Question

What are the first and second derivatives of `g(x) =x^2/(lnx)^2`?

Answer 1

`dy/dx = 2x*(ln(x)-1)/ln^3(x)`
`(dy^2)/dx^2 = (2 (ln^2(x)-3 ln(x)+3))/(ln^4(x))`

Explanation:

We have

`y = x^2/(ln(x))^2`

We can use the product rule, the quotient rule, or the chain rule with logs. To make our life easier I'll use log derivativing.

Take the log of both sides

`ln(y) = ln(x^2/(ln(x))^2)`

Use log properties

`ln(y) = 2ln(x) - 2ln(ln(x))`

Derivate implicity

`1/y*dy/dx = 2/x - 2/ln(x)*1/x`

Multiply by `y`

`dy/dx = (2x)/ln^2(x) - (2x)/ln^3(x)`

Or

`dy/dx = 2x*(ln(x)-1)/ln^3(x)`

To derivate again, do the same thing, now skipping over some steps

`ln(dy/dx) = ln(2x) + ln(ln(x)-1) - 3ln(ln(x))`

`dx/dy * (d^2y)/dx^2 = 2/(2x) + 1/(ln(x)-1)*1/x - 3/ln(x)*1/x`

Multiply both sides by `dy/dx`

`(dy^2)/dx^2 = 2*(ln(x)-1)/(ln^3(x)) + (2x)/ln^3(x) - 6*(ln(x)-1)/ln^4(x)`

Or, using algebra to tie it all together

`(dy^2)/dx^2 = (2 (ln^2(x)-3 ln(x)+3))/(ln^4(x))`