y=[(x+2)*√(1-x²)]/[4x³]
i got as far as
y'/y=ln(x+2)+1/2ln(1-x²)-3ln4x
im not sure if the part is correct since the cubed is only on the x and not the 4.
$\displaystyle y = \frac{(x+2) \sqrt{1-x^2}}{4x^3}$
note that $\displaystyle ln(4x^3)$ can be simplified to $\displaystyle ln(4)+ln(x^3) = 2ln(2)+3ln(x)$
$\displaystyle ln(y) = ln(x+2) + \frac{1}{2}ln(1-x^2) - 2ln(2) - 3ln(x)$
$\displaystyle \frac{y'}{y} = \frac{1}{x+2} - \frac{2x}{2(1-x)^2} - \frac{3}{x}$