# How do you find the derivative of sqrt(x^2-1) / (x^2+1)?

Apr 25, 2018

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

#### Explanation:

Using quotient rule:

$\frac{\left(\sqrt{{x}^{2} - 1}\right) ' \left({x}^{2} + 1\right) - \left(\sqrt{{x}^{2} - 1} \cdot \left({x}^{2} + 1\right) '\right)}{{x}^{2} + 1} ^ 2$

Mind that:
$\sqrt{{x}^{2} - 1} ' = \left({\left({x}^{2} - 1\right)}^{\frac{1}{2}}\right) ' = \frac{1}{2} {\left({x}^{2} - 1\right)}^{\frac{1}{2} - 1} \cdot \left({x}^{2} - 1\right) ' = \frac{1}{2} \cdot \frac{1}{\sqrt{{x}^{2} - 1}} \left(2 x\right) = \frac{x}{\sqrt{{x}^{2} - 1}}$

So the solution is:
((x^2+1)x/sqrt(x^2-1) -sqrt(x^2-1)(2x))/(x^2+1)^2 = x/(sqrt(x^2-1) (x^2+1)^2)(x^2+1-2x^2+2) = -x(x^2-3)/(sqrt(x^2-1) (x^2+1)^2)

Apr 25, 2018

(3x-x^3)/(sqrt(x^2-1)(x^2+1)^2

#### Explanation:

$\text{differentiate using the "color(blue)"quotient rule}$

$\text{Given "y=(g(x))/(h(x))" then}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2 \leftarrow \textcolor{b l u e}{\text{quotient rule}}$

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

$\text{differentiate using the "color(blue)"chain rule}$

$\Rightarrow g ' \left(x\right) = \frac{1}{2} {\left({x}^{2} - 1\right)}^{- \frac{1}{2}} \times \frac{d}{\mathrm{dx}} \left({x}^{2} - 1\right)$

$\textcolor{w h i t e}{\Rightarrow g ' \left(x\right)} = x {\left({x}^{2} - 1\right)}^{- \frac{1}{2}}$

$h \left(x\right) = {x}^{2} + 1 \Rightarrow h ' \left(x\right) = 2 x$

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

$\textcolor{w h i t e}{\Rightarrow} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x {\left({x}^{2} - 1\right)}^{- \frac{1}{2}} \left[{x}^{2} + 1 - 2 \left({x}^{2} - 1\right)\right]}{{x}^{2} + 1} ^ 2$

$\textcolor{w h i t e}{\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}} = \frac{3 x - {x}^{3}}{\sqrt{{x}^{2} - 1} {\left({x}^{2} + 1\right)}^{2}}$