<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] Your mistake here.
$$\sin^{-1} \frac{2x}{1+x^2} = \frac1{\sqrt{1-\left(\frac{2x}{1+x^2}\right)^2}}.$$
Its not $\sin^{-1} \dfrac{2x}{1+x^2} = \dfrac1{\sqrt{1-\frac{2x}{1+x^2}}}$.
After this your quotient rule is fine.
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