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.)
\[\frac{dw}{dt} = \boxed{0}\]
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 :\[\frac{dw}{dt} = \boxed{0}\]