find the derivative of y=x^(lnx)

Question

anonymous · Answer

2x^(ln(x)-1) * ln(x)

anonymous · Answer

i tried to do substitution but couldn't remember if that was just for integrals or not

anonymous · Answer

$$y'=\frac{2y \ln x}{x}$$

jamesj · Answer

@shinigami: that's wrong.

Note first that ln y = ln(x^ln(x)) = ln x . ln x = (ln x)^2

Differentiating both sides we have

1/y . y ' = 2 ln x . (1/x)

So y' = 2y . ln x / x
        =  (2/x) . ln x . ( x^(ln x) )

anonymous · Answer

$$ y^\prime = 2\,{\frac {{x}^{\ln  \left( x \right) }\ln  \left( x \right) }{x}} $$

anonymous · Answer

okay james i understand the first part.of "ln"ing both sides...

anonymous · Answer

why does lnx = (lnx)^2

anonymous · Answer

ln(x) * ln(x) = (ln x)^2 is what he said

jamesj · Answer

I'm saying ln y = (ln x).(ln x) = (ln x)^2.

anonymous · Answer

okay i guess i don't understand why. I'm at:
lny = ln(x^lnx)