Find y'', y''' and y^IV for y = x ln x
Find $y''$, $y'''$ and $y^{\mathrm{IV}}$ for $y = x\ln x$.
Explanation of Method
The logarithm disappears after two steps, leaving a simple power.
Step 1: First derivative
Product rule, with the second term collapsing to $1$: $y' = 1\cdot\ln x + x\cdot\frac1x = \ln x+1$.
Step 2: Second derivative
The constant drops out and the logarithm becomes a power: $y'' = \frac1x = x^{-1}$.
Step 3: Third and fourth derivatives
From here the power rule $\bigl(x^{-k}\bigr)' = -kx^{-k-1}$ does everything: \[y''' = -x^{-2} = -\frac{1}{x^{2}}, \qquad y^{\mathrm{IV}} = 2x^{-3} = \frac{2}{x^{3}}.\]
Step 4: Verification
\[x = 1:\ y = 0,\ y' = 1,\ y'' = 1,\ y''' = -1,\ y^{\mathrm{IV}} = 2\]
\[x = 2:\ y'' = 0.5,\ y''' = -0.25,\ y^{\mathrm{IV}} = 0.25 \;\;\checkmark\]
\[\text{From } y''=x^{-1} \text{ onwards, } y^{(k)} = \frac{(-1)^{k}\,(k-2)!}{x^{\,k-1}} \text{ for } k\ge2 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y'' = \dfrac{1}{x}, \qquad y''' = -\dfrac{1}{x^{2}}, \qquad y^{\mathrm{IV}} = \dfrac{2}{x^{3}}\,}\]
Remark
Remark. Two differentiations turn $x\ln x$ into $1/x$, after which nothing logarithmic survives — this is the general pattern for $x^{m}\ln x$ with integer $m\ge0$: the $(m+1)$-st derivative is a pure power. Problem 294 proves the general statement, that $x^{n-1}\ln x$ has $n$-th derivative $(n-1)!/x$; the present case is $n = 2$.