Question

How do you differentiate `f(x) = abs(5x-2)`?

Answer 1

This graph has two derivatives, and an undefined one at its corner. The two derivatives are simply the positive and the negative slope, so the graph's right half has a derivative of `5` and its left half, `-5`. Its corner would be where `5x - 2` and `-5x + 2` meet.

`5x - 2 = -5x + 2`
`10x = 4`
`x = 2/5`

`pm(5(2/5) - 2) = 2 - 2 = 0`

Therefore, the two halves meet at `(2/5, 0)`, and the derivative is defined for `(-oo, 2/5) uu (2/5, oo)`. At `(2/5, 0)`, there are two different derivatives depending on which side you approach the limit, and so it is considered undefined.

graph{|5x - 2| [-1, 2, -1.215, 4]}

Answer 2

I rewrite it as a piecewise defined function and differentiate each piece.

Explanation:

`f(x) = abs (5x-2) =` `{(5x-2," if ",5x-2 >= 0),(0, " if ", 5x-2 =0),(-(5x-2), " if ", 5x-2 < 0):}`

`= {(5x-2," if ",x >= 2/5),(0, " if ", x =2/5),( -5x+2, " if ", x < 2/5):}`

For `x>2/5`, we get `f'(x) = 5`

For `x < 2/5`, we get `f'(x) = -5`

Because the left and right derivatives are different at `x = 2/5`, the (two-sided) derivative, `f'(2/5)`, does not exist.
(In fact, there is a corner (cusp) at that point.)