differential of ln(1+x) -1

differential of ln(1+x)-1

\(\boxed{\frac{d}{dx}lnf(x)=\frac{f'(x)}{f(x)}}\)

\(\frac{d}{dx}\;[ln(1+x)^{-1}]\\ =\frac{d}{dx}\;[-ln(1+x)]\\ =\frac{-1}{1+x}\\\)

Take ln(1+x) as t.

Find dt in terms of dx.

Replace dx with (1+x)dt.

Solve for d/dt of transformed expression.

The answer is: -[ln(1+x)]^-2 / (1+x)

The difference between Melody's and DarkStrider's answer is:

Melody has interpreted the expression as  ln[(1+x)^-1]

DarkStrider has interpreted it as    [ln(1+x)]^-1

Different expressions give different results!