Find the derivative \( \frac{d w}{d t} \), where \( w=6 x y z, x=4 t^{4}, y=t^{-1} \), and \( z=2 t^{-3} \) \( \frac{\partial w}{\partial x}=6 y z \) (Type an expression using \( x, y \), and \( z \) as the variables.) \( \frac{\partial w}{\partial y}=6 x z \) (Type an expression using \( x, y \), and \( z \) as the variables.) \( \frac{\partial w}{\partial z}=6 x y \) (Type an expression using \( x, y \), and \( z \) as the variables.) \( \frac{d x}{d t}=16 t^{3} \) (Type an expression using \( t \) as the variable.) \( \frac{d y}{d t}=\square \) (Type an expression using \( t \) as the variable.)

To find \( \frac{d y}{d t} \), where \( y = t^{-1} \), we differentiate: \[ \frac{d y}{d t} = -1 t^{-2} = -\frac{1}{t^{2}} \] Now, let's move on to \( z \): We need to find \( \frac{d z}{d t} \) where \( z = 2 t^{-3} \). The derivative is: \[ \frac{d z}{d t} = 2 \cdot (-3) t^{-4} = -\frac{6}{t^{4}} \] Finally, we can compute \( \frac{d w}{d t} \) using the chain rule: \[ \frac{d w}{d t} = \frac{\partial w}{\partial x} \cdot \frac{d x}{d t} + \frac{\partial w}{\partial y} \cdot \frac{d y}{d t} + \frac{\partial w}{\partial z} \cdot \frac{d z}{d t} \] Substituting the values and expressions we have: \[ \frac{d w}{d t} = 6 y z \cdot (16 t^{3}) + 6 x z \cdot \left(-\frac{1}{t^{2}}\right) + 6 x y \cdot \left(-\frac{6}{t^{4}}\right) \] Use the definitions of \( x, y, z \) to simplify and substitute back, if necessary!