3
$\begingroup$

I want to study the derivability of this function

$$f(x)=x\left|{\log{x}}\right|$$

My textbook says the function is defined for $x>0$ (easy to understand for me, the argument of the logarithm must be positive) and it says: "it can certainly be derived for $x\neq 1$". I wonder how my textbook reached this conclusion without deriving the function first. I'm aware derivatives are defined like this:

$$\lim_{h\rightarrow0}{\frac{f(x_0+h)-f(x_0)}{h}}$$

Although I can't understand how we can reach conclusions about derivability just by looking at the function. Any hints?

$\endgroup$
1
  • $\begingroup$ The fact that behaviour of $|x|$ changes at 0 so you would expect whether $|\log(x)|$ also has a different behaviour when it attains it 0 that is at the point $x=1$. $\endgroup$ Aug 30, 2018 at 8:28

4 Answers 4

3
$\begingroup$

For $x<1$ you have $f(x)=-x\log x$ and $f(x)=x \log (x)$ for $x>1$. Now use the product rule.

For $x=1$ the derivative of function does not exist, because $\lim_{x\rightarrow 1^{-}}f'(x)=-1$ and $\lim_{x\rightarrow 1^{+}}f'(x)=1.$

$\endgroup$
2
  • $\begingroup$ What happens to the derivative when $x=1$? I see we all are excluding $x=1$ for some reason but $x\log{(x)}$ when $x=1$ is equal to $0$. What's wrong with that? $\endgroup$
    – Cesare
    Aug 30, 2018 at 7:45
  • $\begingroup$ @Cesare: See edit. $\endgroup$ Aug 30, 2018 at 7:51
1
$\begingroup$

For $x>1$ and $0<x<1$ we have that $x\log x$ and $-x\log x$ are differentiable indeed

$$(x\log x)’=\log x+1$$

otherwise for $x= 1$ we need to check differentiability directly by the definition.

$\endgroup$
1
$\begingroup$

For $x>1$ we have $f(x)=x \log x$. Hence $f$ is a product of differentiable functions on $(1, \infty)$.

For $0<x<1$ we have $f(x)=-x \log x$. Hence $f$ is a product of differentiable functions on $(0,1)$.

$\endgroup$
2
  • $\begingroup$ Really? Have you looked at a plot? $\endgroup$ Aug 30, 2018 at 7:51
  • $\begingroup$ OOps ! Yes you are right. $f$ is not differentiable at $1$. $\endgroup$
    – Fred
    Aug 30, 2018 at 7:54
1
$\begingroup$

One can vizualize:

The graph of $\log x$ cuts $x$-axis at 1. Multiplying by $x$ doesn't influence this fact. Absolute value puts the negative part above the $x$-axis, and the graph becomes broken. Therefore, the function is not smooth at 1 (not derivable).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .