What is the correct answer to this problem..would it be...?

Question

Determine y' when
y= x ^ (1/x)
y' = (y/ x^2) (lnx-1)
???
plz show me how

Answer 1

y= x ^ (1/x)
lny = (1/x)ln(x)
y'(1/y) = [(1/x)(1/x) -ln(x)/x^2]
y' = [(1/x)(1/x) -ln(x)/x^2]/x^(1/x)
y' = [1/x^2 - ln(x)/x^2]/x^(1/x)
y' = [(1-ln(x))/x^2]/x^(1/x)

Answer 2

y = x^(1/x)

taking logs

ln y = (1/x) (ln x) = lnx/x

differentiating

(1/y) y' = x(1/x) -lnx (1)/x^2

= (1 - lnx)/x^2

y' = (y/x^2)(1 - ln x)