## Calculus (3rd Edition)

y'=2xcos($x^{2}$)
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($x^{2}$)