Question

How do you differentiate `y=1/lnx`?

Answer 1

`=- 1/(x (ln x)^{2} )`

Explanation:

you can do this simply as `( (ln x)^{-1})'`

`=- (ln x)^{-2} (ln x)'`

`=- (ln x)^{-2} 1/x`

`=- 1/(x (ln x)^{2} )`

if you want to fiddle about with e and logs i suppose you could say that

`1/y = ln x`

`e^(1/y) = e^ln x = x`

so
`(e^(1/y))' = 1`

and
`( e^(1/y))' = e^(1/y) (1/y)'`

`= e^(1/y) * -(1/y^2) y'`

So `- e^(1/y) (1/y^2) y' = 1`

`y' = -y^2 * 1 / e^(1/y)`

`= -(1/ln x)^2 * 1/x`

`=- 1/(x (ln x)^{2} )`

same but bit more involved and fiddly