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