Differentiate `(log)_x2` with respect to `x` :

Explanation

let `y=(log_x^2)`we know that,
`log_b^a =(log_c^a)/(log_c^b)`
so,`log_x^2=(log_e^2)/(log_e^x)`
now, differentiating `y` we get,
`(dy)/(dx)=d/(dx) [(log_e^2)/(log_e^x)]`
`(dy)/(dx)=d/(dx) [(log_e^2)* (log_e^x)^-1]`
`=log_e^2[-1*(log_e^x)^-2]*d/(dx)(log_e^x)`
`=(log_e^2)/(log_e^x)^2* 1/x`
`=(-log_e^2)/(x*(log_e^x)^2`