if $f(x)=\frac{sqrt{x}-4}{sqrt{x}+4}$, find:\n$f(x)=$\n$f(4)=$\nquestion help: video message…

Explanation:

Step1: Apply 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}-4=x^{\frac{1}{2}} - 4) and (v=\sqrt{x}+4=x^{\frac{1}{2}}+4).

First, find (u^\prime) and (v^\prime). Using the power rule ((x^{n})^\prime=nx^{n - 1}), we have (u^\prime=\frac{1}{2}x^{-\frac{1}{2}}) and (v^\prime=\frac{1}{2}x^{-\frac{1}{2}}).

Step2: Substitute into the quotient rule formula

[ \begin{align*} f^\prime(x)&=\frac{(\frac{1}{2}x^{-\frac{1}{2}})(\sqrt{x}+4)-(\sqrt{x}-4)(\frac{1}{2}x^{-\frac{1}{2}})}{(\sqrt{x}+4)^{2}}\ &=\frac{\frac{1}{2}x^{-\frac{1}{2}}\sqrt{x}+ 2x^{-\frac{1}{2}}-\frac{1}{2}x^{-\frac{1}{2}}\sqrt{x}+2x^{-\frac{1}{2}}}{(\sqrt{x}+4)^{2}}\ &=\frac{4x^{-\frac{1}{2}}}{(\sqrt{x}+4)^{2}}\ &=\frac{4}{x^{\frac{1}{2}}(\sqrt{x}+4)^{2}} \end{align*} ]

Step3: Evaluate (f^\prime(4))

Substitute (x = 4) into (f^\prime(x)). Since (x^{\frac{1}{2}}=\sqrt{4} = 2) and (\sqrt{x}+4=2 + 4=6)

[ f^\prime(4)=\frac{4}{2\times6^{2}}=\frac{4}{2\times36}=\frac{1}{18} ]

Answer:

(f^\prime(x)=\frac{4}{x^{\frac{1}{2}}(\sqrt{x}+4)^{2}})

(f^\prime(4)=\frac{1}{18})