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.) \( \frac{d x}{d t}=\square \) (Type an expression using \( t \) as the variable.) (Ty

To find \( \frac{d w}{d t} \), we will use the chain rule as follows: \[ \frac{d w}{d t} = \frac{\partial w}{\partial x} \frac{d x}{d t} + \frac{\partial w}{\partial y} \frac{d y}{d t} + \frac{\partial w}{\partial z} \frac{d z}{d t} \] We have: \[ \frac{\partial w}{\partial x} = 6 y z, \quad \frac{\partial w}{\partial y} = 6 x z, \quad \frac{\partial w}{\partial z} = 6 x y \] Now, we need to calculate \( \frac{d x}{d t} \), \( \frac{d y}{d t} \), and \( \frac{d z}{d t} \). Given \( x = 4 t^{4} \): \[ \frac{d x}{d t} = 16 t^{3} \] Next, since \( y = t^{-1} \): \[ \frac{d y}{d t} = -t^{-2} \] And for \( z = 2 t^{-3} \): \[ \frac{d z}{d t} = -6 t^{-4} \] Now substituting everything back into the expression for \( \frac{d w}{d t} \): \[ \frac{d w}{d t} = (6 y z) \cdot (16 t^{3}) + (6 x z) \cdot (-t^{-2}) + (6 x y) \cdot (-6 t^{-4}) \] Now to summarize: \(\frac{d x}{d t}=16 t^{3}\)