if ( f(x)=\frac{sqrt{x}-2}{sqrt{x}+2} ), find:\n( f^{prime}(x)= )\n( f^{prime}(2)= )\nquestion help: video…

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

Step2: Substitute into the quotient rule

[ \begin{align*} f^\prime(x)&=\frac{(\frac{1}{2}x^{-\frac{1}{2}})(\sqrt{x}+2)-(\sqrt{x}-2)(\frac{1}{2}x^{-\frac{1}{2}})}{(\sqrt{x}+2)^{2}}\ &=\frac{\frac{1}{2}x^{-\frac{1}{2}}\cdot\sqrt{x}+\frac{1}{2}x^{-\frac{1}{2}}\cdot2-\frac{1}{2}x^{-\frac{1}{2}}\cdot\sqrt{x}+\frac{1}{2}x^{-\frac{1}{2}}\cdot2}{(\sqrt{x}+2)^{2}}\ &=\frac{\frac{1}{2}+\ x^{-\frac{1}{2}}-\frac{1}{2}+x^{-\frac{1}{2}}}{(\sqrt{x}+2)^{2}}\ &=\frac{2x^{-\frac{1}{2}}}{(\sqrt{x}+2)^{2}}\ &=\frac{2}{x^{\frac{1}{2}}(\sqrt{x}+2)^{2}} \end{align*} ]

Step3: Find (f^\prime(2))

Substitute (x = 2) into (f^\prime(x)). (f^\prime(2)=\frac{2}{\sqrt{2}(\sqrt{2}+2)^{2}}). [ \begin{align*} f^\prime(2)&=\frac{2}{\sqrt{2}(2 + 4\sqrt{2}+4)}\ &=\frac{2}{\sqrt{2}(6 + 4\sqrt{2})}\ &=\frac{2}{6\sqrt{2}+8}\ &=\frac{2}{2(3\sqrt{2}+4)}\ &=\frac{1}{3\sqrt{2}+4}\ &=\frac{4 - 3\sqrt{2}}{(4 + 3\sqrt{2})(4-3\sqrt{2})}\ &=\frac{4-3\sqrt{2}}{16-18}\ &=\frac{3\sqrt{2}-4}{2} \end{align*} ]

Answer:

(f^\prime(x)=\frac{2}{x^{\frac{1}{2}}(\sqrt{x}+2)^{2}}), (f^\prime(2)=\frac{3\sqrt{2}-4}{2})