Sea \( g(x)=\sqrt{x} \sin (x) \) \( g^{\prime}(x)=\square \)

To find the derivative of \( g(x) = \sqrt{x} \sin(x) \), we can use the product rule. The product rule states that if you have two functions multiplied together, say \( u(x) \) and \( v(x) \), then the derivative \( (uv)' \) is \( u'v + uv' \). For our function, let \( u(x) = \sqrt{x} \) and \( v(x) = \sin(x) \). Calculating the derivatives: - \( u'(x) = \frac{1}{2\sqrt{x}} \) - \( v'(x) = \cos(x) \) Now, applying the product rule: \[ g'(x) = u'(x)v(x) + u(x)v'(x) = \frac{1}{2\sqrt{x}} \sin(x) + \sqrt{x} \cos(x) \] Thus, we arrive at the final expression: \[ g'(x) = \frac{1}{2\sqrt{x}} \sin(x) + \sqrt{x} \cos(x) \]