\( 1 \leftarrow \quad \) 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 start by calculating \( \frac{d w}{d x} \) and \( \frac{d w}{d y} \), then we also need \( \frac{d x}{d t} \) and \( \frac{d y}{d t} \). 1. First, calculate \( \frac{d w}{d x} \) and \( \frac{d w}{d y} \): \[ \frac{d w}{d x} = \frac{d}{d x}[\cos(8x) \sin(4y)] = -8\sin(8x)\sin(4y) \] \[ \frac{d w}{d y} = \frac{d}{d y}[\cos(8x) \sin(4y)] = 4\cos(8x)\cos(4y) \] 2. Now, calculate \( \frac{d x}{d t} \) and \( \frac{d y}{d t} \): \[ x = \frac{t}{2} \Rightarrow \frac{d x}{d t} = \frac{1}{2} \] \[ y = t^3 \Rightarrow \frac{d y}{d t} = 3t^2 \] 3. Now apply the Chain Rule: \[ \frac{d w}{d t} = \frac{d w}{d x} \cdot \frac{d x}{d t} + \frac{d w}{d y} \cdot \frac{d y}{d t} \] Substituting in the derivatives: \[ \frac{d w}{d t} = \left(-8\sin(8x)\sin(4y)\right) \cdot \left(\frac{1}{2}\right) + \left(4\cos(8x)\cos(4y)\right) \cdot \left(3t^2\right) \] 4. Substitute \( x \) and \( y \): \[ = -4\sin(8\left(\frac{t}{2}\right))\sin(4(t^3)) + 12t^2\cos(8\left(\frac{t}{2}\right))\cos(4(t^3)) \] 5. Simplifying the expression: \[ \frac{d w}{d t} = -4\sin(4t)\sin(4t^3) + 12t^2\cos(4t)\cos(4t^3) \] Thus, the final answer is: \[ \frac{d w}{d t} = -4\sin(4t)\sin(4t^3) + 12t^2\cos(4t)\cos(4t^3) \]