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 \( \frac{\partial z}{\partial y} \), we start with our function of \( z \): \[ z = 3x \cos y \] Using the product rule and the chain rule, we differentiate \( z \) with respect to \( y \): \[ \frac{\partial z}{\partial y} = 3x \cdot \frac{d}{dy}(\cos y) = 3x (-\sin y) = -3x \sin y \] Now, let’s summarize: 1. \( \frac{\partial z}{\partial y} = -3x \sin y \) Now, let’s find \( \frac{d z}{d t} \) utilizing the chain rule: \[ \frac{d z}{d t} = \frac{\partial z}{\partial x} \frac{d x}{d t} + \frac{\partial z}{\partial y} \frac{d y}{d t} \] We already know that: - \( \frac{\partial z}{\partial x} = 3 \cos y \) - \( \frac{d x}{d t} = 5 t^{4} \) - For \( y = 6 t^{4} \), we differentiate to find: \[ \frac{d y}{d t} = 24 t^{3} \] Now plug everything into \( \frac{d z}{d t} \): \[ \frac{d z}{d t} = (3 \cos y)(5 t^{4}) + (-3x \sin y)(24 t^{3}) \] Thus we have an expression for \( \frac{d z}{d t} \) which incorporates everything and can be used for further computation or differentiation as needed.