Answer
$$y'=\theta^2e^{-2\theta}\Big((3-2\theta)\cos5\theta-5\theta\sin5\theta\Big)$$
Work Step by Step
$$y=\theta^3e^{-2\theta}\cos5\theta=(\theta^3e^{-2\theta})(\cos5\theta)$$
The derivative of function $y$ is $$y'=(\theta^3e^{-2\theta})'(\cos5\theta)+(\theta^3e^{-2\theta})(\cos5\theta)'$$
According to the Chain Rule, we have
$$(\theta^3e^{-2\theta})'=(\theta^3)'e^{-2\theta}+\theta^3(e^{-2\theta})'=3\theta^2e^{-2\theta}+\theta^3(e^{-2\theta})(-2\theta)'$$
$$=3\theta^2e^{-2\theta}-2\theta^3e^{-2\theta}$$
$$(\cos5\theta)'=-\sin5\theta(5\theta)'=-5\sin5\theta$$
Therefore, $$y'=(3\theta^2e^{-2\theta}-2\theta^3e^{-2\theta})\cos5\theta-5\theta^3e^{-2\theta}\sin5\theta$$
$$y'=\theta^2e^{-2\theta}(3-2\theta)\cos5\theta-5\theta^3e^{-2\theta}\sin5\theta$$
$$y'=\theta^2e^{-2\theta}\Big((3-2\theta)\cos5\theta-5\theta\sin5\theta\Big)$$
