Question

find the derivative of y=x^(lnx)

Answer 1

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

Answer 2

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

Answer 3

\[y'=\frac{2y \ln x}{x}\]

Answer 4

@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) )

Answer 5

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

Answer 6

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

Answer 7

why does lnx = (lnx)^2

Answer 8

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

Answer 9

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

Answer 10

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