This was a tricky example. Starting with our function h(x) = x^x, we want to take the derivative. We have the power rule for the derivative of things like x^2, x^3, etc. But we don't have a rule for how to take derivatives where the variable is the *exponent*, not the base (2^x, 3^x).

The closest is e^x, where we know d/dx e^x = e^x.

So, if we can rewrite our exponent (x^x) into something that looks like e^foo, we know we can take the derivative.

Because e and natural log are inverse functions, they cancel each other out. So we have:

x = e^[ln(x)]

It's like taking the square of the square root: you get the original number back (e.g., x = [ sqrt(x) ]^2 ).

From here, we say

x^x = [ e^[ln(x)] ]^x

where we replace the first x (the base) with e^[ln(x)], then take that to the x power. This simplifies to:

e^[ln(x) * x]

which is something of the format e^foo. Now, we can finally take the derivative:

d/dx e^foo = d/dx foo * e^foo

where foo = ln(x) * x. The (1 + ln(x)) term is d/dx foo. (We can't just do d/dx e^foo = e^foo, we need the chain rule to find d/dx foo. In the case of plain old e^x, we have d/dx e^x = d/dx x * e^x = 1 * e^x = e^x.)

Hope that helps.

nevermind, I think I figured it out -- 1+ln(x) is the conversion factor. so that's we keep e^ln(x)x (or, as kalid would put it, I think, e^foo) the same because derivative of e^foo is e^foo. Let me know if i"m incorrect. thanks.

"If h=x^x, the final result is:

h`(x) = 1+ln(x).e^ln(x)x "

i'm quite lost here.

What rule is being used here to get this from ... h(x) = e^ln(x)x ?

From what I understand, we evaluated the change on the exponent ln(x)x in terms of changes in x. That's how we get 1+ln(x). Do I have it right so far?

Then we go back to h(x) = e^ln(x)x. What rule was used next? The power rule?

if yes, then I'm quite confused about where e^ln(x)x came back from.

shouldn't it then be h`(x) =1+ln(x).e^ln(x)x -1?

Any help? Thanks.

Okay I get it. (e^x)^2 = e^2x but e^(x^2) doesn't equal e^2x except at x=2.