hw 14 - derivatives of logs section 2.9: problem 7 (1 point) find $f(x)$ if $f(x)=lnsqrt{\frac{2x - 8}{3x +…

Explanation:

Step1: Simplify the function

First, use the property $\ln\sqrt{u}=\frac{1}{2}\ln u$. So, $f(x)=\frac{1}{2}\ln\left(\frac{2x - 8}{3x + 6}\right)$. Then use the property $\ln\frac{a}{b}=\ln a-\ln b$, we get $f(x)=\frac{1}{2}(\ln(2x - 8)-\ln(3x + 6))$.

Step2: Apply the chain - rule for differentiation

The derivative of $\ln(u)$ with respect to $x$ is $\frac{u'}{u}$. For $y = \ln(2x - 8)$, $y'=\frac{2}{2x - 8}$. For $z=\ln(3x + 6)$, $z'=\frac{3}{3x + 6}$.

Step3: Differentiate $f(x)$

$f'(x)=\frac{1}{2}\left(\frac{2}{2x - 8}-\frac{3}{3x + 6}\right)$. Simplify the expression: [ \begin{align*} f'(x)&=\frac{1}{2}\left(\frac{2(3x + 6)-3(2x - 8)}{(2x - 8)(3x + 6)}\right)\ &=\frac{1}{2}\left(\frac{6x+12-(6x - 24)}{(2x - 8)(3x + 6)}\right)\ &=\frac{1}{2}\left(\frac{6x + 12-6x + 24}{(2x - 8)(3x + 6)}\right)\ &=\frac{1}{2}\left(\frac{36}{(2x - 8)(3x + 6)}\right)\ &=\frac{18}{(2x - 8)(3x + 6)}\ &=\frac{18}{6(x - 4)(x + 2)}\ &=\frac{3}{(x - 4)(x + 2)} \end{align*} ]

Answer:

$f'(x)=\frac{3}{(x - 4)(x + 2)}$