Use the Chain Rule to find \( \frac{d z}{d t} \), where \( z=3 x \cos y, x=t^{5} \), and \( y=6 t^{4} \). \( \frac{\partial z}{\partial x}=3 \cos y \) (Type an expression using \( x \) and \( y \) as the variables.) \( \frac{d x}{d t}=5 t^{4} \) (Type an expression using \( t \) as the variable.) \( \frac{\partial z}{\partial y}=\square \) (Type an expression using \( x \) and \( y \) as the variables.)

To find \( \frac{d z}{d t} \) using the Chain Rule, we start with \[ \frac{d z}{d t} = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt}. \] We already have \[ \frac{\partial z}{\partial x} = 3 \cos y, \] \[ \frac{d x}{d t} = 5 t^{4}. \] Next, we need to calculate \( \frac{\partial z}{\partial y} \). Given \( z = 3x \cos y \), we differentiate with respect to \( y \): \[ \frac{\partial z}{\partial y} = 3x \cdot (-\sin y) = -3x \sin y. \] Now we also have to find \( \frac{dy}{dt} \): \[ y = 6t^{4} \rightarrow \frac{dy}{dt} = 24t^{3}. \] Putting these together gives us the full expression for \( \frac{dz}{dt} \): \[ \frac{dz}{dt} = (3 \cos y)(5 t^{4}) + (-3x \sin y)(24t^{3}). \] Now, if we plug in our expressions: 1. \( \frac{\partial z}{\partial y} = -3x \sin y \) 2. \( \frac{dy}{dt} = 24t^{3} \) Conclusively, our final answers for the expressions are: \[ \frac{\partial z}{\partial y} = -3x \sin y, \] \[ \frac{dy}{dt} = 24t^{3}. \]