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### The Derivative of f(x)^g(x)

Date: 09/26/98 at 16:08:35
From: David Bandel
Subject: Derivation

I am aware of the special equations with which one can find the
derivative of the quotient, product, sum, reciprocal of functions.
My question is whether or not there is a similar equation with which I
could find the derivative of f(x)^g(x). This to me would seem like the
most versatile of all equations. If there is such a formula, what is
it?  If there is no formula, then could you refer me to areas where I
could see the work that has been put toward the discovery of the
formula, or at least explain to me why this formula is impossible.

Date: 09/29/98 at 18:12:03
From: Doctor Jeremiah
Subject: Re: Derivation

Hi David:

Actually such a formula must exist.

In this response I am going to use d/dx( y ) to mean the derivative of
y with respect to x.

The formula:

d/dx( x^n ) = x^(n-1) * n

and the formula:

d/dx( a^x ) = a^x * ln(a)

are just special cases of this formula.

To solve d/dx( f(x)^g(x) ) we need to play a little trick to make the
differentiation easier - and nothing is as easy as d/dx( e^x ) = e^x.
So since x = e^( ln(x) ) we will use that to our advantage.

d/dx( f(x)^g(x) ) = d/dx( e^( ln( f(x)^g(x) )))              [1]

Now ln( f(x)^g(x) ) = g(x) * ln( f(x) ) so we plug that in
to get this:

d/dx( f(x)^g(x) ) = d/dx( e^( g(x) * ln( f(x) )))            [2]

Now all we have to do is differentiate the righthand side.
Remember that d/dx( e^f(x) ) = e^f(x) * d/dx( f(x) )

d/dx( e^( g(x) * ln( f(x) ))) = e^( g(x) * ln( f(x) ))
* d/dx( g(x) * ln( f(x) )      [3]

And then simplify (remember equation [2]):

d/dx( f(x)^g(x) ) = f(x)^g(x) * d/dx( g(x) * ln( f(x) )      [4]

Now all we have to do is differentiate the righthand side:
This is the simple multiplication rule.

d/dx( g(x) * ln( f(x) ) = d/dx( g(x) ) * ln( f(x) )
+ g(x) * d/dx( ln( f(x) )            [5]

And then we need to differentiate the natural log:

d/dx( ln( f(x) ) = 1 / f(x) * d/dx( f(x) )                   [6]

So we plug that in to get this:

d/dx( f(x)^g(x) ) = f(x)^g(x) * [ d/dx( g(x) ) * ln( f(x) )
+ g(x) / f(x) * d/dx( f(x) ) ]       [7]

And then distribute f(x)^g(x) to get:

d/dx( f(x)^g(x) ) = f(x)^g(x) * d/dx( g(x) ) * ln( f(x) )
+ f(x)^g(x) * g(x) / f(x) * d/dx( f(x) )   [8]

Then if we simplify:

f(x)^g(x) * g(x) / f(x) = f(x)^g(x) / f(x) * g(x)
f(x)^g(x) * g(x) / f(x) = f(x)^( g(x)-1 ) * g(x)             [9]

We finally end up with:

d/dx( f(x)^g(x) ) = f(x)^g(x) * d/dx( g(x) ) * ln( f(x) )
+ f(x)^( g(x)-1 ) * g(x) * d/dx( f(x) )   [10]

Now to show you where the formula:

d/dx( x^n ) = x^(n-1) * n * d/dx( x )

comes from, all we have to do is set f(x) = x and g(x) = n and we get:

d/dx( x^n ) = x^n * d/dx( n ) * ln( x )
+ x^( n-1 ) * n * d/dx( x )
d/dx( x^n ) = x^n * 0 * ln( x )
+ x^( n-1 ) * n * 1
d/dx( x^n ) = x^( n-1 ) * n

Now to show you where the formula:

d/dx( a^x ) = a^x * ln(a) * d/dx( x )

comes from all we have to do is set f(x) = a and g(x) = x and we get:

d/dx( a^x ) = a^x * d/dx( x ) * ln( a )
+ a^( x-1 ) * x * d/dx( a )
d/dx( a^x ) = a^x * 1 * ln( a )
+ a^( x-1 ) * x * 0
d/dx( a^x ) = a^x * ln( a )

So you can see that the formula is not as easy as the ones for
multiplying or dividing, but it isn't too complicated.

- Doctor Jeremiah, The Math Forum
http://mathforum.org/dr.math/

Associated Topics:
High School Calculus

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