hw 14 - derivatives of logs section 2.9: problem 6 (1 point) find $f(x)$ if $f(x)=lnsqrt{\frac{2x - 6}{3x +…

Explanation:

Step1: Simplify the function

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

Step2: Apply the chain - rule for differentiation

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

Step3: Differentiate $f(x)$

$f'(x)=\frac{1}{2}\left(\frac{2}{2x - 6}-\frac{3}{3x + 8}\right)$. Simplify the right - hand side: [ \begin{align*} f'(x)&=\frac{1}{2}\left(\frac{2(3x + 8)-3(2x - 6)}{(2x - 6)(3x + 8)}\right)\ &=\frac{1}{2}\left(\frac{6x+16-(6x - 18)}{(2x - 6)(3x + 8)}\right)\ &=\frac{1}{2}\left(\frac{6x + 16-6x + 18}{(2x - 6)(3x + 8)}\right)\ &=\frac{1}{2}\left(\frac{34}{(2x - 6)(3x + 8)}\right)\ &=\frac{17}{(2x - 6)(3x + 8)} \end{align*} ]

Answer:

$f'(x)=\frac{17}{(2x - 6)(3x + 8)}$