question 10 of 14, step 1 of 1 determine the derivative of the function g(x)=ln(7√(x + 7)/(x - 7))

Explanation:

Step1: Rewrite the function using exponent - rule

First, rewrite $\sqrt[7]{\frac{x + 7}{x - 7}}$ as $(\frac{x + 7}{x - 7})^{\frac{1}{7}}$. So, $g(x)=\ln((\frac{x + 7}{x - 7})^{\frac{1}{7}})$. By the property of logarithms $\ln(a^b)=b\ln(a)$, we have $g(x)=\frac{1}{7}\ln(\frac{x + 7}{x - 7})$. And by the property $\ln(\frac{a}{b})=\ln(a)-\ln(b)$, $g(x)=\frac{1}{7}(\ln(x + 7)-\ln(x - 7))$.

Step2: Apply the chain - rule for differentiation

The derivative of $\ln(u)$ with respect to $x$ is $\frac{u'}{u}$. For $y = \frac{1}{7}(\ln(x + 7)-\ln(x - 7))$, the derivative of $\ln(x + 7)$ with respect to $x$ is $\frac{1}{x + 7}$, and the derivative of $\ln(x - 7)$ with respect to $x$ is $\frac{1}{x - 7}$. Using the constant - multiple rule and the difference rule of differentiation, $g'(x)=\frac{1}{7}(\frac{1}{x + 7}-\frac{1}{x - 7})$.

Step3: Simplify the expression

[ \begin{align*} g'(x)&=\frac{1}{7}\left(\frac{x - 7-(x + 7)}{(x + 7)(x - 7)}\right)\ &=\frac{1}{7}\left(\frac{x - 7 - x-7}{x^{2}-49}\right)\ &=\frac{1}{7}\left(\frac{-14}{x^{2}-49}\right)\ &=-\frac{2}{x^{2}-49} \end{align*} ]

Answer:

$g'(x)=-\frac{2}{x^{2}-49}$