If someone can check this:
differentiate y = x^(e^x)

approach:
lny (dy) = e^x * (lnx) (dx)
1/y (dy) = e^x * 1/x + e^x (lnx) (dx) via product rule
dy/dx = e^x *1/x + e^x(lnx) y
= [e^x *1/x + e^x(lnx)] x^(e^x)

Question

If someone can check this:
differentiate y = x^(e^x)

approach:
lny (dy) = e^x * (lnx) (dx)
1/y (dy) = e^x * 1/x + e^x (lnx) (dx)     via product rule
dy/dx = e^x *1/x + e^x(lnx) y
= [e^x *1/x + e^x(lnx)]  x^(e^x)

anonymous · Answer

Do you mean differentiate y = x^(e^x)

anonymous · Answer

yes, ill fix that

abb0t · Answer

$$y' = \frac{ dy }{ dx }$$
why did you separate dy and dx?

abb0t · Answer

You're solving for dy/dx.

anonymous · Answer

that's how we were taught in class... iderno

anonymous · Answer

can't I treat them like terms and move them around?

abb0t · Answer

You can, but I wouldn't for this since you're solving for y'

abb0t · Answer

but your answer is correct.

anonymous · Answer

wolfram alpha tells me otherwise

abb0t · Answer

wolfram often simplifies the solution fully. so the answer might appear different, but is often the same solution.