Question

Find derivative of lnx^lnx

Answer 1

So let y equal that! :)
We are going to find y' by implicit differentiation.
\[y=(\ln(x))^{\ln(x)}\]
Take ln( ) of both sides!
\[\ln(y)=\ln(x) \ln(\ln(x))\]
We did that because we don't like (or I don't like) the power of a function of x (this does not include constants)
I guess I could said I don't like non-constant powers.
And also we it like this so we can find the derivative.
Now we take derivative of both sides!

Answer 2

http://www.wolframalpha.com/input/?i=derive+ln%28x%29^%28ln%28x%29%29

Answer 3

sorry i accidently hit the back button :(
i will start over
\[\frac{y'}{y}=\frac{1}{x} \ln(\ln(x))+\ln(x) (\ln(\ln(x))'\]
I haven't differentiated that one part because I think it may look confusing on how to differentiate.
So I'm going to give more steps on that part!
So we know the chain rule:
\[w=f(g(x))=> w'=f'(g(x))g'(x)\]
So f(x)=ln(x) and g(x)=ln(x)
=> f'(x)=1/x and g'(x)=1/x
So this implies
\[(\ln(\ln(x))'=\frac{1}{\ln(x)} \cdot \frac{1}{x}\]
So this means we have
\[\frac{y'}{y}=\frac{\ln(\ln(x))}{x}+\ln(x) \cdot \frac{1}{x \ln(x)}\]
Last step multiply y on both sides and replace y with (ln(x))^(ln(x))

Answer 4

Awesome!, thank you.