Deriative of ${\frac{(\ln x)^{2}}{\sqrt{x}}}$ is not correct

Question

Using the chain and quotient rules, I get: $${\frac{2*\ln x}{x*\sqrt{x}}-\frac{(\ln x)^{2}}{2*\sqrt{x}}}$$

Answer:$(0,0)$ - min, $(1,0)$- min, $(e^4$,$\frac{16}{e^2})$ - max

Can anyone please describe how I should continue, because I don't know if I should (clearly yes because of the answers) include $0$ in the set of solutions, and by meaning of domain, in the original function x can't be negative...

Answer 1

Actually we get by the Quotient rule

$$\frac{2\ln(x)\frac{1}{x}\sqrt{x}-(\ln(x))^2\frac{1}{2}x^{-1/2}}{x}$$

Answer 2

I got the following: $$u={\frac{(\ln x)^{2}}{\sqrt{x}}}$$ $$u'={\frac{2\ln x}{x\sqrt{x}}}-{(\ln x)^{2}}\frac 12x^{-3/2}$$ $$u'={\frac{2\ln x}{x\sqrt{x}}}-\frac {{(\ln x)^{2}}} {2x\sqrt x}$$ $$u'={\frac{\ln x}{x\sqrt{x}}}\left (2-\frac 12\ln x \right )$$

Answer 3

Well $[(\ln x)^2]' = 2(\ln x)\frac 1x$ by chain rule.

And $[\sqrt{x}]' = \frac 12 \frac 1{\sqrt x}$

so $[\frac {[(\ln x)^2]}{\sqrt{x}}]' =$

$\frac {[(\ln x)^2]'\sqrt{x} - [(\ln x)^2][\sqrt{x}]'}{[\sqrt{x}]^2}=$ (by quotient rule)

$\frac { 2(\ln x)\frac 1x\sqrt{x} - (\ln x)^2 \frac 12 \frac 1{\sqrt x}}{x}$

which can be simplified to

$\frac {\ln x(2-\frac {\ln x}2)}{x\sqrt{x}}$

That equals $0$ if $x = 1$. Or if $\ln x = 4$ or $x= e^4$.

So $\frac {\ln^2 1}{\sqrt 1} = 0$ for $x=0$ so $(1,0)$ is an extrema.

And $\frac {\ln^2 e^4}{\sqrt{e^4}} = \frac {16}{e^2}$ for $x = e^4$ so $(e^4,\frac {16}{e^2})$ is another extrema.

I have absolutely no idea why you got your derivative over a numerater of $1$ nor do I have any idea why you thought $x = 0$ could be a solution to $\frac{\frac{2*\ln x}{x*\sqrt{x}}-\frac{(\ln x)^{2}}{2*\sqrt{x}}}{1} = 0$. It's clearly undefined for $x \le 0$.

Answer 4

Let $x=e^u$ therefore $$f(x)=u^2e^\dfrac{-u}{2}$$and we have $$\dfrac{\partial f}{\partial x}=\dfrac{\partial f}{\partial u}\dfrac{\partial u}{\partial x}=(2ue^\dfrac{-u}{2}-\dfrac{u^2}{2}e^\dfrac{-u}{2})\dfrac{1}{x}$$by resubstituting $$f'(x)=\left(2\ln x-\dfrac{\ln^2x}{2}\right)\dfrac{1}{x\sqrt x}$$

Answer 5

With a function like this it can be convenient to use the (formal) logarithmic derivative: if $$ f(x)=\frac{(\ln x)^2}{\sqrt{x}} $$ then $$ \ln f(x)=\ln\frac{(\ln x)^2}{\sqrt{x}}= \ln((\ln x)^2)-\ln(\sqrt{x})=2\ln\ln x-\frac{1}{2}\ln x $$ Therefore $$ \frac{f'(x)}{f(x)}=2\frac{1}{\ln x}\frac{1}{x}-\frac{1}{2}\frac{1}{x}= \frac{4-\ln x}{2x\ln x} $$ and finally $$ f'(x)=\frac{(\ln x)^2}{\sqrt{x}}\frac{4-\ln x}{2x\ln x}=\frac{(4-\ln x)\ln x}{2x\sqrt{x}} $$

Your function is only defined for $x>0$. The derivative vanishes where $4-\ln x=0$ or $\ln x=0$, that is, at $x=1$ or $x=e^4$.

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Some care should be used with this method, but in essence it works without the need of checking for the existence of the single steps. Only the final result is what we should care about.