Geometric proof of chain rule with the derivative of $\sin(2x)$ 
I'm following this post https://math.stackexchange.com/a/2169/612996 as my example and I've figured out how it works for $\sin(\theta)$,
During my first try: I keep on missing the factor of $2$ when it's $\sin(2\theta)$. I always get $\cos(2\theta)$
In my work, I've made the angle $2\theta$ and then changed everything that has $\theta$ to $2\theta$, but I've kept the increase in the angle $\Delta\theta$. I think I could reverse-engineer the answer so I use $\Delta 2 \theta$ and get the right answer, but I don't know why that should work and not $\Delta\theta$ since isn't it just a small increase in the angle anyway? (Or is it a small increase that is relative to the angle, and that's why there's a 2?)
During my second try: I understand the algebraic way of doing the chain rule but I want to do it with some sort of geometric intuition.
 A: You are taking the derivative with respect to $x$ or $\theta$ not $2 x$ or $2\theta$
So you want $$\frac{d}{d\theta} \sin(2\theta) =\lim\limits_{\Delta \theta \to 0} \frac{\sin(2(\theta+\Delta \theta))-\sin(2\theta) }{\Delta \theta}  \\= \lim\limits_{\Delta \theta \to 0} \frac{2\sin\left(\frac{2\theta+2\Delta \theta -2 \theta}{2}\right)\cos\left(\frac{2\theta+2\Delta \theta +2 \theta}{2}\right) }{\Delta \theta} \\= \lim\limits_{\Delta \theta \to 0} \frac{2\sin\left(\Delta\theta\right)\cos\left(2\theta+\Delta \theta\right) }{\Delta \theta}\\= \lim\limits_{\Delta \theta \to 0} \frac{2\Delta\theta\cos\left(2\theta\right) }{\Delta \theta} \\=2 \cos(2\theta)$$
A: In your picture, you want to change all the $\theta$'s into $2\theta$'s, right?  Then the whole argument for $\sin \theta$ goes through, but instead for $\sin 2\theta$; especially you have $\Delta 2\theta$ instead of $\Delta \theta.$
So if $\theta$ increases by $\Delta \theta$, what is the increase in $\Delta 2\theta$?  Answer:  It's $2\Delta\theta.$  So I think that your sticking point comes down to:
$$\Delta 2\theta = 2\Delta \theta.$$
