1st and 2nd derivatives of y = xe^(ex)?

Question

please help, im lost.

i tried the multiplicity rule and ended up with (x)(ex)(e^(ex-1)) + (e^(ex))(1) for the 1st derivative, which i think is wrong.

Answer 1

I assume that the last ex you wrote is e^x too, right?
y' = e^(e^x) + x [e^(e^x)] (e^x)
y' = e^(e^x) [1+ x e^x]

Y''= e^(e^x)[ xe^x + e^x] + e^(e^x) e^x [1 + xe^x]
y''= e^(e^x)[ xe^x + e^x + 1 + xe^x]

Answer 2

for the first derivative, you use the product rule along with the chain rule. so you have:
e^(ex)+x*e^(ex)*e
for the second derivative, just repeat the process:
e^(ex)*e+e*(e^(ex)+x*e^(ex)*e)

Answer 3

y''=xe^(x+1)+e^(x+1)+e^(ex+1)

Answer 4

first derivative is
((e)(x) + 1) e^((e)(x))
second derivative is
e^((e)(x) + 1)((e)(x) + 2)