Question

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

Answer 1

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

Explanation:

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))`