Question

What is the derivative of `y=1/log_2(x)`?

Answer 1

`y'=-log(2)/(x(logx)^2)`

Explanation:

Note that:

`1/log_2x=log(2)/log(x)`

We then have using the chain rule:

`y=kf(g(x))`
`y'=kf'(g(x))g'(x)`
`f(u)=1/u`
`g(x)=logx`

These have derivatives:
`f'(u)=-1/u^2`
`g'(x)=1/x`

The full derivative is therefore:
`y'=-log(2)/(x(logx)^2)`