Question

if ( f(x)=\frac{sqrt{x}-6}{sqrt{x}+6} ), find:
( f^{prime}(x)=)
( 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 \(v = \sqrt{x}+6=x^{\frac{1}{2}}+6\), \(v^\prime=\frac{1}{2}x^{-\frac{1}{2}}\).

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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 \(39-12\sqrt{3}\):

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Answer:

\(f^\prime(x)=\frac{6}{\sqrt{x}(\sqrt{x}+6)^{2}}\), \(f^\prime(3)=\frac{26\sqrt{3}-24}{363}\)