Second derivative of x ln(x)

Using the product rule on $f(x) = x\ln(x)$: $f'(x) = (1)\ln(x) + x\cdot\frac{1}{x} = \ln(x) + 1$.

Differentiating $f'(x) = \ln(x) + 1$ term by term gives $f''(x) = \frac{1}{x} + 0 = \frac{1}{x}$.

Write $f''(x) = \frac{1}{x} = x^{-1}$ so it is ready for the power rule.

Differentiating gives $f'''(x) = -1\cdot x^{-2} = -\frac{1}{x^2}$.

The third derivative is negative one over x squared. Differentiating using the power rule gives positive two over x cubed. For positive x, this value is positive. So the fourth derivative is positive.