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### Rate of Change Constant?

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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
explanation.

Mariam
```

```
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.

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
http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus
High School Functions

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