How do you find the derivative of sin^2x+cosxsin2x+cosx?

1 Answer
Gió
Oct 30, 2015

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

Explanation:

For the sin^2(x)sin2(x) I would use the Chain Rule deriving the ()^2()2 first and sinsin after, getting:
f`(x)=2sin(x)cos(x)-sin(x)=f`(x)=2sin(x)cos(x)−sin(x)=
=sin(x)[2cos(x)-1]=sin(x)[2cos(x)−1]

(Alternately, if you want, you can write: sin^2(x)=sin(x)sin(x)sin2(x)=sin(x)sin(x) and use the Product Rule).

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