Find the derivative $\frac{d w}{d t}$, where $w=2 x y z, x=6 t^{4}, y=5 t^{-1}$, and $z=3 t^{-3}$ \[ \frac{\partial w}{\partial x}= \] (Type an expression using $x, y$, and $z$ as the variables.) \[ \frac{\partial w}{\partial y}= \] (Type an expression using $x, y$, and $z$ as the variables.) \[ \frac{\partial w}{\partial z}= \] (Type an expression using $x, y$, and $z$ as the variables.) \[ \frac{\mathrm{dx}}{\mathrm{dt}}= \] (Type an expression using $t$ as the variable.) \[ \frac{d y}{d t}= \] (Type an expression using $t$ as the variable.) \[ \frac{d z}{d t}= \] (Type an expression using $t$ as the variable.) \[ \frac{d w}{d t}= \] (Type an expression using $t$ as the variable.)

Step 1 ：First, we calculate the partial derivatives of \(w\) with respect to \(x\), \(y\), and \(z\).

Step 2 ：\[\frac{\partial w}{\partial x} = 2yz\]

Step 3 ：\[\frac{\partial w}{\partial y} = 2xz\]

Step 4 ：\[\frac{\partial w}{\partial z} = 2xy\]

Step 5 ：Next, we calculate the derivatives of \(x\), \(y\), and \(z\) with respect to \(t\).

Step 6 ：\[\frac{dx}{dt} = 24t^3\]

Step 7 ：\[\frac{dy}{dt} = -5t^{-2}\]

Step 8 ：\[\frac{dz}{dt} = -9t^{-4}\]

Step 9 ：Finally, we use the chain rule to find the derivative of \(w\) with respect to \(t\).

Step 10 ：\[\frac{dw}{dt} = \frac{\partial w}{\partial x}\frac{dx}{dt} + \frac{\partial w}{\partial y}\frac{dy}{dt} + \frac{\partial w}{\partial z}\frac{dz}{dt}\]

Step 11 ：Substitute the above values into the equation, we get

Step 12 ：\[\frac{dw}{dt} = 2yz(24t^3) + 2xz(-5t^{-2}) + 2xy(-9t^{-4})\]

Step 13 ：\[\frac{dw}{dt} = 48yt^3z - 10xzt^{-2} - 18xyt^{-4}\]

Step 14 ：Substitute \(x=6t^4\), \(y=5t^{-1}\), and \(z=3t^{-3}\) into the equation, we get

Step 15 ：\[\frac{dw}{dt} = 48(5t^{-1})(t^3)(3t^{-3}) - 10(6t^4)(3t^{-3})(t^{-2}) - 18(6t^4)(5t^{-1})(t^{-4})\]

Step 16 ：\[\frac{dw}{dt} = 720 - 180 - 540\]

Step 17 ：\[\frac{dw}{dt} = \boxed{0}\]