\( \mid \) 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 derivatives and expressions requested, we start with the given functions. 1. **Derivatives of \( w \)**: - Given \( w = 6xyz \): - \( \frac{\partial w}{\partial x} = 6yz \) - \( \frac{\partial w}{\partial y} = 6xz \) - \( \frac{\partial w}{\partial z} = 6xy \) 2. **Finding \( \frac{d w}{d t} \)**: By using the chain rule, we have: \[ \frac{d w}{d t} = \frac{\partial w}{\partial x} \frac{d x}{d t} + \frac{\partial w}{\partial y} \frac{d y}{d t} + \frac{\partial w}{\partial z} \frac{d z}{d t} \] And we know: \[ x = 4t^4 \quad \Rightarrow \quad \frac{d x}{d t} = 16t^3 \] Now let's express \( \frac{d y}{d t} \) and \( \frac{d z}{d t} \): - For \( y = t^{-1} \): \[ \frac{d y}{d t} = -t^{-2} \] - For \( z = 2t^{-3} \): \[ \frac{d z}{d t} = -6t^{-4} \] 3. **Putting everything together**: Now, substituting these into the derivative expression yields: \[ \frac{d w}{d t} = \left(6yz\right)(16t^3) + \left(6xz\right)(-t^{-2}) + \left(6xy\right)(-6t^{-4}) \] From here, you can plug in the expressions for \( x, y, \) and \( z \) in terms of \( t \) to calculate \( \frac{d w}{d t} \). In conclusion, the simplified expressions are: - \( \frac{\partial w}{\partial x} = 6yz \) - \( \frac{\partial w}{\partial y} = 6xz \) - \( \frac{\partial w}{\partial z} = 6xy \) - \( \frac{d x}{d t} = 16t^3 \)