Question: differentiate [log^2 (e^x + x - 1)] .
Let u = ln(v), wher v = e^x + x - 1. Then
y = u^2
dy/dx = (dy/du)*(du/dv)*(dv/dx)
and
dy/du = 2u,
du/dv = 1/v,
dv/dx = e^x + 1.
Then,
dy/dx = (2u)*(1/v)*(e^x + 1) = (2ln(v))*(1/(e^x + x - 1))*(e^x + 1)
dy/dx = ln[(e^x + x - 1)^2](e^x + 1)/(e^x + x - 1)