Y = ln (1 + 2x)

Y = ln (1 + 2x)

First derivative f '(x) = 2 / (1 + 2x)
Second derivative f '' (x) = - 4 / (1 + 2x) & # 178;
The third derivative f '' '(x) = 16 / (1 + 2x) & # 179;
Fourth derivative f (4) (x) = - 96 / (1 + 2x) ^ 4

Urgently seeking the derivative of y = ln (2x + 1) (detailed process)
Wrong, it should be y = ln [1 / (2x + 1)]

Y = ln [1 / (2x + 1)] y '= (2x + 1) * [1 / (2x + 1)]' = - (2x + 1) * [1 / (2x + 1) & # 178;] * (2x + 1) '= - 2 (2x + 1) * [1 / (2x + 1) & # 178;] = - 2 / (2x + 1) or y = ln [1 / (2x + 1)] = - ln (2x + 1) y' = - 1 / (2x + 1) * (2x + 1) '= - 2 / (2x + 1)