# What is the second derivative of f(x) = ln x/x^2 ?

Dec 19, 2015

$\frac{6 \ln \left(x\right) - 5}{{x}^{4}}$

#### Explanation:

By the quotient rule, the first derivative is $f ' \left(x\right) = \frac{{x}^{2} \cdot \frac{1}{x} - \ln \left(x\right) \cdot 2 x}{{x}^{4}} = \frac{x - 2 x \ln \left(x\right)}{{x}^{4}} = \frac{1 - 2 \ln \left(x\right)}{{x}^{3}}$. Use the quotient rule again to find the second derivative:

$f ' ' \left(x\right) = \frac{{x}^{3} \cdot - \frac{2}{x} - \left(1 - 2 \ln \left(x\right)\right) \cdot 3 {x}^{2}}{{x}^{6}}$

$= \frac{- 2 {x}^{2} - 3 {x}^{2} + 6 {x}^{2} \ln \left(x\right)}{{x}^{6}} = \frac{6 \ln \left(x\right) - 5}{{x}^{4}}$

For extra interest, the fact that $f ' \left(x\right) = \frac{1 - 2 \ln \left(x\right)}{{x}^{3}}$ implies that $f$ is increasing for $0 < x < {e}^{\frac{1}{2}} \approx 1.65$ and decreasing for $x > {e}^{\frac{1}{2}}$, with a local maximum value of $f \left({e}^{\frac{1}{2}}\right) = \frac{\frac{1}{2}}{{\left({e}^{\frac{1}{2}}\right)}^{2}} = \frac{1}{2 e} \approx 0.18$ at $x = {e}^{\frac{1}{2}} \approx 1.65$. (Make sure you check all this!)

The fact that $f ' ' \left(x\right) = \frac{6 \ln \left(x\right) - 5}{{x}^{4}}$ implies that $f$ is concave down for $0 < x < {e}^{\frac{5}{6}} \approx 2.30$ and concave up for $x > {e}^{\frac{5}{6}}$, with an inflection point at $\left(x , y\right) = \left({e}^{\frac{5}{6}} , f \left({e}^{\frac{5}{6}}\right)\right) = \left({e}^{\frac{5}{6}} , \frac{\frac{5}{6}}{{\left({e}^{\frac{5}{6}}\right)}^{2}}\right) = \left({e}^{\frac{5}{6}} , \frac{5}{6 {e}^{\frac{5}{3}}}\right) \approx \left(2.30 , 0.16\right)$. (Make sure you check all this!)

Here's the graph to allow you to see these features.

graph{ln(x)/(x^2) [-0.562, 4.438, -0.77, 1.73]}