91Ó°ÊÓ

First, we need to find the derivatives of the scalar function \(g(t)\) and the vector function \(\mathbf{v}(t)\). Let's begin with \(g(t)\): $$g(t) = 2 \sqrt{t} = 2t^{\frac{1}{2}}$$ Now, find the derivative of \(g(t)\): $$g'(t) = \frac{d}{dt} (2t^{\frac{1}{2}}) = t^{-\frac{1}{2}}$$ Next, find the derivative of the vector function \(\mathbf{v}(t)\): $$\mathbf{v}(t) = \left\langle t^{2}, -2t, 1\right\rangle$$ The derivative of \(\mathbf{v}(t)\) is: $$\mathbf{v}'(t) = \left\langle \frac{d}{dt}(t^{2}), \frac{d}{dt}(-2t), \frac{d}{dt}(1) \right\rangle = \left\langle 2t, -2, 0\right\rangle$$

Now, we will use the product rule to find the derivative of \(g(t)\mathbf{v}(t)\): $$(g(t) \mathbf{v}(t))' = g'(t) \mathbf{v}(t) + g(t) \mathbf{v}'(t)$$ Plug in the values we found for \(g'(t)\), \(\mathbf{v}(t)\), and \(\mathbf{v}'(t)\): $$(g(t) \mathbf{v}(t))' = t^{-\frac{1}{2}} \left\langle t^{2}, -2t, 1\right\rangle + 2t^{\frac{1}{2}} \left\langle 2t, -2, 0\right\rangle$$

Now, it's time to calculate the actual derivative: $$(g(t) \mathbf{v}(t))' = \left\langle t^{-\frac{1}{2}}t^{2}, -2t^{-\frac{1}{2}}t, t^{-\frac{1}{2}}\right\rangle + \left\langle 2t^{\frac{3}{2}}, -4t^{\frac{1}{2}}, 0\right\rangle$$ Simplify: $$(g(t) \mathbf{v}(t))' = \left\langle t^{\frac{3}{2}}, -2t^{\frac{1}{2}}, t^{-\frac{1}{2}}\right\rangle + \left\langle 2t^{\frac{3}{2}}, -4t^{\frac{1}{2}}, 0\right\rangle$$ Finally, add the corresponding components of the two vectors: $$(g(t) \mathbf{v}(t))' = \left\langle t^{\frac{3}{2}} + 2t^{\frac{3}{2}}, -2t^{\frac{1}{2}} - 4t^{\frac{1}{2}}, t^{-\frac{1}{2}} + 0\right\rangle$$ Simplify the result: $$(g(t) \mathbf{v}(t))' = \left\langle 3t^{\frac{3}{2}}, -6t^{\frac{1}{2}}, t^{-\frac{1}{2}}\right\rangle$$ The derivative of the function \(g(t) \mathbf{v}(t)\) is: $$\boxed{(g(t) \mathbf{v}(t))' = \left\langle 3t^{\frac{3}{2}}, -6t^{\frac{1}{2}}, t^{-\frac{1}{2}}\right\rangle}$$