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.)

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We are given
$$
w = 6xyz,\quad x = 4t^4,\quad y = t^{-1},\quad z = 2t^{-3}.
$$

Using the chain rule, we have
$$
\frac{dw}{dt} = \frac{\partial w}{\partial x}\frac{dx}{dt} + \frac{\partial w}{\partial y}\frac{dy}{dt} + \frac{\partial w}{\partial z}\frac{dz}{dt},
$$
with
$$
\frac{\partial w}{\partial x} = 6yz,\quad \frac{\partial w}{\partial y} = 6xz,\quad \frac{\partial w}{\partial z} = 6xy.
$$
We are also given:
$$
\frac{dx}{dt} = 16t^3,\quad \frac{dy}{dt} = -t^{-2}.
$$
We need to find:
$$
\frac{dz}{dt} \quad \text{where} \quad z=2t^{-3}.
$$

Differentiate $$ z = 2t^{-3} $$ with respect to $$ t $$:
$$
\frac{dz}{dt} = 2 \cdot \frac{d}{dt}\left(t^{-3}\right) = 2(-3t^{-4}) = -6t^{-4}.
$$

Thus, 
$$
\frac{dz}{dt} = -6t^{-4}.
$$

Now, substitute the partial derivatives and the derivatives of $$ x $$, $$ y $$, and $$ z $$ into the chain rule formula:
$$
\frac{dw}{dt} = (6yz)(16t^3) + (6xz)(-t^{-2}) + (6xy)(-6t^{-4}).
$$

Step 1. Compute each term individually:

1. First term:
   $$
   (6yz)(16t^3) = 96\, yzt^3.
   $$
   Substituting $$ y = t^{-1} $$ and $$ z = 2t^{-3} $$:
   $$
   96 \cdot \left(t^{-1}\right) \cdot \left(2t^{-3}\right) t^3 = 96 \cdot 2 \cdot t^{-1-3+3} = 192\,t^{-1}.
   $$

2. Second term:
   $$
   (6xz)(-t^{-2}) = -6\, xz\,t^{-2}.
   $$
   Substituting $$ x = 4t^4 $$ and $$ z = 2t^{-3} $$:
   $$
   -6 \cdot (4t^4) \cdot (2t^{-3}) t^{-2} = -6 \cdot 8\, t^{4-3-2} = -48\,t^{-1}.
   $$

3. Third term:
   $$
   (6xy)(-6t^{-4}) = -36\, xy\,t^{-4}.
   $$
   Substituting $$ x = 4t^4 $$ and $$ y = t^{-1} $$:
   $$
   -36 \cdot (4t^4) \cdot (t^{-1}) t^{-4} = -36 \cdot 4\, t^{4-1-4} = -144\,t^{-1}.
   $$

Step 2. Combine the three terms:
$$
\frac{dw}{dt} = 192\,t^{-1} - 48\,t^{-1} - 144\,t^{-1}.
$$

Simplify:
$$
192\,t^{-1} - 48\,t^{-1} = 144\,t^{-1},
$$
$$
144\,t^{-1} - 144\,t^{-1} = 0.
$$

Thus, 
$$
\frac{dw}{dt} = 0.
$$
$$ \frac{dw}{dt} = 0 $$

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