Question

How do you differentiate `y=log_x2`?

Answer 1

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

Explanation:

As `y=log_x2`, we have `y=ln2/lnx`

Hence assuming `g(f(x))=1/(f(x))`, where `f(x)=lnx`

`(dy)/(dx)=ln2xx(dg(x))/(df(x))xx(df)/(dx)`

`=ln2xx(-1/(lnx)^2)xx1/x`

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