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.

By using the chain rule, we have
$$
\left(\arctan\frac{{2x}}{{1+x^2}}\right)'=\left(\frac{{2x}}{{1+x^2}}\right)'\times\frac1{1+(\frac{2x}{1+x^2})^2}
$$ or
$$
\left(\arctan\frac{{2x}}{{1+x^2}}\right)'=\frac{2 \left(1-x^2\right)}{\left(1+x^2\right)^2}\times\frac1{1+(\frac{2x}{1+x^2})^2}
$$ giving
$$
\left(\arctan\frac{{2x}}{{1+x^2}}\right)'=\frac{2(1-x^2)}{1+6 x^2+x^4}.
$$
