Let \( z(x, y)=-9 e^{x} \sin (y) \) where \( x=t^{5} \) \& \( y=6 \pi t \). Calculate \( \frac{d z}{d t} \) by first finding \( \frac{d x}{d t} \) \& \( \frac{d y}{d t} \) and using the chain rule. \( \frac{d x}{d t}= \) \( \frac{d y}{d t}= \) Now use the chain rule to calculate the following: \( \frac{d z}{d t}= \)

To solve for \( \frac{d z}{d t} \), we first need to compute \( \frac{d x}{d t} \) and \( \frac{d y}{d t} \). Starting with \( x = t^5 \): \[ \frac{d x}{d t} = 5t^4 \] Next, for \( y = 6\pi t \): \[ \frac{d y}{d t} = 6\pi \] Now, applying the chain rule for \( z(x, y) = -9 e^{x} \sin(y) \), we have: \[ \frac{d z}{d t} = \frac{\partial z}{\partial x} \frac{d x}{d t} + \frac{\partial z}{\partial y} \frac{d y}{d t} \] First, we calculate the partial derivatives. For \( z \): \[ \frac{\partial z}{\partial x} = -9 e^{x} \sin(y) \] \[ \frac{\partial z}{\partial y} = -9 e^{x} \cos(y) \] Now substituting in the chain rule: \[ \frac{d z}{d t} = \left(-9 e^{x} \sin(y)\right) (5t^4) + \left(-9 e^{x} \cos(y)\right) (6\pi) \] \[ = -9 e^{t^5} \sin(6\pi t) (5t^4) - 9 e^{t^5} \cos(6\pi t) (6\pi) \] So, the final result is: \[ \frac{d z}{d t} = -9 e^{t^5} \left(5t^4 \sin(6\pi t) + 6\pi \cos(6\pi t)\right) \]