Find the derivative of the function.
-$$\mathrm { y } = \theta ^ { 4 } \mathrm { e } ^ { - 2 \theta } \cos 4 \theta$$
A) $- 56 \theta ^ { 3 } \mathrm { e } ^ { - 2 \theta } \sin 4 \theta$
B) $\theta ^ { 3 } \mathrm { e } ^ { - 2 \theta } ( 4 \cos 4 \theta - 2 \cos 4 \theta + 4 \cos 4 \theta )$
C) $- 4 \theta ^ { 3 } \mathrm { e } ^ { - 2 \theta } \sin 4 \theta$
D) $\theta ^ { 3 } \mathrm { e } ^ { - 2 \theta } ( 4 \cos 4 \theta - 2 \theta \cos 4 \theta - 4 \theta \sin 4 \theta )$

Question

Quizplus · Accepted Answer

We need to use the product rule and the chain rule to find the derivative of the given function.

Let $u = 	heta^4$ and $v = e^{-2	heta}\cos4	heta$. Then, we have:

\begin{align*}
y &= uv \
\frac{dy}{d	heta} &= u\frac{dv}{d	heta} + v\frac{du}{d	heta}
\end{align*}

Using the product rule and the chain rule, we get:

\begin{align*}
u\frac{dv}{d	heta} &= 	heta^4(-2e^{-2	heta}\cos4	heta - 4e^{-2	heta}\sin4	heta) \
v\frac{du}{d	heta} &= e^{-2	heta}\cos4	heta \cdot 4	heta^3 \
&= 4	heta^3 e^{-2	heta}\cos4	heta \
\frac{dy}{d	heta} &= 	heta^4(-2e^{-2	heta}\cos4	heta - 4e^{-2	heta}\sin4	heta) + 4	heta^3 e^{-2	heta}\cos4	heta \
&= 4	heta^3 e^{-2	heta}\cos4	heta - 2	heta^4 e^{-2	heta}\cos4	heta - 4	heta^4 e^{-2	heta}\sin4	heta \
&= 	heta^3 e^{-2	heta}(4\cos4	heta - 2	heta\cos4	heta - 4	heta\sin4	heta)
\end{align*}

Therefore, the answer is D, $	heta^3 e^{-2	heta}(4\cos4	heta - 2	heta\cos4	heta - 4	heta\sin4	heta)$.

Find the Derivative of the Function y=θ4e−2θcos⁡4θ\mathrm { y } = \theta ^ { 4 } \mathrm { e } ^ { - 2 \theta } \cos 4 \thetay=θ4e−2θcos4θ

Find the Derivative of the Function $\mathrm { y } = \theta ^ { 4 } \mathrm { e } ^ { - 2 \theta } \cos 4 \theta$