If $y=\tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)+\tan ^{-1}\left(\frac{4 x-4 x^3}{1-6 x^2+x^4}\right)$, then $\frac{d y}{d x}$ is equal to
- A$\frac{2}{1+x^2}$
- B$\frac{4}{1+x^2}$
- C$\frac{6}{1+x^2}$
- D$\frac{7}{1+x^2}$
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