Functions that are always less than their derivatives

Mike Brown 08/20/2017. 12 answers, 3.142 views
calculus integration differential-equations derivatives inequality

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]

3 Comments
1 BallpointBen 07/28/2017
Off the top of my head, any bounded, monotonically increasing function in the bottom half-plane.
1 Robin Saunders 07/29/2017
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.
Hamsteriffic 07/30/2017
+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!

12 Answers


Ixion 07/29/2017.

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.


Aidan Connelly 07/29/2017.

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:

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

5 comments
6 Hagen von Eitzen 07/28/2017
You assume $f(x)>0$ in the beginning
2 MPW 07/28/2017
@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 $\hat{f}(x)>0$.
Robin Saunders 07/29/2017
Ixion's answer gives the full generalization by allowing $\frac{df}{dx} - f(x)$ to be any function which is everywhere-positive.
Adayah 07/29/2017
@RobinSaunders No, he assumes continuity of $f'(x)$.
Robin Saunders 07/29/2017
I'm pretty sure that condition isn't actually needed.

Peter 07/28/2017.

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


dromastyx 07/28/2017.

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 $x\in \mathbb{R}$.


M. Winter 07/28/2017.

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


steven gregory 07/28/2017.

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 $y = Ce^x - \sin x - \cos x -3$

3 comments
Robin Saunders 07/29/2017
Ixion's answer generalizes this to $y'(x) = y(x) + f(x)$ for any $f(x) > 0$.
steven gregory 07/29/2017
@RobinSaunders - should I delete my answer?
Robin Saunders 07/30/2017
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.

Eric Towers 07/30/2017.

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.

Thiago Nascimento 07/28/2017.

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

1 comments
7 GEdgar 07/28/2017
More generally, any negative function with positive derivative...

Joshua Kidd 07/28/2017.

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


Adayah 07/29/2017.

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.


HelloGoodbye 07/30/2017.

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.

5 comments
1 Adayah 07/29/2017
The last statement is wrong, since not every differentiable periodic function has bounded derivative.
HelloGoodbye 07/30/2017
@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.
Adayah 07/30/2017
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.
HelloGoodbye 07/30/2017
@Adayah Do you have any example of such a function?
HelloGoodbye 07/30/2017
@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?

Henk Koppelaar 08/02/2017.

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

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