Question

What is the derivative of `f(x)=tan(1/lnx)`?

Answer 1

`=-(sec^2(1/lnx))/(xln^2x)`

Explanation:

According to the chain rule:

`f'(x)=sec^2(1/lnx)d/dx[1/lnx]`

`=sec^2(1/lnx)d/dx[ln^-1x]`

`=sec^2(1/lnx)xx-ln^-2x xx d/dx[lnx]`

`=sec^2(1/lnx)(1/-ln^2x)(1/x)`

`=-(sec^2(1/lnx))/(xln^2x)`