<s>[INST] Differentiation Calculus: $\tan^{-1} \text{Problem}$ Well, Today at Math Revision exam I have to answer for $\frac{dy}{dx}$
Question:$$ y= \arctan\frac{{2x}}{{1+x^2}}$$
I got the answer $$ \frac{2}{1+(\frac{2x}{1+x})^2}\frac{cos(2\tan^{-1}x)}{1+x^2}$$
i think this not correct.
I have to know the right solution with steps.
 [/INST] 
$$ y= \arctan\frac{{2x}}{{1+x^2}}$$

You need to use the chain rule here,
$$ y'=\frac{\frac{d}{dx}\left(\frac{2x}{1+x^2}\right)}{1+\frac{4x^2}{(1+x^2)^2}}=\frac{(x^2+1)\frac{d}{dx}(x)-x\frac{d}{dx}(1+x^2)}{(x^2+1)^2}\frac{2}{1+\frac{4x^2}{(1+x^2)^2}}=\frac{2(1-x^2)}{(1+x^2)^2\left(1+\frac{4x^2}{(1+x^2)^2}\right)}=\color{red}{\frac{2-2x^2}{x^4+6x^2+1}}$$
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