Question

How do you find the derivative of `1/logx`?

Answer 1

It is `=-1/(xlog(x)^2)`.

Explanation:

With the chain rule.

`1/log(x)=log(x)^-1`

So you apply the chain rule to the power and then to the log:

`d/dxlog(x)^-1=-1*log(x)^-2*d/dxlog(x)`

`=-1/(log(x)^2) 1/x`

`=-1/(xlog(x)^2)`.