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

Apr 12, 2018

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

#### Explanation:

$f \left(x\right) = \frac{1}{\sqrt{1 - {x}^{2}}} = {\left(\sqrt{1 - {x}^{2}}\right)}^{-} 1 = {\left(1 - {x}^{2}\right)}^{- \frac{1}{2}}$
$\frac{d}{\mathrm{dx}} f \left(x\right) = - \frac{1}{2} {\left(1 - {x}^{2}\right)}^{- \frac{1}{2} - 1} \left(0 - 2 x\right)$
$= \frac{2 x}{2} {\left(1 - {x}^{2}\right)}^{- \frac{3}{2}}$
$= \left(x\right) {\left(1 - {x}^{2}\right)}^{- \frac{3}{2}}$

Apr 12, 2018

The derivative is $\frac{x \sqrt{1 - {x}^{2}}}{1 - {x}^{2}} ^ 2$.

#### Explanation:

Using the quotient rule:

$\textcolor{w h i t e}{=} \frac{d}{\mathrm{dx}} \left[\frac{1}{\sqrt{1 - {x}^{2}}}\right]$

$= \frac{\frac{d}{\mathrm{dx}} \left[1\right] \cdot \sqrt{1 - {x}^{2}} - 1 \cdot \frac{d}{\mathrm{dx}} \left[\sqrt{1 - {x}^{2}}\right]}{\sqrt{1 - {x}^{2}}} ^ 2$

$= \frac{0 \cdot \sqrt{1 - {x}^{2}} - 1 \cdot \frac{d}{\mathrm{dx}} \left[\sqrt{1 - {x}^{2}}\right]}{1 - {x}^{2}}$

$= \frac{- \frac{d}{\mathrm{dx}} \left[\sqrt{1 - {x}^{2}}\right]}{1 - {x}^{2}}$

$= \frac{- \frac{1}{2 \sqrt{1 - {x}^{2}}} \cdot \frac{d}{\mathrm{dx}} \left[1 - {x}^{2}\right]}{1 - {x}^{2}}$

$= \frac{- \frac{1}{2 \sqrt{1 - {x}^{2}}} \cdot - 2 x}{1 - {x}^{2}}$

$= \frac{\frac{x}{\sqrt{1 - {x}^{2}}}}{1 - {x}^{2}}$

$= \frac{\frac{x \sqrt{1 - {x}^{2}}}{1 - {x}^{2}}}{1 - {x}^{2}}$

$= \frac{x \sqrt{1 - {x}^{2}}}{1 - {x}^{2}} ^ 2$

That's the derivative. Hope this helped!