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. We start with: 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/ |
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