if $f(x)=\frac{sqrt{x}-2}{sqrt{x}+2}$, find: $f(x)=$

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = \sqrt{x}-2=x^{\frac{1}{2}}-2$, $v=\sqrt{x}+2=x^{\frac{1}{2}}+2$.

Step2: Find $u'$ and $v'$

Differentiate $u$ with respect to $x$: $u'=\frac{d}{dx}(x^{\frac{1}{2}}-2)=\frac{1}{2}x^{-\frac{1}{2}}$. Differentiate $v$ with respect to $x$: $v'=\frac{d}{dx}(x^{\frac{1}{2}}+2)=\frac{1}{2}x^{-\frac{1}{2}}$.

Step3: Substitute into quotient - rule formula

[ \begin{align*} f'(x)&=\frac{(\frac{1}{2}x^{-\frac{1}{2}})(\sqrt{x}+2)-(\sqrt{x}-2)(\frac{1}{2}x^{-\frac{1}{2}})}{(\sqrt{x}+2)^{2}}\ &=\frac{\frac{1}{2}x^{-\frac{1}{2}}\sqrt{x}+\ x^{-\frac{1}{2}}-\frac{1}{2}x^{-\frac{1}{2}}\sqrt{x}+x^{-\frac{1}{2}}}{(\sqrt{x}+2)^{2}}\ &=\frac{2x^{-\frac{1}{2}}}{(\sqrt{x}+2)^{2}}\ &=\frac{2}{ \sqrt{x}(\sqrt{x}+2)^{2}} \end{align*} ]

Answer:

$\frac{2}{\sqrt{x}(\sqrt{x}+2)^{2}}$