Use the Chain Rule to find \( \frac{d Q}{d t} \), where \( Q=\sqrt{2 x^{2}+2 y^{2}+2 z^{2}}, x=\sin t, y=\cos t \), and \( z=\sin t \). \( \frac{\partial Q}{\partial x}=\square \) (Type an expression using \( x, y \), and \( z \) as the variables.)

To find \( \frac{dQ}{dt} \) using the Chain Rule, we first need to calculate \( \frac{\partial Q}{\partial x} \). The expression for \( Q \) is: \[ Q = \sqrt{2x^2 + 2y^2 + 2z^2} \] Calculating \( \frac{\partial Q}{\partial x} \): Using the chain rule, we first express the derivative of \( Q \) with respect to its argument: \[ \frac{\partial Q}{\partial x} = \frac{1}{2\sqrt{2x^2 + 2y^2 + 2z^2}} \cdot \frac{\partial}{\partial x}(2x^2 + 2y^2 + 2z^2) \] Calculating the inner derivative: \[ \frac{\partial}{\partial x}(2x^2 + 2y^2 + 2z^2) = 4x \] Now substituting back into the expression for \( \frac{\partial Q}{\partial x} \): \[ \frac{\partial Q}{\partial x} = \frac{4x}{2\sqrt{2x^2 + 2y^2 + 2z^2}} = \frac{2x}{\sqrt{2x^2 + 2y^2 + 2z^2}} \] So the value of \( \frac{\partial Q}{\partial x} \) is: \[ \frac{\partial Q}{\partial x} = \frac{2x}{\sqrt{2(x^2 + y^2 + z^2)}} \]