Critical points of $y=x^2 (\ln x)^3$.

Question

What are the critical points of

$$y = x^2 \ln^3(x)\;?$$

The derivative is: $y'=2\ln^3(x) +3\ln^2(x)$ (after simplifying the expression)

Which is the minumum?

It is clear that the answer is $x=1$ while comparing to $0.$

Though the real answer is $\dfrac 1 { \sqrt{e^3}}$.

Where is my mistake?

Answer 1

Actually, the derivative is $$y'=2x \ln^3(x)+3x^2\ln^2(x)\,\frac1x.$$

Answer 2

You are wrong about $y'$; it is $3x\log^2(x)+2x\log^3(x)$. Therefore, $y'=0$ if and only if $x=1$ or $\log x=-\dfrac32$. However, $y$ has no local minimum at $1$, since its Taylor series there is$$(x-1)^3+\frac{1}{2} (x-1)^4-\frac{1}{4} (x-1)^5+\cdots$$On the other hand, if $\log x=-\dfrac32$, then $x=\dfrac1{\sqrt{e^3}}$, and$$y''\left(\frac1{\sqrt{e^3}}\right)=\frac92.$$Therefore $y$ has a local minimum at $\dfrac1{\sqrt{e^3}}$.

Answer 3

Upon taking the first derivative carefully, you get $$ y' = 2x\ln^3(x) + 3x\ln^2(x). $$ My advice: Don't do any premature division by $x$ yet! Instead, factor out whatever you can: $$ y'=x\ln^2(x)[2\ln(x) + 3]. $$ Now set each factor equal to $0$ and solve independently. The first gives $x=1$ (your solution) while the second gives $x=e^{-3/2}$.

These are our only two critical points. Now run the first derivative test around each critical point to see who is a local min (if any). For this, using the factored form of $y'$ (the second one) is easier. Around $x=1$ you get signs $+ +$ so there is no min/max there. Around $x=e^{-3/2}$ you DO get a sign change $-+$ which indicates a local min.