Question

How do you find the first and second derivative of `1/lnx`?

Answer 1

For the first derivative start by rewriting

`1/lnx=(lnx)^-1`

now take the derivative using power rule and chain rule

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

For the second derivative use the quotient rule. keep negative sign out in front so you do not lose track of it

`(d^2y)/dx^2=-[(x(lnx)^2(0)-((lnx)^2(1)+2(lnx)(1/x)x))/((x(lnx)^2)^2]]`

`(d^2y)/dx^2=-[(0-(lnx)^2-2ln(x))/(x^2(lnx)^4)]`

`(d^2y)/dx^2=((lnx)^2+2lnx)/(x^2(lnx)^4)`

`(d^2y)/dx^2=(lnx(lnx+2))/(x^2(lnx)^4)`

`(d^2y)/dx^2=(lnx+2)/(x^2(lnx)^3)`