Use the Chain Rule to find $\frac{d w}{d t}$, where $w=\sin 8 x \cos 2 y, x=\frac{t}{4}$, and $y=t^{5}$.
\[
\frac{\partial w}{\partial x}=
\]
(Type an expression using $x$ and $y$ as the variables.)
\[
\frac{d x}{d t}=
\]
(Type an expression using $t$ as the variable.)
\[
\frac{\partial w}{\partial y}=
\]
(Type an expression using $x$ and $y$ as the variables.)
\[
\frac{d y}{d t}=
\]
(Type an expression using $t$ as the variable.)
\[
\frac{\mathrm{dw}}{\mathrm{dt}}=
\]
(Type an expression using $t$ as the variable.)

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

Step 2 ：\[\frac{\partial w}{\partial x} = 8\cos 8x \cos 2y\]

Step 3 ：\[\frac{\partial w}{\partial y} = -2\sin 8x \sin 2y\]

Step 4 ：Next, we calculate the derivatives of \(x\) and \(y\) with respect to \(t\).

Step 5 ：\[\frac{dx}{dt} = \frac{1}{4}\]

Step 6 ：\[\frac{dy}{dt} = 5t^{4}\]

Step 7 ：Finally, we use the Chain Rule to find \(\frac{dw}{dt}\).

Step 8 ：\[\frac{dw}{dt} = \frac{\partial w}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial w}{\partial y} \cdot \frac{dy}{dt}\]

Step 9 ：Substitute the above values into the equation.

Step 10 ：\[\frac{dw}{dt} = 8\cos 8x \cos 2y \cdot \frac{1}{4} - 2\sin 8x \sin 2y \cdot 5t^{4}\]

Step 11 ：\[\frac{dw}{dt} = 2\cos 8x \cos 2y - 10t^{4}\sin 8x \sin 2y\]

Step 12 ：Substitute \(x = \frac{t}{4}\) and \(y = t^{5}\) into the equation.

Step 13 ：\[\frac{dw}{dt} = 2\cos 2t \cos 2t^{5} - 10t^{4}\sin 2t \sin 2t^{5}\]

Step 14 ：\[\frac{dw}{dt} = \boxed{2\cos 2t \cos 2t^{5} - 20t^{4}\sin 2t \sin 2t^{5}}\]