if ( f(x)=\frac{sqrt{x}-2}{sqrt{x}+2} ), find:
( f^{prime}(x)= )
( f^{prime}(2)= )
question help: video message instructor
submit question jump to answer

Question

Accepted Answer

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

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step1: Use the quotient rule

Step2: Substitute into the quotient rule

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

Answer: