Question

What are the first two derivatives of `1/ln(x)`?

Answer 1

Hello,

Answer If `f(x) = 1/(ln(x))`, then
`f'(x) = - 1/(x \cdot ln(x)^2)` and `f''(x) = (ln(x) + 2)/(x^2 \cdot ln(x)^3)`

First, use the rule `(1/u)' = - (u')/(u^2)` with `u(x) = ln(x)` (therefore `u'(x) = 1/x`). You get `f'(x)`.

Second, use the same rule but with `u(x)=x \cdot ln(x)^2`. Here,
`u'(x) = 1\cdot ln(x)^2 + x \cdot 2\ln(x) \cdot \frac{1}{x}`
`u'(x) = ln(x)^2 + 2\ln(x)`.

Finally, simplify by `ln(x)` to obtain `f''(x)`.