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

To find \( \frac{dz}{dt} \) using the Chain Rule, we can start by computing the necessary derivatives. Let's first express \( z \) in terms of \( t \) through \( x \) and \( y \): Given: - \( z = 3x \sin y \) - \( x = t^4 \) - \( y = 3t^2 \) Now, find the partial derivatives: 1. Calculate \( \frac{dz}{dx} \) and \( \frac{dz}{dy} \): \[ \frac{dz}{dx} = 3 \sin y \] \[ \frac{dz}{dy} = 3x \cos y \] 2. Calculate \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \): \[ \frac{dx}{dt} = 4t^3 \] \[ \frac{dy}{dt} = 6t \] Now, apply the Chain Rule: \[ \frac{dz}{dt} = \frac{dz}{dx} \cdot \frac{dx}{dt} + \frac{dz}{dy} \cdot \frac{dy}{dt} \] Substituting the derivatives we found: \[ \frac{dz}{dt} = (3 \sin(3t^2)) \cdot (4t^3) + (3t^4) \cdot (6t) \cos(3t^2) \] Thus, combining these gives us: \[ \frac{dz}{dt} = 12t^3 \sin(3t^2) + 18t^5 \cos(3t^2) \] So, the final expression is: \[ \frac{dz}{dt} = 12t^3 \sin(3t^2) + 18t^5 \cos(3t^2) \]