Question

How do you differentiate `y=ln(1/x)-1/lnx`?

Answer 1

`y'=-1/x+1/(x(lnx)^2)`

Explanation:

Let's rewrite the function a bit, mostly to get rid of `ln(1/x).`

`y=(ln1-lnx)-(lnx)^-1`

Since `ln1=0:`

`y=-lnx-(lnx)^-1`

`y'=-1/x-(-(lnx)^-2*d/dxlnx)` (Chain Rule)

`y'=-1/x+1/(x(lnx)^2)`