Answer
$$\frac{dz}{dt} =(4t^{-2} - 20t^3) sin(x+4y)$$
Work Step by Step
According to the Chain Rule:
$$\frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} +\frac{\partial z}{\partial y} \frac{dy}{dt} $$
$$ \frac{\partial z}{\partial x} = \frac{\partial (cos(x + 4y))}{\partial x} = -sin(x+4y)(1) = -sin(x + 4y)$$ $$\frac{dx}{dt} = \frac{d(5t^4)}{dt} = 20t^3$$ $$\frac{\partial z}{\partial y} = \frac{\partial (cos(x + 4y))}{\partial y} = -sin(x+4y)(4) = -4sin(x + 4y) $$ $$ \frac{dy}{dt} = \frac{d(1/t)}{dt} = -t^{-2}$$
Therefore:
$$\frac{dz}{dt} =( -sin(x+4y))(20t^3) + (-4sin(x + 4y))(-t^{-2})$$ $$\frac{dz}{dt} =-20t^3 sin(x+4y) + 4t^{-2}sin(x + 4y)$$ $$\frac{dz}{dt} =(4t^{-2} - 20t^3) sin(x+4y)$$
