# How do you find the derivative using quotient rule and chain rule for 1/sqrt(1-x^2)?

Apr 22, 2015

Differentiating $\ln \left(1 + x\right)$

Say: $p = \ln \left(1 + x\right) = \ln u$

$\frac{\mathrm{dp}}{\mathrm{du}} = \frac{1}{u} = \frac{1}{1 + x}$

$u = 1 + x$, $\frac{\mathrm{du}}{\mathrm{dx}} = 1$

Therefore:

$\frac{\mathrm{dp}}{\mathrm{dx}} = \frac{1}{1 + x}$

Differentiating $\ln \left(1 - x\right)$

Say:

$q = \ln \left(1 - x\right) = \ln u$

$\frac{\mathrm{dq}}{\mathrm{du}} = \frac{1}{u} = \frac{1}{1 - x}$

$u = 1 - x$, $\frac{\mathrm{du}}{\mathrm{dx}} = - 1$

Therefore:

$\frac{\mathrm{dq}}{\mathrm{dx}} = - \frac{1}{1 - x}$

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

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

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

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

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

$\ln y = - \frac{1}{2} \ln \left(\left(1 + x\right) \left(1 - x\right)\right)$

$\ln y = - \frac{1}{2} \left\{\ln \left(1 + x\right) + \ln \left(1 - x\right)\right\}$

$\ln y = - \frac{1}{2} \ln \left(1 + x\right) - \frac{1}{2} \ln \left(1 - x\right)$

$\frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2} \left\{\frac{1}{1 + x}\right\} - \frac{1}{2} \left\{- \frac{1}{1 - x}\right\}$

$\frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \left(1 - x\right)} - \frac{1}{2 \left(1 + x\right)}$

$y \cdot \frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = y \left\{\frac{1}{2 \left(1 - x\right)} - \frac{1}{2 \left(1 + x\right)}\right\}$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{\left(1 + x\right) \left(1 - x\right)}} \left\{\frac{1}{2 \left(1 - x\right)} - \frac{1}{2 \left(1 + x\right)}\right\}$

An explanation provided by Harivogind S. as to why the result above is the same as the result he has provided...

$\left\{\frac{1}{2 \left(1 - x\right)} - \frac{1}{2 \left(1 + x\right)}\right\} = \frac{\left(1 + x\right) - \left(1 - x\right)}{2 \cdot \left(1 - x\right) \cdot \left(1 + x\right)}$ - common denominator.

$= \frac{2 x}{2 \cdot \left(1 - {x}^{2}\right)}$
$\left\{\frac{1}{2 \left(1 - x\right)} - \frac{1}{2 \left(1 + x\right)}\right\} = \frac{x}{1 - {x}^{2}}$
and,
$\frac{1}{\sqrt{\left(1 + x\right) \left(1 - x\right)}} = \frac{1}{\sqrt{\left(1 - {x}^{2}\right)}}$
Therefore -
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{\left(1 + x\right) \left(1 - x\right)}} \left\{\frac{1}{2 \left(1 - x\right)} - \frac{1}{2 \left(1 + x\right)}\right\} = \frac{1}{\sqrt{1 - {x}^{2}}} \cdot \left[\frac{x}{1 - {x}^{2}}\right]$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{1 - {x}^{2}} ^ \left(\frac{3}{2}\right)$

*Quotient rule was not required in these workings as your fraction didn't contain both a numerator and denominator that were both functions of x.