I'm going to calculate the derivative of $y=\frac{\ln(x)}{x^3}$ and $y=\ln(\sqrt{\frac{x+2}{x-2}})$
Solution:
I will use the quotient rule to find the derivative of $y=\frac{\ln(x)}{x^3}$
$y'= \frac{x^3 \ln(x)' - \ln(x) (x^3)'}{(x^3)^2}$
$=\frac{x^3 \frac{1}{x} - \ln(x)(3x^2)}{x^6}$
$=\frac{x^2-3x^2\ln(x)}{x^6}$
$=\frac{x^2(1-3\ln(x))}{x^6}$
$=\frac{1-3\ln(x)}{x^4}$
Now I will find the derivative of $y=\ln(\sqrt{\frac{x+2}{x-2}})$. A very important first step is that we use properties of logarithmic functions to simply the expression, this will make calculating the derivative much easier.
$y=\ln(\sqrt{\frac{x+2}{x-2}})$
$=\ln((\frac{x+2}{x-2})^{\frac{1}{2}})$
$=\frac{1}{2}\ln(\frac{x+2}{x-2})$
$=\frac{1}{2}(\ln(x+2)-\ln(x-2))$
Cool. Now let's calculate $y'$.
$y'=\frac{1}{2}(\ln(x+2)-\ln(x-2))'$
$=\frac{1}{2}(\ln(x+2)'-\ln(x-2)')$
$=\frac{1}{2}(\frac{1}{x+2}(x+2)'-\frac{1}{x-2}(x-2)')$
$=\frac{1}{2}(\frac{1}{x+2}(1)-\frac{1}{x-2}(1))$
$=\frac{1}{2}(\frac{1}{x+2}-\frac{1}{x-2})$
This will probably count as the correct answer, but lets make it look nicer by finding a common denominator:
$=\frac{1}{2}(\frac{x-2}{(x+2)(x-2)}-\frac{x+2}{(x-2)(x+2)})$
$=\frac{1}{2}(\frac{-4}{x^2-4})$
$=\frac{-2}{x^2-4}$