Let $z(x,y)=-6e^{x}sin(y)$ where $x=t^2$&$y=8t.$

Calculate $\frac{dz}{dt}$ by first finding $\frac{dx}{dt}$&$\frac{dy}{dt}$ and using the chain rule.

$\frac{dx}{dt}=$

$\frac{dy}{dt}=$

Now use the chain rule to calculate the following:

$\frac{dz}{dt}=$

Hint: Recall the chain rule: $\frac{dz}{dt}=(\frac{\text{d}\text{e}\text{l}\text{z}}{\text{d}\text{e}\text{l}\text{x}}*\frac{dx}{dt})+(\frac{\text{d}\text{e}\text{l}\text{z}}{\text{d}\text{e}\text{l}\text{y}}*\frac{dy}{dt})$