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:<br /><br />\[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \]<br /><br />In this case, let \( u = 2x^2 \). Then \( y = \sin(u) \).<br /><br />First, we find the derivative of \( y \) with respect to \( u \):<br /><br />\[ \frac{dy}{du} = \cos(u) \]<br /><br />Next, we find the derivative of \( u \) with respect to \( x \):<br /><br />\[ \frac{du}{dx} = \frac{d}{dx}(2x^2) = 4x \]<br /><br />Now, applying the chain rule:<br /><br />\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \cos(u) \cdot 4x \]<br /><br />Substitute back \( u = 2x^2 \):<br /><br />\[ \frac{dy}{dx} = \cos(2x^2) \cdot 4x \]<br /><br />Therefore, the derivative of \( y = \sin(2x^2) \) is:<br /><br />\[ y' = 4x \cos(2x^2) \]<br /><br />So, the correct answer is:<br /><br />\[ 4x \cos(2x^2) \]