<s>[INST] EXERCISE VERIFICATION: Find where $f(x):=|x|+|x+1|$ is differentiable and calculate its derivative Could someone verify my exercise?
a) $f(x):=|x|+|x+1|$

First, analyse the roots of each absolute value, where they go to zero:
$$|x|:=\left\{\begin{matrix}
 & x& x>0 \\ 
 & x- &  x<0\\ 
 & 0 & x=0
\end{matrix}\right.$$
$$|x+1|:=\left\{\begin{matrix}
 & x+1& x>-1 \\ 
 & -(x+1) &  x<-1\\ 
 & 0 & x=-1
\end{matrix}\right.$$
So the principal function will be:
$$f(x):=|x|+|x+1|:=\left\{\begin{matrix}
 & x+x+1 &=2x+1& x\ge0 \\ 
 & -x+x+1&=1 &  -1\le x<0\\ 
 & -x-x-1 &=-(2x+1) & x<-1
\end{matrix}\right.$$
So, the derivatives wil be:
for $x\ge0$:
$$\lim_{x\to c} \dfrac{f(x)-f(c)}{x-c}=\lim_{x\to c} \dfrac{2x+1-2c-1}{x-c}=\lim_{x\to c} \dfrac{2(x-c)}{x-c}=\lim_{x\to c}2= 2$$
for $-1\le x<0$:
$$\lim_{x\to c} \dfrac{f(x)-f(c)}{x-c}=\lim_{x\to c} \dfrac{1-1}{x-c}=\lim_{x\to c} \dfrac{0}{x-c}=\infty$$ so, $f(x)$ isn´t differentiable at $.-1\le x<0$
for $x<-10$:
$$\lim_{x\to c} \dfrac{f(x)-f(c)}{x-c}=\lim_{x\to c} \dfrac{-(2x+1)+(2c+1)}{x-c}=\lim_{x\to c} \dfrac{-2(x-c)}{x-c}=\lim_{x\to c}-2= -2$$
So, $f(x)$ isn´t differentiable at $-1\le x<0$
 [/INST] Since $|x|=\sqrt{x^2}$,  we have $|x|'=\dfrac{x}{|x|}$ for $x\neq0$, hence $f'(x)=\dfrac{x}{|x|}+\dfrac{x+1}{|x+1|}$ for $x\notin\{-1,0\}$.
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