# What is the derivative of sqrt(x^2-1) / (x^2+1)?

##### 1 Answer
Jan 15, 2016

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

#### Explanation:

The derivative can be found by using the quotient rule, which states that if $f \left(x\right) = g \frac{x}{h \left(x\right)}$ then
$f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - h ' \left(x\right) g \left(x\right)}{{h}^{2} \left(x\right)}$

In this case $g \left(x\right) = {\left({x}^{2} - 1\right)}^{\frac{1}{2}}$
and $h \left(x\right) = \left({x}^{2} + 1\right)$

Using the chain rule, $g ' \left(x\right) = \frac{1}{2} {\left({x}^{2} - 1\right)}^{- \frac{1}{2}} \cdot 2 x = \frac{x}{\sqrt{{x}^{2} - 1}}$
$h ' \left(x\right) = 2 x$

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

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

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