# Derivative of function is bigger??!

Today I had a lesson about derivatives and our professor showed us a derivative that is bigger than his function. This doesn't seem legit, can somebody help me with that?

Any explanation will be appreciated.

By the way can somebody tell what is so special about the number e?

• Seems a legitimate interrogation about the concept of derivative... why the downvote Aug 16 '13 at 18:52
• Why should it not seem legit in the first place? Aug 16 '13 at 19:22
• Note that a function with a small value can be changing fast. For example a formula 1 car on a flying lap will pass point zero travelling at high speed. For another example the derivative of $e^{2x}$ is $2e^{2x}$ - at each point the rate of change is twice as much as the magnitude of the function. Aug 16 '13 at 20:14
• The number $e$ has many properties - see en.wikipedia.org/wiki/E_(mathematical_constant) Aug 16 '13 at 20:16
• BTW, it some context may be not quite correct to compare a function with its derivative. Suppose you have a function of spacial coordinate $f(x)$, where $x$ has dimensions of length ($L$), and $f$ is dimensionless. Then $f'(x)$ would have dimensions of $L^{-1}$, and since dimensions of $f(x)$ and $f'(x)$ are different, you can't really compare them (results will depend on units of length) Aug 16 '13 at 20:55

Draw your favorite function $f$ on a piece of paper. Roughly draw its derivative $f'$ on the same piece of paper. Now, move your function $f$ down (along the $y$-axis), and call this new function $f_{down}$. We have $f_{down}' = f'$. Hence, the derivatives of the down function and the original function are the same (but meanwhile, we have made our original function "smaller" via moving it down).

Indeed, the derivative measures the rate of change of a function and has nothing to do with the "height" of our function. In particular, if our original function is bounded and also has a bounded derivative, i.e. $|f| \leq b$ and $|f'| \leq a$ for some real numbers $a$ and $b$, we can always create a function whose derivative is always larger than itself using the same procedure described above.

• What a wonderful construction!! Thanks man. Mar 30 '18 at 13:09

Let $f(x)=-e^{-x}$; then $f$ is always negative, but $f'(x)=e^{-x}$ is always positive, so $f'(x)>f(x)$ for all $x\in\mathbb{R}$.

EDIT: What's really going on here is that we're taking a function $f$ which is always positive, but always decreasing (this can happen by just making sure that the function decreases at a slower and slower rate; that is, that the second derivative of the function is "positive enough"). Since it's always decreasing, it's derivative is always negative, and hence strictly less than the value of the function itself. You can show that no polynomial function has this property; however, there are still polynomial functions which are always greater than their derivatives - a fun exercise.

Or, look at $f(x)=e^x-12$.

Or any of a number of other examples . . .

EDIT: towards your second question, "What's so special about $e$," let me give a very short answer: $e$ happens to have the property that the derivative of $e^x$ with respect to $x$ is just $e^x$ again. This makes $e$ a very useful number for lots of examples in calculus - such as mine, above. There is much, much more to this story: why should there be a number with that property, at all? What other properties does $e$ have? How else is $e$ useful in mathematics? But I suspect this question has already been asked (and well answered) on this site.

It does not really make sense to directly compare the value of the derivative with the value of the function. It is like comparing how fast a car is travelling with the position of the car on the road. If you really want to compare somehow, it makes more sense to compare the speed of the car, which is the distance travelled by the car per some unit of time, with the distance between two points on the road, that is how far the car has travelled (not the position of the car). Note that we must agree on the unit of time to measure speed, as in km/hour or km/sec, otherwise just by changing the unit of time you can make the numerical value of the speed as large as you want, making the comparison meaningless.

In formulas, what I am saying is that $f'(x)$ is the derivative of $g(x)=f(x)+C$ for any constant $C$, so by choosing $C$ very large positive or very large negative, you can get any of the two inequalities $g'(x)<g(x)$ and $g'(x)>g(x)$ for any fixed $x$. But the difference $g(x)-g(y)$ (for any two points $x$ and $y$) stays the same no matter what the constant $C$ is, so it is a more meaningful quantity to be compared with $g'$. This is still not enough, because if we consider the new function $h(x)=g(ax)$, then $h(x)-h(y)=g(ax)-g(ay)$ and $h'(x)=ag'(ax)$, so that by choosing $a$ very large or very small, one can make any of the two inequalities $|h'(x)|<|h(x)-h(y)|$ and $|h'(x)|>|h(x)-h(y)|$ true. A way to fix this would be to require $x-y=1$, or to consider functions that are defined on the interval $[a,b]$, etc. I think you know what I am doing here: It introduces a "standard length" in the $x$-direction, which is like introducing a unit of time in the car example.

In any case, assuming that we choose one of the legitimate ways of comparing, if you were blindfolded and to pick a function from a zoo of functions, there is a very large chance that the derivative of the function will be much much larger than the function. This is because a function can oscillate very fast even though having a very limited range of values. An example is $\varepsilon\sin(nx/\varepsilon)$ with small $\varepsilon$ and large $n$. Another crude explanation is that you are allowed to divide by a number that is as small as you like in the definition of derivative, but you are dividing by 1 or something moderately sized in the other expression $g(x)-g(y)$.