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

Feb 21, 2015

Hello,

Answer If $f \left(x\right) = \frac{1}{\ln \left(x\right)}$, then
$f ' \left(x\right) = - \frac{1}{x \setminus \cdot \ln {\left(x\right)}^{2}}$ and $f ' ' \left(x\right) = \frac{\ln \left(x\right) + 2}{{x}^{2} \setminus \cdot \ln {\left(x\right)}^{3}}$

First, use the rule $\left(\frac{1}{u}\right) ' = - \frac{u '}{{u}^{2}}$ with $u \left(x\right) = \ln \left(x\right)$ (therefore $u ' \left(x\right) = \frac{1}{x}$). You get $f ' \left(x\right)$.

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

Finally, simplify by $\ln \left(x\right)$ to obtain $f ' ' \left(x\right)$.