<s>[INST] Find the derivative of $f=\arcsin\left(\frac{2x}{1+x^2}\right)$ I'm trying to find the derivative of $f=\arcsin\left(\frac{2x}{1+x^2}\right)$. I think I'm mistaken and perhaps using the chain rule incorrectly. 
Let $g(x) = \frac{2x}{1+x^2}$ and let $h(x) = \arcsin x$
According to the chain rule - 
$$f'(x) = \frac{1}{\sqrt{1-\frac{2x}{1+x^2}}}⋅((2⋅(1+x^2 )-2x⋅2x)/(1+x^2 )^2 ) = \cdots \frac{-2(x^4-1)}{x-1}$$
Is this a correct usage of the chain rule?
 [/INST] With the chain rule:
$$f'(x)=\frac{1}{\sqrt{1-\left(\frac{2x}{1+x^2}\right)^2}}\left(\frac{2(1+x^2)-2x(2x)}{(1+x^2)^2}\right)=\frac{1+x^2}{\sqrt{(1+x^2)^2-4x^2}}\left(\frac{2-2x^2}{(1+x^2)^2}\right)$$
$$=\frac{2-2x^2}{1+x^2}\frac{1}{|1-x^2|}=\frac{2}{1+x^2}\frac{1-x^2}{|1-x^2|}$$
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