Calculus, derivatives

Find the derivative of y = x^(e^x). Show your work.

Can someone please help me with this?

y = x^(e^x)

Since there's an x in the exponent, you'll need to bring that down. Take the natural log of both sides:

ln(y) = ln[x^(e^x)]

Now, by the laws of natural logarithms, ln(x^(something)) = (something)(ln(x)).

ln(y) = (e^x)(lnx)

Now you can differentiate both sides. The derivative of ln(y) with respect to x will be (1/y)*y'.
Use the product rule on the stuff on the right.

(y' / y) = (e^x)(1/x) + (lnx)(e^x)

Simplify a bit, take out that factor of (e^x)..

(y' / y) = (e^x)[1/x + lnx]

Now multiply both sides by y to get the final answer, (dy/dx) or y'. Refer to the original problem, y = x^(e^x)

y' = y(e^x)[1/x + lnx]
y' = (e^x) * x^(e^x) * [1/x + lnx]

Hope that helps.

let y = x^e^x

ln y = (e^x) ln x

take derivative both sides
y'/y = e^x lnx + e^x/x

y'= y[( e^x)lnx+ e^x/x]

derivative of y = (x^e^x)e^x lnx + (x^e^x)e^x/x + C

x^(e^x)
lnx(x^(e^x))(e^x)