# How do you find the derivative of the function f(x)=abs(x+1)?

Dec 27, 2016

$\setminus \setminus \setminus f ' \left(x\right) = \left\{\begin{matrix}- 1 & x < - 1 \\ \text{undefined} & x = - 1 \\ 1 & x > - 1\end{matrix}\right.$

#### Explanation:

The graph of $y = f \left(x\right)$ is as follows:
graph{|x+1| [-10, 10, -5, 5]}

$f \left(x\right) = | x + 1 |$ can be written as :

$\setminus \setminus \setminus \setminus \setminus \setminus f \left(x\right) = \left\{\begin{matrix}- \left(x + 1\right) & x + 1 < 0 \\ 0 & x + 1 = 0 \\ + \left(x + 1\right) & x + 1 > 0\end{matrix}\right.$

$\therefore f \left(x\right) = \left\{\begin{matrix}- x - 1 & x < - 1 \\ 0 & x = - 1 \\ x + 1 & x > - 1\end{matrix}\right.$

And so:

$\setminus \setminus \setminus f ' \left(x\right) = \left\{\begin{matrix}- 1 & x < - 1 \\ \text{undefined} & x = - 1 \\ 1 & x > - 1\end{matrix}\right.$

If you use the formal definition of the derivative as a limit you can easily establish that at $x = - 2$ then he LH limit (of -1) does not equal the RH limit (of +1) and so the derivative limit at $x = - 1$ is undefined, therefore $f ' \left(- 1\right)$ is undefined . (ie Continuity $N o t \implies$ differentiability)
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