$\text { If } y=\operatorname{Tan}^{-1}\left(\frac{x}{1+2 x^2}\right)+\operatorname{Tan}^{-1}\left(\frac{x}{1+6 x^2}\right) \text {, then } \frac{d y}{d x}=$

• A

  $\frac{4}{16 x^2+1}-\frac{3}{9 x^2+1}$
• Correct answer
  B

  $\frac{3}{9 x^2+1}-\frac{1}{x^2+1}$
• C

  $\frac{3}{9 x^2+1}-\frac{2}{4 x^2+1}$
• D

  $\frac{1}{9 x^2+1}-\frac{1}{x^2+1}$

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