MCQ
If $\text{y}=\frac{1+\frac{1}{\text{x}^{2}}}{1-\frac{1}{\text{x}^{2}}}$ then $\frac{\text{dy}}{\text{dx}}$ is equal to:
- ✓$\frac{-4\text{x}}{(\text{x}^{2}-1)^{2}}$
- B$\frac{-4\text{x}}{(\text{x}^{2}-1)^{2}}$
- C$\frac{1-\text{x}^{2}}{4\text{x}}$
- D$\frac{4\text{x}}{\text{x}^{2}-1}$