Question

attempt 2: 9 attempts remaining. if $f(x) = \frac{\sqrt{x} - 7}{\sqrt{x} + 7}$, find $f(x)$. then find $f(6)$. $f(x) = $ $f(6) = $ video example: solving a similar problem

Explanation:

Step1: Rewrite the function

First, rewrite \( f(x)=\frac{\sqrt{x} - 7}{\sqrt{x}+7} \) as \( f(x)=\frac{x^{\frac{1}{2}} - 7}{x^{\frac{1}{2}}+7} \). We will use the quotient rule to find the derivative. The quotient rule states that if \( y = \frac{u}{v} \), then \( y'=\frac{u'v - uv'}{v^{2}} \), where \( u = x^{\frac{1}{2}}-7 \) and \( v = x^{\frac{1}{2}}+7 \).

Step2: Find \( u' \) and \( v' \)

Find the derivative of \( u \) with respect to \( x \): \( u'=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}} \).
Find the derivative of \( v \) with respect to \( x \): \( v'=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}} \).

Step3: Apply the quotient rule

Using the quotient rule \( f'(x)=\frac{u'v - uv'}{v^{2}} \), substitute \( u \), \( u' \), \( v \), and \( v' \):
\[

$$\begin{align*} f'(x)&=\frac{\frac{1}{2\sqrt{x}}(\sqrt{x}+7)-(\sqrt{x}-7)\frac{1}{2\sqrt{x}}}{(\sqrt{x}+7)^{2}}\\ &=\frac{\frac{1}{2\sqrt{x}}\sqrt{x}+\frac{7}{2\sqrt{x}}-\frac{1}{2\sqrt{x}}\sqrt{x}+\frac{7}{2\sqrt{x}}}{(\sqrt{x}+7)^{2}}\\ &=\frac{\frac{1}{2}+\frac{7}{2\sqrt{x}}-\frac{1}{2}+\frac{7}{2\sqrt{x}}}{(\sqrt{x}+7)^{2}}\\ &=\frac{\frac{14}{2\sqrt{x}}}{(\sqrt{x}+7)^{2}}\\ &=\frac{\frac{7}{\sqrt{x}}}{(\sqrt{x}+7)^{2}}\\ &=\frac{7}{\sqrt{x}(\sqrt{x}+7)^{2}} \end{align*}$$

\]

Step4: Evaluate \( f'(6) \)

Substitute \( x = 6 \) into \( f'(x) \):
\[

$$\begin{align*} f'(6)&=\frac{7}{\sqrt{6}(\sqrt{6}+7)^{2}}\\ &=\frac{7}{\sqrt{6}(6 + 14\sqrt{6}+49)}\\ &=\frac{7}{\sqrt{6}(55 + 14\sqrt{6})}\\ &=\frac{7}{55\sqrt{6}+84}\\ &=\frac{7(55\sqrt{6}-84)}{(55\sqrt{6}+84)(55\sqrt{6}-84)}\\ &=\frac{385\sqrt{6}-588}{55^{2}\times6-84^{2}}\\ &=\frac{385\sqrt{6}-588}{18150 - 7056}\\ &=\frac{385\sqrt{6}-588}{11094}\\ \end{align*}$$

\]
Simplify the fraction (or we can also rationalize earlier steps, but let's check the derivative again for possible miscalculations. Wait, when we did the quotient rule, let's re - check step 3:

Wait, \( u=\sqrt{x}-7=x^{\frac{1}{2}} - 7 \), \( u'=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}} \)

\( v=\sqrt{x}+7=x^{\frac{1}{2}}+7 \), \( v'=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}} \)

Then \( u'v - uv'=\frac{1}{2\sqrt{x}}(\sqrt{x}+7)-(\sqrt{x}-7)\frac{1}{2\sqrt{x}}=\frac{1}{2\sqrt{x}}(\sqrt{x}+7-\sqrt{x}+7)=\frac{1}{2\sqrt{x}}\times14=\frac{7}{\sqrt{x}} \)

And \( v^{2}=(\sqrt{x}+7)^{2}=x + 14\sqrt{x}+49 \)

So \( f'(x)=\frac{7}{\sqrt{x}(x + 14\sqrt{x}+49)} \)

Now substitute \( x = 6 \):

\( \sqrt{6}\approx2.449 \), \( x + 14\sqrt{x}+49=6+14\times2.449 + 49=6 + 34.286+49=89.286 \)

\( \sqrt{x}(x + 14\sqrt{x}+49)\approx2.449\times89.286\approx218.6 \)

\( f'(6)=\frac{7}{218.6}\approx0.032 \) (approximate value). But let's do it symbolically:

\( f'(x)=\frac{7}{\sqrt{x}(\sqrt{x}+7)^{2}} \), so \( f'(6)=\frac{7}{\sqrt{6}(\sqrt{6}+7)^{2}} \)

Rationalize the denominator:

Multiply numerator and denominator by \( \sqrt{6} \):

\( f'(6)=\frac{7\sqrt{6}}{6(\sqrt{6}+7)^{2}} \)

Expand \( (\sqrt{6}+7)^{2}=6 + 14\sqrt{6}+49=55 + 14\sqrt{6} \)

So \( f'(6)=\frac{7\sqrt{6}}{6(55 + 14\sqrt{6})}=\frac{7\sqrt{6}}{330+84\sqrt{6}} \)

Multiply numerator and denominator by \( 330 - 84\sqrt{6} \):

\[

$$\begin{align*} f'(6)&=\frac{7\sqrt{6}(330 - 84\sqrt{6})}{(330)^{2}-(84\sqrt{6})^{2}}\\ &=\frac{2310\sqrt{6}-7\sqrt{6}\times84\sqrt{6}}{108900-84^{2}\times6}\\ &=\frac{2310\sqrt{6}-7\times84\times6}{108900 - 42336}\\ &=\frac{2310\sqrt{6}-3528}{66564}\\ &=\frac{385\sqrt{6}-588}{11094}\\ &=\frac{385\sqrt{6}-588}{11094}\approx\frac{385\times2.449-588}{11094}\approx\frac{942.865 - 588}{11094}\approx\frac{354.865}{11094}\approx0.032 \end{align*}$$

\]

Answer:

\( f'(x)=\frac{7}{\sqrt{x}(\sqrt{x}+7)^{2}} \) and \( f'(6)=\frac{7}{\sqrt{6}(\sqrt{6}+7)^{2}}\approx\frac{385\sqrt{6}-588}{11094}\approx0.032 \) (exact form: \( \frac{7\sqrt{6}}{6(55 + 14\sqrt{6})} \) or simplified as above)