$ we get $$e^{-x}f(x)>0$$ for all $$x>0$$. Hence $$f(x)>0$$ for all $$x>0$$.

Take $f(x)=e^{\alpha x}$. Then for $\alpha >1$ we have $f'(x)>f(x)$ and for $\alpha <1$ we have $f'(x)<f(x)$.

How about if you look at it as a differential equation. Say

$y' = y + 1$

which has solution $y=Ce^x -1$

Or $y'=y+x^2+1$

which has solution $y=Ce^x - (x^2+2x+3)$

Or $y'=y+2\sin x+3$

which has solution $y = Ce^x - \sin x - \cos x -3$

• Ixion's answer generalizes this to $y'(x) = y(x) + f(x)$ for any $f(x) > 0$. – Robin Saunders Jul 29 '17 at 10:00
• @RobinSaunders - should I delete my answer? – steven gregory Jul 29 '17 at 23:35
• I don't know much about Stack Exchange etiquette, but my guess would be that since you posted your answer first and it contains specific examples not in the other answer, it should be fine to leave it. – Robin Saunders Jul 30 '17 at 12:01

The inequality $$f'(x) > f(x)$$ is equivalent to $$\left[ f(x) e^{-x} \right]' > 0.$$

So the general solution is to take any differentiable function $g(x)$ with $g'(x) > 0$ and put $f(x) = g(x) e^x$.

Note that nothing is assumed about $f$ except differentiability, which is necessary to ask the question in the first place.

A very simple example is $f(x) = -1 < 0 = f'(x)$. Relevant to your edit: this isn't exponential at all.

Other examples that aren't immediately exponential:

• $\frac{-\pi}{2} + \arctan x$ is everywhere negative and everywhere strictly monotonically increasing, so is everywhere less than its derivative.
• $-1 + \mathrm{erf}(x)$ is also everywhere negative and everywhere strictly monotonically increasing. (These are very similar, since they are shifted copies of the CDFs of the (standard/normalized) Cauchy and Gaussian distributions.)
• $\frac{1}{2}\left( x - \sqrt{x^2 + 4} \right)$ is the lower branch of a hyperbola having the $x$-axis and the line $y = x$ as asymptotes. It is everywhere negative and everywhere strictly monotonically increasing.

Another simple example would be $f(x) = -e^{-x}$, $f'(x) = e^{-x}$

See, $-\frac{1}{x}, \frac{1}{x^{2}} \ in \ [0, \infty]$

• More generally, any negative function with positive derivative... – GEdgar Jul 28 '17 at 12:22

For any differential function $f$ for which both $f(x)$ and $f'(x)$ are limited to finite ranges, $f'(x) - f(x)$ is also limited to a finite range, so there is a $c$ for which $f'(x) - f(x) > -c\ \forall\ x$. Therefore, a function $g(x) = f(x) - c$ can be formed for which $g'(x) - g(x) - c> -c\ \forall\ x$ or $g'(x) > g(x)\ \forall\ x$.

For example, this holds for many differential periodic functions.

• The last statement is wrong, since not every differentiable periodic function has bounded derivative. – Adayah Jul 29 '17 at 15:59
• @Adayah You're right. I was considering periodic functions that were differentiable at every point in $\mathbb{R}$, but I realize that a function only has to be differentiable at all points in its domain to be considered differentiable. I've updated my answer. – HelloGoodbye Jul 30 '17 at 13:04
• I mean, a function $f : \mathbb{R} \to \mathbb{R}$ may be periodic and differentiable in every point $a \in \mathbb{R}$ and still have unbounded derivative. – Adayah Jul 30 '17 at 13:09
• @Adayah Do you have any example of such a function? – HelloGoodbye Jul 30 '17 at 16:25
• @Adayah I mean, if a function $f$ is differentiable everywhere, its derivative $f'$ must exist everywhere, and $f'$ must be continuous (because if it contains any discontinuity, $f'$ cannot exist at that point). That makes it impossible for $f'$ to be unbounded, right? – HelloGoodbye Jul 30 '17 at 16:47

Mike an answer to your additional question "Are there physical examples of this?" is enabled by dromastyx.

His example shows hyperbolic functions which describe accurately the physical phenomenon of 'solitons'.

Solitons are solitary waves such as sun flares, Tsunamis etc. An example of finding such waves hidden in known equations is :

http://rsos.royalsocietypublishing.org/content/2/7/140406.review-history

Here's a proof using the mean value theorem.

Let $$m = \operatorname{inf} \{ x > 0 \colon f(x)\leq 0\}$$. This infimum exists, since the reals are complete and the set is bounded below by $$0$$. Since $$f'(0)>0$$, there exists some $$\varepsilon >0$$ such that $$f(x) > f(0)$$ for $$x\in (0,\varepsilon)$$. This means that if there is some $$x>0$$ with $$f(x) \leq 0$$, then $$\varepsilon$$ is also a lower bound. (This is common lemma In the context of the mean value theorem).

Thus it suffices to show that $$m=0$$. Suppose $$m>0$$. By the mean value theorem, there is some point $$y\in (0,m)$$ with $$f'(y) \leq 0$$. By our assumptions, we then have $$f(y) < f'(y) \leq 0$$. Thus $$y$$ contradicts the minimality of $$m$$.

• You have to show that the set $\{m\;|\;m>0 \text{ and } f(m)\le 0\}$ has a minimal element, and that this minimal element is non-zero. This is not obvious, because it might not actually contain its greatest lower bound. It's not hard to prove by continuity that this can't happen, but you have to prove it. – TonyK Jan 31 '19 at 10:45
• But the set $\{x>0:f(x)\le 0\}$ turns out to be empty, so it doesn't have an infimum. (Some people take the view that the infimum of the empty set is $+\infty$, but nobody claimes that it is zero.) – TonyK Feb 2 '19 at 22:10