Answer
y'=2sin(x)cos(x)
Work Step by Step
The chain rule states that if y=g(h(x)), then y'=g'(h(x))h'(x). Since y=$sin^{2}$(x), which we'll begin by rewriting as $(sin(x))^{2}$, we can set g(x)=$x^{2}$ (outside function) and h(x)=sin(x) (inside function), and use the chain rule to find y'. Therefore,
y'=g'(sin(x))h'(x)
We know that
$\frac{d}{dx}$[$x^{2}$]=2x, using the power rule
$\frac{d}{dx}$[sin(x)]=cos(x), it'd be a good idea to memorize this derivative
Now that we know that g'(x)=2x and h'(x)=cos(x), we can find y'.
y'=2(sin(x))*(cos(x))
=2sin(x)cos(x)
