differentiate
$y = \frac{2t\sin(2t)}{1 + 2t}$

Question

Accepted Answer

# Explanation:
## Step1: Recall the quotient rule
The quotient rule states that if $y = \frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 2t\sin(2t)$ and $v=1 + 2t$.
## Step2: Find $u^\prime$ using the product rule
The product rule: if $u = ab$ ($a = 2t$, $b=\sin(2t)$), then $u^\prime=a^\prime b+ab^\prime$.
$a^\prime = 2$, $b^\prime = 2\cos(2t)$. So $u^\prime=2\sin(2t)+2t\times2\cos(2t)=2\sin(2t) + 4t\cos(2t)$.
$v^\prime=2$.
## Step3: Apply the quotient rule
$y^\prime=\frac{(2\sin(2t)+4t\cos(2t))(1 + 2t)-2t\sin(2t)\times2}{(1 + 2t)^{2}}$
$$
\begin{align*}
y^\prime&=\frac{2\sin(2t)+4t\sin(2t)+4t\cos(2t)+8t^{2}\cos(2t)-4t\sin(2t)}{(1 + 2t)^{2}}\
&=\frac{2\sin(2t)+4t\cos(2t)+8t^{2}\cos(2t)}{(1 + 2t)^{2}}\
&=\frac{2\sin(2t)+4t\cos(2t)(1 + 2t)}{(1 + 2t)^{2}}\
&=\frac{2\sin(2t)}{(1 + 2t)^{2}}+\frac{4t\cos(2t)}{1 + 2t}
\end{align*}
$$

# Answer:
$y^\prime=\frac{2\sin(2t)}{(1 + 2t)^{2}}+\frac{4t\cos(2t)}{1 + 2t}$

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the quotient rule

Step2: Find \(u^\prime\) using the product rule

Step3: Apply the quotient rule

Answer: