What am i doing wrong? Log differentiation

[A_Munk3y]

Homework Statement

Use logarithmic differentiation to find the derivative of the function.
y=x^ln(x)


also, if anyone could help me with this...
i have (for a diff problem) 1/(x²-1)^1/2 * 1/2(x²-1)^-1/2*2x
i know that is right, but i don't know how to get from that, to x/(x²-1)
i keep getting x/(x²-1)^-1 because i thought that (x²-1)^-1/2/(x²-1)^1/2 = (x²-1)^-1

The Attempt at a Solution

=> ln(y)=ln(x^ln(x))
=> ln(y)=ln(x)*ln(x)
1/y*y'=1/x*ln(x)+1/x*ln(x)
y'=1/x*ln(x)+1/x*ln(x)*x^ln(x)

 

[stevenb]

=> ln(y)=ln(x^ln(x))
=> ln(y)=ln(x)*ln(x)
1/y*y'=1/x*ln(x)+1/x*ln(x)
y'=1/x*ln(x)+1/x*ln(x)*x^ln(x)

Looks like you have a typo there with missing parentheses.

y'=(1/x*ln(x)+1/x*ln(x))*x^ln(x)

or y'=2*ln(x)*x^ln(x)-1

 

[A_Munk3y]

yea, i see the typo. Thanks for pointing that out, but

how did you get from the first equation to the 2nd?
where did the 1/x's go and how did you get x^ln(x)-1

 

[stevenb]

how did you get from the first equation to the 2nd?
where did the 1/x's go and how did you get x^ln(x)-1

Oh, that is because 1/x=x^-1 and then generally, x^a*x^b=x^a+b.

 

[A_Munk3y]

Ahh! i see :)
Alright thanks for the help man!
I really appreciate it