let f(x) = (sqrt(x) - 3)/(sqrt(x) + 3). then, f(x) = f(5) = submit answer next item

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = \sqrt{x}-3=x^{\frac{1}{2}}-3$, $u'=\frac{1}{2}x^{-\frac{1}{2}}$, $v=\sqrt{x}+3=x^{\frac{1}{2}}+3$, and $v'=\frac{1}{2}x^{-\frac{1}{2}}$.

Step2: Calculate $f'(x)$

[ \begin{align*} f'(x)&=\frac{(\frac{1}{2}x^{-\frac{1}{2}})(\sqrt{x}+3)-(\sqrt{x}-3)(\frac{1}{2}x^{-\frac{1}{2}})}{(\sqrt{x}+3)^{2}}\ &=\frac{\frac{1}{2}x^{-\frac{1}{2}}\sqrt{x}+\frac{3}{2}x^{-\frac{1}{2}}-\frac{1}{2}x^{-\frac{1}{2}}\sqrt{x}+\frac{3}{2}x^{-\frac{1}{2}}}{(\sqrt{x}+3)^{2}}\ &=\frac{3x^{-\frac{1}{2}}}{(\sqrt{x}+3)^{2}}=\frac{3}{2\sqrt{x}(\sqrt{x}+3)^{2}} \end{align*} ]

Step3: Calculate $f'(5)$

Substitute $x = 5$ into $f'(x)$: [ \begin{align*} f'(5)&=\frac{3}{2\sqrt{5}(\sqrt{5}+3)^{2}}\ &=\frac{3}{2\sqrt{5}(5 + 6\sqrt{5}+9)}\ &=\frac{3}{2\sqrt{5}(14 + 6\sqrt{5})}\ &=\frac{3}{28\sqrt{5}+60} \end{align*} ]

Answer:

$f'(x)=\frac{3}{2\sqrt{x}(\sqrt{x}+3)^{2}}$, $f'(5)=\frac{3}{28\sqrt{5}+60}$