What is the derivative of $f (x) = \frac{{(1 + ln x)}^{2}}{1 - ln x}$ $f(x) = (1+lnx)^2/(1-lnx)$ ?

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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:

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

What is the derivative of f(x) = (1+lnx)^2/(1-lnx)f(x)=(1+lnx)21−lnxf(x) = (1+lnx)^2/(1-lnx)?