\( 1 \leftarrow \quad \) 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.)

To find the derivative \( \frac{d w}{d t} \), we can use the chain rule. Since \( w = 6xyz \), we'll also need to find \( \frac{dx}{dt} \), \( \frac{dy}{dt} \), and \( \frac{dz}{dt} \) based on the given \( x, y, \) and \( z \). First, we have: - \( x = 4t^{4} \) so \( \frac{dx}{dt} = 16t^{3} \). Next, we differentiate \( y \) and \( z \): - \( y = t^{-1} \) so \( \frac{dy}{dt} = -t^{-2} \). - \( z = 2t^{-3} \) so \( \frac{dz}{dt} = -6t^{-4} \). Now, let's apply the chain rule: \[ \frac{d w}{d t} = \frac{\partial w}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial w}{\partial y} \cdot \frac{dy}{dt} + \frac{\partial w}{\partial z} \cdot \frac{dz}{dt}. \] Now we can plug into the given definitions: 1. \( \frac{\partial w}{\partial x} = 6yz \) becomes \( 6(t^{-1})(2t^{-3}) = 12t^{-4} \). 2. \( \frac{\partial w}{\partial y} = 6xz \) becomes \( 6(4t^{4})(2t^{-3}) = 48t^{1} \). 3. \( \frac{\partial w}{\partial z} = 6xy \) becomes \( 6(4t^{4})(t^{-1}) = 24t^{3} \). Now substituting everything back: \[ \frac{d w}{d t} = 12t^{-4} \cdot 16t^{3} + 48t^{1} \cdot (-t^{-2}) + 24t^{3} \cdot (-6t^{-4}). \] Calculating this gives: 1. \( 192t^{-1} \) 2. \( -48t^{-1} \) 3. \( -144t^{-1} \) So combining them we get: \[ \frac{d w}{d t} = 192t^{-1} - 48t^{-1} - 144t^{-1} = 0. \] Now, we fill in the blanks: \(\frac{dx}{dt}=16 t^{3}\)