if ( f(x)=\frac{sqrt{x}-6}{sqrt{x}+6} ), find:\n( f^{prime}(x)= )\n( f^{prime}(3)= )\nquestion help: video…

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: [ \begin{align*} &\frac{1}{2\sqrt{x}}(\sqrt{x}+6)-\frac{1}{2\sqrt{x}}(\sqrt{x}-6)\ =&\frac{1}{2}+\frac{3}{\sqrt{x}}-\frac{1}{2}+\frac{3}{\sqrt{x}}\ =&\frac{6}{\sqrt{x}} \end{align*} ]

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: [ \begin{align*} \frac{2\sqrt{3}(39-12\sqrt{3})}{(39 + 12\sqrt{3})(39-12\sqrt{3})}&=\frac{78\sqrt{3}-72}{39^{2}-(12\sqrt{3})^{2}}\ &=\frac{78\sqrt{3}-72}{1521 - 432}\ &=\frac{78\sqrt{3}-72}{1089}\ &=\frac{26\sqrt{3}-24}{363} \end{align*} ]

Answer:

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

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