# How do you find the derivative of sqrt((x^2-1) / (x^2+1))?

It is $f ' \left(x\right) = \frac{2 x}{\left(\sqrt{{x}^{2} - 1}\right) \cdot {\left({x}^{2} + 1\right)}^{\frac{3}{2}}}$

#### Explanation:

We have that

$f \left(x\right) = \sqrt{\frac{{x}^{2} - 1}{{x}^{2} + 1}}$ hence

$f ' \left(x\right) = \frac{1}{2} \cdot \frac{1}{\sqrt{\frac{{x}^{2} - 1}{{x}^{2} + 1}}} \cdot \left(\frac{{x}^{2} - 1}{{x}^{2} + 1}\right) '$

$f ' \left(x\right) = \frac{1}{2} \cdot \frac{\sqrt{{x}^{2} + 1}}{\sqrt{{x}^{2} - 1}} \cdot \left(\frac{2 x \left({x}^{2} + 1\right) - \left({x}^{2} - 1\right) 2 x}{{x}^{2} + 1}\right)$

$f ' \left(x\right) = \frac{1}{2} \cdot \frac{\sqrt{{x}^{2} + 1}}{\sqrt{{x}^{2} - 1}} \cdot \left(4 \frac{x}{{x}^{2} + 1} ^ 2\right)$

$f ' \left(x\right) = \frac{2 x}{\left(\sqrt{{x}^{2} - 1}\right) \cdot {\left({x}^{2} + 1\right)}^{\frac{3}{2}}}$