Where did I go wrong on trying to solve this question on an exam?

Question

I took an exam yesterday, and I almost for a fact know I got this question wrong. I couldn't figure it out, since my answer wasn't an answer choice, so I ended up guessing. An explanation of what I did wrong and how to properly solve this would be great appreciated! The question: Evaluate $f'(\frac{\pi}{6})$, where $f(x)=\tan^{-1}(\sin 2x)$. So these are the steps that I did. First using chain rule$$\frac{d}{dx}\tan^{-1}(\sin 2x)=\frac{1}{1+(\sin 2x)^2}\cdot\frac{d}{dx}\sin(2x)$$ $$\frac{d}{dx}\sin(2x)=\cos(2x)\cdot2$$so $$\frac{d}{dx}\tan^{-1}(\sin 2x)=\frac{1}{1+(\sin 2x)^2}*2\cos(2x)$$ Plugging in $(\frac{\pi}{6})$ I ended with $$$$ $$\frac{d}{dx}\tan^{-1}\left(\sin\frac{\pi}{3}\right)=\frac{1}{1+(\sin\frac{\pi}{3})^2}\cdot2\cos\left(\frac{\pi}{3}\right)$$I ended with some answer that didn't match up with any of the answer choices, did I go about solving this wrong? If so, how would I solve it correctly? Thanks in advance.

Answer 1

We use the chain rule, and get $$f'(x) = \frac{1}{1 + (\sin (2x))^2} \cdot \cos (2x) \cdot 2 $$

See the chain rule in action: the derivative of $\tan^{-1} x $ is $\frac{1}{1 + x^2}$. The derivative of $\sin x$ is $\cos x$ and the derivative of $2x$ is $2$. Look at how everything fits.

Now: $$f' \left(\frac{\pi}{6}\right) = \frac{1}{1 + \sin^2 \frac{\pi}{3}} \cdot \cos \frac{\pi}{3} \cdot 2 = \frac{1}{1 + \frac{3}{4}} \cdot \frac{1}{2} \cdot 2 = \frac{4}{7} $$