# How do you find the derivative of F(x) = sqrt( (x-8)/(x^2-2) )?

Oct 3, 2017

$F \left(x\right) = \sqrt{\frac{x - 8}{{x}^{2} - 2}}$
$F ' \left(x\right) = \frac{1}{2 \sqrt{\frac{x - 8}{{x}^{2} - 2}}} \left(\frac{\left({x}^{2} - 2\right) \frac{d}{\mathrm{dx}} \left(x - 8\right) - \left(x - 8\right) \frac{d}{\mathrm{dx}} \left({x}^{2} - 2\right)}{{x}^{2} - 2} ^ 2\right)$

$F ' \left(x\right) = \frac{1}{2 \sqrt{\frac{x - 8}{{x}^{2} - 2}}} \left(\frac{\left({x}^{2} - 2\right) \left(1\right) - \left(x - 8\right) \left(2 x\right)}{{x}^{2} - 2} ^ 2\right)$

$F ' \left(x\right) = \frac{\sqrt{{x}^{2} - 2}}{2 \sqrt{\left(x - 8\right)}} \left(\frac{{x}^{2} - 2 - 2 {x}^{2} + 16 x}{{x}^{2} - 2} ^ 2\right)$

$F ' \left(x\right) = \frac{\sqrt{{x}^{2} - 2}}{2 \sqrt{\left(x - 8\right)}} \left(\frac{- {x}^{2} + 16 x - 2}{{x}^{2} - 2} ^ 2\right)$

Oct 3, 2017

$F ' \left(x\right) = - \frac{\left({x}^{2} - 16 x + 2\right)}{2 {\left(x - 8\right)}^{\frac{1}{2}} {\left({x}^{2} - 2\right)}^{\frac{3}{2}}} .$

#### Explanation:

Let, $y = F \left(x\right) = \sqrt{\frac{x - 8}{{x}^{2} - 2}} = {\left\{\frac{x - 8}{{x}^{2} - 2}\right\}}^{\frac{1}{2}} .$

$\therefore \ln y = \frac{1}{2} \ln \left(\frac{x - 8}{{x}^{2} - 2}\right) = \frac{1}{2} \left\{\ln \left(x - 8\right) - \ln \left({x}^{2} - 2\right)\right\} .$

$\therefore \frac{d}{\mathrm{dx}} \left\{\ln y\right\} = \frac{1}{2} \frac{d}{\mathrm{dx}} \left\{\ln \left(x - 8\right) - \ln \left({x}^{2} - 2\right)\right\} .$

By the Chain Rule, then, we have,

$\therefore \frac{d}{\mathrm{dy}} \left(\ln y\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

$= \frac{1}{2} \left[\frac{1}{x - 8} \cdot \frac{d}{\mathrm{dx}} \left(x - 8\right) - \frac{1}{{x}^{2} - 2} \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} - 2\right)\right\} .$

$\therefore \frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \left\{\frac{1}{x - 8} \cdot 1 - \frac{1}{{x}^{2} - 2} \cdot 2 x\right\} ,$

$= \frac{1}{2} \left\{\frac{\left({x}^{2} - 2\right) - 2 x \left(x - 8\right)}{\left(x - 8\right) \left({x}^{2} - 2\right)}\right\} ,$

$= \frac{1}{2} \left\{\frac{- {x}^{2} + 16 x - 2}{\left(x - 8\right) \left({x}^{2} - 2\right)}\right\} .$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{y \left({x}^{2} - 16 x + 2\right)}{2 \left(x - 8\right) \left({x}^{2} - 2\right)} .$

But, $y = {\left\{\frac{x - 8}{{x}^{2} - 2}\right\}}^{\frac{1}{2}} .$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - {\left\{\frac{x - 8}{{x}^{2} - 2}\right\}}^{\frac{1}{2}} \cdot \frac{{x}^{2} - 16 x + 2}{2 \left(x - 8\right) \left({x}^{2} - 2\right)} , i . e . ,$

$\frac{\mathrm{dy}}{\mathrm{dx}} = F ' \left(x\right) = - \frac{\left({x}^{2} - 16 x + 2\right)}{2 {\left(x - 8\right)}^{\frac{1}{2}} {\left({x}^{2} - 2\right)}^{\frac{3}{2}}} .$

Enjoy Mayhs.!