Question

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

Answer 1

`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`

Explanation:

`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`