# Question #c75cc

Jan 27, 2016

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

#### Explanation:

We rewrite the expression as:

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

Using the quotient rule:

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

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

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

$= - x {\left({a}^{2} - {x}^{2}\right)}^{- \frac{1}{2}} {\left({a}^{2} + {x}^{2}\right)}^{- \frac{3}{2}} \left[\left({a}^{2} + {x}^{2}\right) + \left({a}^{2} - {x}^{2}\right)\right] =$

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

So, the final expression is:

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