How do I differentiate sin 2 ⁡ x?I thought that because this is true: sin...

You can only use the rule: $\frac{d}{dx}{x}^n=n{x}^{n-1}$ if x is just x and not a function. In the case for ${\sin }^{2}x$ then as you write it in the form $(\sin x{)}^2$ you can see that we can't use the power rule because it's not just an x term being raised to a constant power, it's a function being raised to a constant power. To differentiate ${\sin }^{2}x$ one must use the Chain Rule because $\sin x$ is a function of x within another function (the function that is squaring $\sin x$) $x^2$. The chain rule is: $\frac{d}{dx}f(g(x))=g^{\prime }(x)\times (f^{\prime }(g(x))$ so you can apply that rule for this case with $f(x)=x^2$ and $g(x)=\sin x$

You are using the chain rule incorrectly:
$\frac{d}{dx}f(g(x))=\frac{df}{dx}(g(x))\frac{dg}{dx}(x)$
Now $f(x)=x^2$ and $g(x)=\sin (x)$, therefore
$\frac{d}{dx}(\sin (x){)}^2=\stackrel{f^{\mathrm{\prime }}(g(x))}{\overbrace{2\sin (x)}}\cdot \stackrel{g^{\mathrm{\prime }}(x)}{\overbrace{\cos (x)}}$
And to explain your other point of confusion: $\sin (2x)=2\sin (x)\cos (x)$ is a well-known trigonometric formula