# How do you find the second derivative of Y=(ln^2[x])/x?

Dec 21, 2016

$y ' ' = \frac{2 \left(- {\left(\ln x\right)}^{2} - 3 \ln x + 1\right)}{{x}^{3}}$

#### Explanation:

$y = \frac{{\left(\ln x\right)}^{2}}{x}$

$y ' = \frac{\left(x\right) \left(2\right) \left(\ln x\right) \left(\frac{1}{x}\right) - {\left(\ln x\right)}^{2} \left(1\right)}{{x}^{2}}$

$y ' = \frac{2 \left(\ln x\right) - {\left(\ln x\right)}^{2}}{{x}^{2}}$

$y ' = \frac{\left(\ln x\right) \left(2 - \ln x\right)}{{x}^{2}}$

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

$y ' ' = \frac{x \left(\frac{- \ln x + 2 - \ln x}{x}\right) - 2 \left(2 \ln x - {\left(\ln x\right)}^{2}\right)}{{x}^{3}}$

$y ' ' = \frac{2 - 2 \ln x - 4 \ln x - 2 {\left(\ln x\right)}^{2}}{{x}^{3}}$

$y ' ' = \frac{2 \left(- {\left(\ln x\right)}^{2} - 3 \ln x + 1\right)}{{x}^{3}}$