Question

How do you find the derivative of `sin^2x+cosx`?

Answer 1

I found: `f`(x)=sin(x)[2cos(x)-1]`

Explanation:

For the `sin^2(x)` I would use the Chain Rule deriving the `()^2` first and `sin` after, getting:
`f`(x)=2sin(x)cos(x)-sin(x)=`
`=sin(x)[2cos(x)-1]`

(Alternately, if you want, you can write: `sin^2(x)=sin(x)sin(x)` and use the Product Rule).