find y' for y=sin(2x^2) 4xcos(2x^2) xcos(4x^2) 4sin(2x^2) 2xcos(2x^2)

To find the derivative of \( y = \sin(2x^2) \), we will use the chain rule. The chain rule states that if you have a composite function \( y = f(g(x)) \), then the derivative is given by: In this case, let . Then \( y = \sin(u) \).First, we find the derivative of with respect to : Next, we find the derivative of with respect to : Now, applying the chain rule: Substitute back : Therefore, the derivative of \( y = \sin(2x^2) \) is: So, the correct answer is: