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.) \( \frac{\partial w}{\partial y}=\square \) (Type an expression using \( x \) and \( y \) as the variables.)

To find \( \frac{\partial w}{\partial y} \), we start by differentiating \( w \) with respect to \( y \): \[ \frac{\partial w}{\partial y} = -2 \sin(12x) \sin(2y) \] Now, we can compute \( \frac{d w}{d t} \) using the Chain Rule: \[ \frac{d w}{d t} = \frac{\partial w}{\partial x} \frac{d x}{d t} + \frac{\partial w}{\partial y} \frac{d y}{d t} \] From the information provided, we substitute \( \frac{\partial w}{\partial x} \), \( \frac{d x}{d t} \), and our derived \( \frac{\partial w}{\partial y} \). Thus, we need to calculate \( \frac{d y}{d t} \): Since \( y = t^5 \), \[ \frac{d y}{d t} = 5t^4 \] Now substituting everything into the expression: \[ \frac{d w}{d t} = 12 \cos(12x) \cos(2y) \cdot \frac{1}{4} + (-2 \sin(12x) \sin(2y)) \cdot (5t^4) \] Combining these, we get: \[ \frac{d w}{d t} = 3 \cos(12x) \cos(2y) - 10t^4 \sin(12x) \sin(2y) \]