   Associated Topics || Dr. Math Home || Search Dr. Math ### 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.

Thank you in advance,
Mariam
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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
http://mathforum.org/dr.math/
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