Find y''' for y = x²ln x
Find $y'''$ for $y = x^{2}\ln x$.
Explanation of Method
Three steps, after which only a reciprocal remains.
Step 1: First derivative
\[ y' = 2x\ln x+x^{2}\cdot\frac1x = 2x\ln x+x. \]
Step 2: Second derivative
Product rule on the first term, and $\bigl(x\bigr)' = 1$: \[y'' = 2\ln x+2x\cdot\frac1x+1 = 2\ln x+2+1 = 2\ln x+3.\]
Step 3: Third derivative
The constant $3$ vanishes and only the logarithm's derivative survives: $y''' = \frac2x$.
Step 4: Verification
\[x = 1:\ y = 0,\ y' = 1,\ y'' = 3,\ y''' = 2 \;\;\checkmark\]
\[x = e:\ y'' = 2+3 = 5, \qquad y''' = \frac{2}{e} \approx 0.7358 \;\;\checkmark\]
\[\text{From here on: } y^{\mathrm{IV}} = -\frac{2}{x^{2}},\ y^{(5)} = \frac{4}{x^{3}} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y''' = \dfrac{2}{x}\,}\]
Remark
Remark. The answer $2/x = 2!/x$ is the case $n = 3$ of the general formula of Problem 294: the $n$-th derivative of $x^{n-1}\ln x$ is $\dfrac{(n-1)!}{x}$. Problem 275 is the same statement for $n = 2$. Each extra power of $x$ buys exactly one more differentiation before the logarithm is gone.