What is the equation of the normal line of $f(x)=1-x/lnx$ at $x=2$ ?

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Answer 1 · 2016-01-13T10:07:22

$y=-((ln 2)^2)/(1-ln 2)*x+ ( 2(ln 2)^2)/(1-ln 2) + 1 -2/ln 2$

OR $y=-1.565744172711 x+ 1.2460982636442$

$f(x) = 1 -x/ln x$
$f(2)=1-2/ln 2$
$f' (x) = -1((ln x *(1) -x(1/x) (1))/(ln x)^2)$

$f' (2)= (1-ln 2)/(ln 2)^2$

What is the equation of the normal line of f(x)=1-x/lnxf(x)=1-x/lnx at x=2x=2?