Question: differentiate ln [(1 - 3x^2) / (1+ 3x^2)] ^1/2
Here we have multiple applications of the chain rule, and one application of the quotient rule.
Let y = ln[w^(1/2)], where
w = u/v and u = (1 - 3x^2) and v = 1 + 3x^2. Then,
using du/dx = -6x and dv/dx = 6x,
dw/dx = (v.du/dx - u/dv/dx)/v^2 = [(1 + 3x^2).(-6x) - (1 - 3x^2).(6x)]/(1 + 3x^2)^2
dw/dx = [-6x - 18x^3 - 6x + 18x^3]/(1 + 3x^2)^2 = -12x/(1+3x^2)^2
dw/dx = -12x/(1+3x^2)^2
Continuing ...
y = ln[w^(1/2)] = (1/2)ln(w)
dy/dw = (1/2w)
and,
dy/dx = (dy/dw).(dw/dx) = (1/2w).(-12x/(1+3x^2)^2)
dy/dx = -6x/{[(1 - 3x^2) / (1+ 3x^2)] * (1 + 3x^2)^2}
dy/dx = -6x/{(1 - 3x^2) * (1 + 3x^2)}
dy/dx = -6x/(1 - 9x^4)