Differentiate y=2x(3-x) (dy)/(dx)=3x-2x^2 (dy)/(dx)=6-2x (dy)/(dx)=6-4x (dy)/(dx)=6x-4x^2

To differentiate the function \( y = 2x(3 - x) \), we can use the product rule. The product rule states that if you have a function \( y = u(x)v(x) \), then the derivative \( \frac{dy}{dx} \) is given by:<br /><br />\[ \frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) \]<br /><br />Here, let \( u(x) = 2x \) and \( v(x) = 3 - x \).<br /><br />First, we find the derivatives of \( u(x) \) and \( v(x) \):<br /><br />\[ u'(x) = 2 \]<br />\[ v'(x) = -1 \]<br /><br />Now, apply the product rule:<br /><br />\[ \frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) \]<br />\[ \frac{dy}{dx} = 2(3 - x) + 2x(-1) \]<br />\[ \frac{dy}{dx} = 6 - 2x - 2x \]<br />\[ \frac{dy}{dx} = 6 - 4x \]<br /><br />So, the correct answer is:<br /><br />\[ \frac{dy}{dx} = 6 - 4x \]