y=e^cosx + cos(e^x) what is the derivative?

Question

I got y'= e^cosx - sin(e^x). Is this wrong? Please show a lot of work.

Answer 1

y=e^cos(x) + cos(e^x)
y'=-sin(x)e^cos(x) + -sin(e^x)e^x
=-(sin(x)e^cos(x)+sin(e^x)e^x)

Answer 2

y=e^cosx + cos(e^x)
You need to use the chain rule in both parts
So lets make

y = u + v where u = e^a where a = cosx and v = cosb where b = e^x

Then y' = u' + v' = du/da*da/dx + dv/db*db/dx
= e^a * (-sinx) + -sinb.(e^x)
= -sinx e^(cosx) - e^x sin(e^x)

Answer 3

yes that is wrong you need to use the chain rule on both terms

Like this:

y'=(e^cos(x))*(-sin(x))-sin(e^x)*(e^x)