Use the Chain Rule to find \( \frac{d w}{d t} \), where \( w=\cos (8 x) \sin (4 y), x=\frac{t}{2} \), and \( y=t^{3} \). \( \frac{d w}{d t}=\square \) (Type an expression using \( t \) as the variable. Use parentheses to clearly denote the argument of each function.)

To find \( \frac{d w}{d t} \) using the Chain Rule, we begin by calculating the partial derivatives of \( w \) with respect to \( x \) and \( y \): 1. \( \frac{\partial w}{\partial x} = -8 \sin(8x) \sin(4y) \) 2. \( \frac{\partial w}{\partial y} = 4 \cos(8x) \cos(4y) \) Next, we find the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \): - \( x = \frac{t}{2} \) implies \( \frac{dx}{dt} = \frac{1}{2} \) - \( y = t^3 \) implies \( \frac{dy}{dt} = 3t^2 \) Now, we apply the Chain Rule: \[ \frac{d w}{d t} = \frac{\partial w}{\partial x} \frac{dx}{dt} + \frac{\partial w}{\partial y} \frac{dy}{dt} \] Substituting our expressions from above: \[ \frac{d w}{d t} = \left(-8 \sin\left(8\left(\frac{t}{2}\right)\right) \sin\left(4(t^3)\right)\right) \left(\frac{1}{2}\right) + \left(4 \cos\left(8\left(\frac{t}{2}\right)\right) \cos\left(4(t^3)\right)\right) (3t^2) \] Simplifying, we get: \[ \frac{d w}{d t} = -4 \sin(4t) \sin(4t^3) + 12t^2 \cos(4t) \cos(4t^3) \] So, the expression for \( \frac{d w}{d t} \) is: \[ \frac{d w}{d t} = -4 \sin(4t) \sin(4t^3) + 12t^2 \cos(4t) \cos(4t^3) \]