Use the Chain Rule to find \( \frac{d w}{d t} \), where \( w=\sin (12 x) \cos (2 y), x=\frac{t}{4} \), and \( y=t^{5} \). \( \frac{\partial w}{\partial x}=12 \cos (12 x) \cos (2 y) \) (Type an expression using \( x \) and \( y \) as the variables.) \( \frac{d x}{d t}=\frac{1}{4} \) (Type an expression using \( t \) as the variable.) \( \frac{\partial w}{\partial y}=-2 \sin (12 x) \sin (2 y) \) (Type an expression using \( x \) and \( y \) as the variables.) \( \frac{d y}{d t}=\square \) (Type an expression using \( t \) as the variable.)

First, we need to find \( \frac{d y}{d t} \). Given that \( y = t^5 \), we differentiate it with respect to \( t \): \[ \frac{d y}{d t} = 5t^4 \] Now we can apply the Chain Rule to find \( \frac{d w}{d t} \): \[ \frac{d w}{d t} = \frac{\partial w}{\partial x} \cdot \frac{d x}{d t} + \frac{\partial w}{\partial y} \cdot \frac{d y}{d t} \] Substituting in our expressions: \[ \frac{d w}{d t} = \left(12 \cos (12 x) \cos (2 y)\right) \cdot \left(\frac{1}{4}\right) + \left(-2 \sin (12 x) \sin (2 y)\right) \cdot \left(5 t^4\right) \] Thus, the final expression for \( \frac{d w}{d t} \) is: \[ \frac{d w}{d t} = 3 \cos (12 x) \cos (2 y) - 10 t^4 \sin (12 x) \sin (2 y) \]