Vtyjkyuk

I want to study the derivability of this function

$$f(x)=xleft|logxright|$$

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 $xneq 1$". I wonder how my textbook reached this conclusion without deriving the function first. I'm aware derivatives are defined like this:

$$lim_hrightarrow0fracf(x_0+h)-f(x_0)h$$

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

For $x<1$ you have $f(x)=-xlog x$ 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
$lim_xrightarrow 1^-f'(x)=-1$ and $lim_xrightarrow 1^+f'(x)=1.$

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

$$(xlog x)’=log x+1$$

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

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 $(0,1)$.

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).