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

Feb 23, 2015

For the first derivative start by rewriting

$\frac{1}{\ln} x = {\left(\ln x\right)}^{-} 1$

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

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \left[\frac{x {\left(\ln x\right)}^{2} \left(0\right) - \left({\left(\ln x\right)}^{2} \left(1\right) + 2 \left(\ln x\right) \left(\frac{1}{x}\right) x\right)}{{\left(x {\left(\ln x\right)}^{2}\right)}^{2}}\right]$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \left[\frac{0 - {\left(\ln x\right)}^{2} - 2 \ln \left(x\right)}{{x}^{2} {\left(\ln x\right)}^{4}}\right]$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{{\left(\ln x\right)}^{2} + 2 \ln x}{{x}^{2} {\left(\ln x\right)}^{4}}$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{\ln x \left(\ln x + 2\right)}{{x}^{2} {\left(\ln x\right)}^{4}}$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{\ln x + 2}{{x}^{2} {\left(\ln x\right)}^{3}}$