Question

if (f(x)=\frac{sqrt{x}-9}{sqrt{x}+9}), find (f(x)). then find (f(6)).
(f(x)=)
(f(6)=)

Explanation:

Step1: Apply the quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Let $u = \sqrt{x}-9=x^{\frac{1}{2}}-9$ and $v=\sqrt{x}+9=x^{\frac{1}{2}}+9$. Then $u'=\frac{1}{2}x^{-\frac{1}{2}}$ and $v'=\frac{1}{2}x^{-\frac{1}{2}}$.

Step2: Calculate $f'(x)$

\[

$$\begin{align*} f'(x)&=\frac{(\frac{1}{2}x^{-\frac{1}{2}})(x^{\frac{1}{2}} + 9)-(\frac{1}{2}x^{-\frac{1}{2}})(x^{\frac{1}{2}} - 9)}{(x^{\frac{1}{2}}+9)^{2}}\\ &=\frac{\frac{1}{2}x^{-\frac{1}{2}}\times x^{\frac{1}{2}}+\frac{9}{2}x^{-\frac{1}{2}}-\frac{1}{2}x^{-\frac{1}{2}}\times x^{\frac{1}{2}}+\frac{9}{2}x^{-\frac{1}{2}}}{(x^{\frac{1}{2}} + 9)^{2}}\\ &=\frac{\frac{1}{2}+\frac{9}{2}x^{-\frac{1}{2}}-\frac{1}{2}+\frac{9}{2}x^{-\frac{1}{2}}}{(x^{\frac{1}{2}}+9)^{2}}\\ &=\frac{9x^{-\frac{1}{2}}}{(x^{\frac{1}{2}} + 9)^{2}}=\frac{9}{2\sqrt{x}(\sqrt{x}+9)^{2}} \end{align*}$$

\]

Step3: Calculate $f'(6)$

Substitute $x = 6$ into $f'(x)$.
\[

$$\begin{align*} f'(6)&=\frac{9}{2\sqrt{6}(\sqrt{6}+9)^{2}}\\ &=\frac{9}{2\sqrt{6}(6 + 18\sqrt{6}+81)}\\ &=\frac{9}{2\sqrt{6}(87 + 18\sqrt{6})}\\ &=\frac{9}{174\sqrt{6}+216} \end{align*}$$

\]

Answer:

$f'(x)=\frac{9}{2\sqrt{x}(\sqrt{x}+9)^{2}}$; $f'(6)=\frac{9}{174\sqrt{6}+216}$