The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Rate of Change Constant?

Date: 03/20/2002 at 03:27:01
From: Mariam
Subject: Why are (diff(e^x,x)=e^x, sin(x)&sinh(x) similar) ?

Hello Dr. Math,

First I want to thank you for this service.

My first quesion is: int(e^x,x) = e^x diff(e^x,x)=e^x - these are the 
formulas, but what I am looking for is not the proof but the verbal 
explanation. The rate of change of e^x is e^x. Does this mean that the 
rate of change is constant? I don't mean constant number, but in the 
same way.

The second question is, why are sinx,cosx,... and sinhx,coshx,.... 
similar? I don't want the derivation, but the reason, the verbal 

Thank you in advance,

Date: 03/22/2002 at 15:15:56
From: Doctor Douglas
Subject: Re: Why are (diff(e^x,x)=e^x, sin(x)&sinh(x) similar) ?

Hi, Mariam.

Thanks for submitting your question to the Math Forum.

The basic reason that all of these functions are similar is that there 
is a function which, under some operation, remains the same.  

Consider the function f(x) = x. If you compose this function with 
itself, you end up with what you started with, namely f(f(x)) = x.  
Now consider g(x)= -x. Again, if you compose this function with 
itself, you obtain g(g(x)) = x. And if you consider p(x) = ix and 
q(x) = -ix, iterating these gives you p(p(x)) = i^2*x = -x and 
q(q(x)) = -x. For p and q you need to take four iterations to return 
to x: p(p(p(p(x)))) = q(q(q(q(x)))) = x.

Instead of function composition, your question deals with the 
operations of differentiation and integration. There is one nontrivial 
(i.e. not equal to zero everywhere) function f(x) such that if you 
differentiate once, you get the same thing:

   diff(e^x,x) = e^x 

If you differentiate twice, you still get the same thing:

   diff(diff(e^x,x),x) = diff(e^x,x) = e^x

Or, by integrating everything once with respect to x (and setting the
constants equal to zero),

   diff(e^x,x) = e^x = integral(e^x,x)

In other words, this result is due to the fact that the exponential
function, e^x, has derivative equal to itself, and that it remains the
same under repeated differentiation.  

Now, just as with the original functions g above, it's possible to 
have the first operation give the negative, so that two operations in 
succession give us back what we started:

   diff(e^-x,x)         = -e^-x
   diff(diff(e^-x,x),x) = -diff(e^-x,x) = +e^-x

Since sinh(x) and cosh(x) are linear combinations of e^x and e^-x, 
two differentiations will give us back what we started with.  

And as with the original functions p and q, it's possible to 
differentiate the complex exponentials e^(ix) and e^(-ix):

   diff(e^(ix),x)       = i*e(ix)
   diff(e^(-ix),x)      = -i*e(-ix)

The functions sin(x) and cos(x) are linear combinations of these 
complex exponentials, so just as with p and q above, after two 
differentiations they give us the negative, and after four 
differentiations they give us what we started with.  

This is why the exponential function is so important: it is the 
function that "remains the same" [or at least is very similar in 
the case of e^-x or e^(ix)] under the operation of differentiation.  

- Doctor Douglas, The Math Forum   
Associated Topics:
High School Calculus
High School Functions

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.