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

To find \( \frac{d z}{d t} \), we differentiate \( z = 2 t^{-3} \): \[ \frac{d z}{d t} = -6 t^{-4} \] Now you have all the necessary components! For \( w = 6xyz \), by applying the chain rule: \[ \frac{d w}{d t} = \frac{\partial w}{\partial x} \cdot \frac{d x}{d t} + \frac{\partial w}{\partial y} \cdot \frac{d y}{d t} + \frac{\partial w}{\partial z} \cdot \frac{d z}{d t} \] Inserting your previously calculated derivatives and expressions for \( x, y, z \): 1. \( \frac{\partial w}{\partial x} = 6yz \) 2. \( \frac{\partial w}{\partial y} = 6xz \) 3. \( \frac{\partial w}{\partial z} = 6xy \) Therefore, \[ \frac{d w}{d t} = (6 y z) \cdot (16 t^{3}) + (6 x z) \cdot (-t^{-2}) + (6 x y) \cdot (-6 t^{-4}) \] Now, substitute \( x, y, z \) with their respective expressions in terms of \( t \) to find the final expression!