# How do you differentiate f(x) = ln((1-x^2)^(-1/2) )? using the chain rule?

Dec 10, 2017

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

#### Explanation:

If $f \left(x\right) = \ln \left(g \left(x\right)\right)$
Then $f ' \left(x\right) = \frac{g ' \left(x\right)}{g \left(x\right)}$

$g \left(x\right) = {\left(1 - {x}^{2}\right)}^{- \frac{1}{2}} = {\left(h \left(x\right)\right)}^{n}$
$g ' \left(x\right) = n \cdot h ' \left(x\right) \cdot h {\left(x\right)}^{n - 1}$
$h ' \left(x\right) = - 2 x$
$g ' \left(x\right) = - \frac{1}{2} \cdot - 2 x \cdot {\left(1 - {x}^{2}\right)}^{- \frac{3}{2}} = x {\left(1 - {x}^{2}\right)}^{- \frac{3}{2}}$

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