# How do you find the derivative of (x)/sqrt(x^2-4)?

Jan 8, 2017

-4/(sqrt((x^2-4)^3)

#### Explanation:

Differentiate using a combination of $\textcolor{b l u e}{\text{quotient and chain rule}}$

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

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2} \textcolor{w h i t e}{\frac{2}{2}} |}}} \leftarrow \text{quotient rule}$

$g \left(x\right) = x \Rightarrow g ' \left(x\right) = 1$

$h \left(x\right) = \sqrt{{x}^{2} - 4} = {\left({x}^{2} - 4\right)}^{\frac{1}{2}}$

$\Rightarrow h ' \left(x\right) = \frac{1}{2} {\left({x}^{2} - 4\right)}^{- \frac{1}{2}} . \frac{d}{\mathrm{dx}} \left({x}^{2} - 4\right) = x {\left({x}^{2} - 4\right)}^{- \frac{1}{2}}$

$f ' \left(x\right) = \frac{{\left({x}^{2} - 4\right)}^{\frac{1}{2}} .1 - x . x {\left({x}^{2} - 4\right)}^{- \frac{1}{2}}}{{x}^{2} - 4}$

$= \frac{{\left({x}^{2} - 4\right)}^{- \frac{1}{2}} \left[{x}^{2} - 4 - {x}^{2}\right]}{{x}^{2} - 4}$

=(-4)/(x^2-4)^(3/2)=-4/(sqrt((x^2-4)^3)