Calculate derivate of an function

Question

We consider the fonction $f$ defined by $$ f(x) = \begin{cases} x \log|x|-x, &x \neq 0\\ 0, &x =0. \end{cases} $$ The question is calculate $f'(x)$.

We have for all $x \neq 0$: $$f'(x)= (x(\log|x|-1))' = \log|x|-1 - \frac{x}{|x|}.$$ I have difficulties to finite calculate, the result have to be $\log|x|$ but how we found this result? Please.

Answer 1

Pedestrian approach:

1) $y:=\ln |x|$ , $x \not =0.$

a) $x >0:$

$y=\ln |x| =\ln x; $

$y' = 1/x.$

b) $x <0:$

$y= \ln |x| = \ln (-x);$

$y' = (1/(-x))(-1) = 1/x$, chain rule.

c) $x \not =0:$

$f'(x)= \log |x| +x/x -1= \log |x|.$

d) $x=0:$

$\lim_{h \rightarrow 0} \dfrac{(h \ln|h| -h) -f(0)}{h}=$

$\lim_{h \rightarrow 0}(\ln |h| -1).$

Does not exist.

Answer 2

You are wrong about the derivative of $\log|x|$; it is $\frac1x$. THerefor$$x\neq0\implies f'(x)=x\times\frac1x+\log|x|=1+\log|x|.$$And $f'(0)$ doesn't exist (this follows from the definition of derivative).