Question

How do you differentiate `y = 1 / log_2 x`?

Answer 1

`(dy)/(dx)=-ln2/(x(lnx)^2)`

Explanation:

`log_2x=lnx/ln2`

Therefore `y=ln2/lnx=ln2*1/lnx`

and `(dy)/(dx)=ln2*(-1/(lnx)^2)*1/x`

= `-ln2/(x(lnx)^2)`