how to do it

y + e^y = x²

Differentiating this with respect to x we get:

dy/dx + e^y*dy/dx = 2x   (Use the chain rule for de^y/dx: i.e.  de^y/dy*dy/dx)

(1 + e^y)*dy/dx = 2x

dy/dx = 2x/(1 + e^y)

f(x) = sin(x)*cos³(x) 

df(x)/dx = sin(x)*dcos³(x)/dx + cos³(x)*dsin(x)/dx

df(x)/dx = sin(x)*3cos²(x)*(-sin(x)) + cos³(x)*cos(x)

df(x)/dx = cos²(x)*(-3sin²(x) + cos²(x))

Now since cos(2x) = cos²(x) - sin²(x) = 1 - 2sin²(x) = 2cos²(x) - 1 we could write

df(x)/dx = (cos(2x) + 1)*(2cos(2x) - 1)/2

Thanks Alan,

I was waiting for you to answer that first one.