Question

Find dy/dx.

y = x^ln x, x > 0

Answer 1

recall that
d/dx x^n = n x^(n-1)
d/dx a^x = lna a^x

so, if y = u^v,
y' = v u^(v-1) u' + lnu u^v v'

Here you have
u = x
v = lnx

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

or, do it like this

y = x^lnx
lny = lnx lnx = (lnx)^2
1/y y' = 2lnx * 1/x
y' = 2lnx/x y = 2lnx/x x^lnx

Answer 2

I don't understand how you went from y' = lnx x^(lnx -1) + lnx x^lnx (1/x) to lnx/x x^lnx + lnx/x x^lnx. Where did the -1 go? May you please explain?

Answer 3

Is the lnx/x suppose to be (lnx)/(x) or ln(x/x)?

Answer 4

(lnx)/(x)

x^(lnx-1) = x^(lnx)/x