Find y' for y = x³ln 1/x, x > 0
Find $y'$ for $y = x^{3}\ln\dfrac{1}{x}$, $x \gt 0$.
Explanation of Method
Simplify the reciprocal inside the logarithm, then use the product rule.
Step 1: Simplify the logarithm
Using $\ln\dfrac{1}{x} = \ln x^{-1} = -\ln x$: $y = -x^{3}\ln x$.
The minus sign can now be carried through the whole computation.
Step 2: Apply the product rule
With $u = x^{3}$, $u' = 3x^{2}$, $v = \ln x$, $v' = \dfrac{1}{x}$: \[y' = -\bigl(u'v + uv'\bigr) = -\left(3x^{2}\ln x + x^{3}\cdot\frac{1}{x}\right) = -\bigl(3x^{2}\ln x + x^{2}\bigr).\]
Step 3: Factor
\[ y' = -x^{2}\bigl(3\ln x + 1\bigr). \]
Step 4: Verification
\[x = 1:\quad \ln 1 = 0 \;\Longrightarrow\; y = 0,\quad y' = -1\cdot(1+0) = -1\]
\[\frac{y(1+10^{-5})-y(1-10^{-5})}{2\times 10^{-5}} \approx -1.0 \;\;\checkmark\]
\[\text{Problem 70 with } n = 3, \text{ sign reversed:}\quad -x^{2}(1+3\ln x) \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = -x^{2}\bigl(1 + 3\ln x\bigr), \qquad x > 0\,}\]
Remark
Remark. This is Problem 70 with $n = 3$ and an overall minus sign. The derivative vanishes at $\ln x = -\tfrac13$, i.e. $x = e^{-1/3} \approx 0.717$, where $y$ attains its maximum on $(0,\infty)$.