First and Second Derivative of function

Question

Hello, I need help finding the first and second derivative of y=ln(1+lnx)

Answer 1

y = ln(1+lnx)

y' = [1/(lnx +1)][lnx +1]'

y' = (1/x +0)/lnx+1)

y' = 1/[x(lnx)+1]

dy/dx = 1[x(lnx) +1]

y" = - (lnx +2)/x^2(lnx +1)^2

i'm fairly certain of both y' and y'" It's a little tricky, but 2 other experts confirm these answers.

Answer 2

Solution:

y = Ln(1+Lnx), dy/dx = 1/(1+Lnx). d( 1+Lnx) , dy/dx = y' = 1/(1+Lnx). (1/x) = 1/x(1+lnx) = 1/(x + x.lnx )

y" = du/dx.v - dv/dx.u / v² . Let u =1 and v = x+xLnx , du/dx = 0, and dv/dx = 1+Lnx+(1/x).x = 1+Lnx+1 = 2+Lnx

y" = (0)(x+xLnx) - (2+Lnx).1 / (x+xLnx)² = -(2+Lnx) / ( x+ xLnx)² = -(2+Lnx)/ ( x²+ 2x²Lnx+x²Ln²x)