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By definition of absolute value, you have
[tex]f(x) = |x+1| = \begin{cases}x+1&\text{if }x+1\ge0 \\ -(x+1)&\text{if }x+1<0\end{cases}[/tex]
or more simply,
[tex]f(x) = \begin{cases}x+1&\text{if }x\ge-1\\-x-1&\text{if }x<-1\end{cases}[/tex]
On their own, each piece is differentiable over their respective domains, except at the point where they split off.
For x > -1, we have
(x + 1)' = 1
while for x < -1,
(-x - 1)' = -1
More concisely,
[tex]f'(x) = \begin{cases}1&\text{if }x>-1\\-1&\text{if }x<-1\end{cases}[/tex]
Note the strict inequalities in the definition of f '(x).
In order for f(x) to be differentiable at x = -1, the derivative f '(x) must be continuous at x = -1. But this is not the case, because the limits from either side of x = -1 for the derivative do not match:
[tex]\displaystyle \lim_{x\to-1^-}f'(x) = \lim_{x\to-1}(-1) = -1[/tex]
[tex]\displaystyle \lim_{x\to-1^+}f'(x) = \lim_{x\to-1}1 = 1[/tex]
All this to say that f(x) is differentiable everywhere on its domain, except at the point x = -1.