Question

How do you differentiate `y= log _5 x`?

Answer 1

`(dy)/(dx)=1/(xln5)`

Explanation:

`y=log_5x=lnx/ln5->`change of base

`=1/(ln5)lnx`

`(dy)/(dx)=1/ln5*1/x->1/ln5` is a constant, so we don't change it

`=1/(xln5)`