If $y=\operatorname{Tan}^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)+\operatorname{Tan}^{-1}\left(\frac{7 x}{1-12 x^2}\right)$, then at $x=0, \frac{d y}{d x}=$

• A

  $6$
• B

  $7$
• C

  $9$
• Correct answer
  D

  $10$

Hence, the correct answer is option: 'D'.