Question

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

Answer 1

`\ \ \ f'(x) = { (-1, x<-1), ("undefined", x=-1), (1, x> -1) :}`

Explanation:

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

`f(x)=|x+1|` can be written as :

`\ \ \ \ \ f(x) = { (-(x+1), x+1<0), (0, x+1=0), (+(x+1), x+1> 0) :}`

`:. f(x) = { (-x-1, x<-1), (0, x=-1), (x+1, x> -1) :}`

And so:

`\ \ \ f'(x) = { (-1, x<-1), ("undefined", x=-1), (1, x> -1) :}`

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'(-1)` is undefined . (ie Continuity `Not =>` differentiability)