Question

if ( f(x)=\frac{sqrt{x}-6}{sqrt{x}+6} ), find:
( f^{prime}(x)= )
( f^{prime}(3)= )
question help: video message instructor

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 \(u=\sqrt{x}-6=x^{\frac{1}{2}}-6\) and \(v = \sqrt{x}+6=x^{\frac{1}{2}}+6\).

First, find \(u'\) and \(v'\):
\(u'=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}\), \(v'=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}\)

Then, \(f'(x)=\frac{(\frac{1}{2\sqrt{x}})(\sqrt{x}+6)-(\sqrt{x}-6)(\frac{1}{2\sqrt{x}})}{(\sqrt{x}+6)^{2}}\)

Step2: Simplify the numerator

Expand the numerator:

$$
LATEXBLOCK0
$$

So \(f'(x)=\frac{\frac{6}{\sqrt{x}}}{(\sqrt{x}+6)^{2}}=\frac{6}{\sqrt{x}(\sqrt{x}+6)^{2}}\)

Step3: Find \(f'(3)\)

Substitute \(x = 3\) into \(f'(x)\):
\(f'(3)=\frac{6}{\sqrt{3}(\sqrt{3}+6)^{2}}\)

Simplify \(\frac{6}{\sqrt{3}(\sqrt{3}+6)^{2}}=\frac{6\sqrt{3}}{3(3 + 12\sqrt{3}+ 36)}=\frac{2\sqrt{3}}{39 + 12\sqrt{3}}\)

Rationalize the denominator:

$$
LATEXBLOCK1
$$

Answer:

\(f'(x)=\frac{6}{\sqrt{x}(\sqrt{x}+6)^{2}}\)

\(f'(3)=\frac{26\sqrt{3}-24}{363}\)