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

Explanation:

Step1: Use the quotient rule

The quotient rule states that if (y = \frac{u}{v}), then (y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}). Let (u=\sqrt{x}-6=x^{\frac{1}{2}}-6), (u^\prime=\frac{1}{2}x^{-\frac{1}{2}}), and (v = \sqrt{x}+6=x^{\frac{1}{2}}+6), (v^\prime=\frac{1}{2}x^{-\frac{1}{2}}). [ \begin{align*} f^\prime(x)&=\frac{(\frac{1}{2}x^{-\frac{1}{2}})(\sqrt{x}+6)-(\sqrt{x}-6)(\frac{1}{2}x^{-\frac{1}{2}})}{(\sqrt{x}+6)^{2}}\ &=\frac{\frac{1}{2}x^{-\frac{1}{2}}\cdot\sqrt{x}+\frac{6}{2}x^{-\frac{1}{2}}-\frac{1}{2}x^{-\frac{1}{2}}\cdot\sqrt{x}+\frac{6}{2}x^{-\frac{1}{2}}}{(\sqrt{x}+6)^{2}} \end{align*} ]

Step2: Simplify the numerator

Simplify the numerator: (\frac{1}{2}x^{-\frac{1}{2}}\cdot\sqrt{x}=\frac{1}{2}), (\frac{6}{2}x^{-\frac{1}{2}} = 3x^{-\frac{1}{2}}), (-\frac{1}{2}x^{-\frac{1}{2}}\cdot\sqrt{x}=-\frac{1}{2}). The numerator becomes (3x^{-\frac{1}{2}}+3x^{-\frac{1}{2}}=6x^{-\frac{1}{2}}). So (f^\prime(x)=\frac{6x^{-\frac{1}{2}}}{(\sqrt{x}+6)^{2}}=\frac{6}{x^{\frac{1}{2}}(\sqrt{x}+6)^{2}}=\frac{6}{\sqrt{x}(\sqrt{x}+6)^{2}})

Step3: Evaluate (f^\prime(3))

Substitute (x = 3) into (f^\prime(x)): (f^\prime(3)=\frac{6}{\sqrt{3}(\sqrt{3}+6)^{2}}). Rationalize (\frac{6}{\sqrt{3}} = 2\sqrt{3}), and ((\sqrt{3}+6)^{2}=3 + 12\sqrt{3}+36=39+12\sqrt{3}). Then (f^\prime(3)=\frac{2\sqrt{3}}{39 + 12\sqrt{3}}). Multiply numerator and denominator by (39-12\sqrt{3}): [ \begin{align*} f^\prime(3)&=\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^\prime(x)=\frac{6}{\sqrt{x}(\sqrt{x}+6)^{2}}), (f^\prime(3)=\frac{26\sqrt{3}-24}{363})