Answer
$r' = 2\theta\cos(\theta^2)\cos(2\theta) -2\sin(\theta^2)\sin(2\theta)$
Work Step by Step
To solve this use product rule and the chain rule, which is as follows:
The given function is: $r=\sin(\theta^2)\cos(2\theta)$
Let's denote $u = \sin(\theta^2)$ and $v = \cos(2\theta)$. Then, the function can be rewritten as: $r = uv$.
Now, let's find the derivative of $r$ with respect to $\theta$ using the product rule: $$r' = u'v + uv'$$ First, let's find the derivative of $u$ with respect to $\theta$, using the chain rule: $$u' = 2\theta\cos(\theta^2)$$ Next, let's find the derivative of $v$ with respect to $\theta$ using the chain rule: $$v' = -2\sin(2\theta)$$ Now, substitute the derivatives of $u$ and $v$ back into the product rule formula: $$\begin{aligned}
r'& = 2\theta\cos(\theta^2)\cos(2\theta) +\sin(\theta^2)(-2\sin(2\theta))\\
&=2\theta\cos(\theta^2)\cos(2\theta) -2\sin(\theta^2)\sin(2\theta).
\end{aligned}$$ Therefore, the derivative of the given function is $$r' = 2\theta\cos(\theta^2)\cos(2\theta) -2\sin(\theta^2)\sin(2\theta).$$
