What is the derivative of #f(x) = (1+lnx)^2/(1-lnx)#?

Question

1084 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-30T15:27:39

#f^' (x) = (dy)/dx = (2(1+lnx))/(x(1-lnx)) +(1+lnx)/(x(1-lnx))#

let #u = (1+ln x)^2#
Let #v= (1-ln x)#

Note: #(d)/dx (lnx) = 1/x #

so #(du)/dx = 2(1+ln x)(1/x) =2/x(1+ln x)#

and #(dv)/dx = - 1/x#

Using #(dy)/dx = (v (du)/dx + u (dv)/dx)/v^2#

# (dy)/dx =((1-lnx) times (2/x)(1+lnx))/(1-lnx)^(2) - ((1+lnx)^2(-1/x))/(1-lnx)^2#

Cancelling out:

#f^' (x) = (dy)/dx = (2(1+lnx))/(x(1-lnx)) +(1+lnx)/(x(1-lnx))#