I am stuck on this homework problem. if someone can please help me out showing all the steps I would really appreciate it!
find the derivative:
y=x^lnx
find the derivative:
y=x^lnx
-
y = x ^(ln x)
ln y = ln [x^(ln x)]
ln y = ln x (ln x) = (ln x)^2
y' / y = 2 ln x / x
y ' = 2y ln x / x
but y = x^(ln x) ==>
y ' = 2 x^(ln x) ln x / x
ln y = ln [x^(ln x)]
ln y = ln x (ln x) = (ln x)^2
y' / y = 2 ln x / x
y ' = 2y ln x / x
but y = x^(ln x) ==>
y ' = 2 x^(ln x) ln x / x
-
Hello....
lny = (lnx)^2
(1/y)y' = 2lnx/x
y' = 2x^(lnx)·lnx/x = 2x^(lnx)·x(-1)·lnx = 2x^(lnx - 1)lnx
or
y = x^lnx = e^ln(x^lnx) = e^[(lnx)^2]
y' = 2lnx·(1/x)·e^[(lnx)^2] = 2lnx/x·x^lnx = 2x^(lnx - 1)lnx
lny = (lnx)^2
(1/y)y' = 2lnx/x
y' = 2x^(lnx)·lnx/x = 2x^(lnx)·x(-1)·lnx = 2x^(lnx - 1)lnx
or
y = x^lnx = e^ln(x^lnx) = e^[(lnx)^2]
y' = 2lnx·(1/x)·e^[(lnx)^2] = 2lnx/x·x^lnx = 2x^(lnx - 1)lnx