# Derivatives of Absolute Value Functions

Tutorial on how to find derivatives of functions in calculus (Differentiation) involving absolute value functions.

## Example 1Find the first derivative f '(x), if f is given byf(x) = |x - 1|
## Solution to Example 1Noting that | u | = √ ( u ^{2} ), let u = x - 1 so that f(x) may be written asf(x) = y = √( u ^{2} )
We now use the chain rule f '(x) = (dy / du) (du / dx) = (1/2) (2 u) / √ (u ^{2}) (du / dx)
= u . u' / | u | = u . 1 / √ (u ^{2}) = (x - 1) / |x - 1|
We note that if x > 1, |x - 1| = x - 1 and f '(x) = 1. If x < 1, |x - 1| = -(x - 1) and f '(x) = -1. f'(x) does not exist at x = 1.
## Example 2Find the first derivative of f given byf(x) = - x + 2 + |- x + 2|
## Solution to Example 2f(x) is made up of the sum of two functions. Let u = - x + 2 so that f'(x) = -1 + u u' / |u| = -1 + (- x + 2)(-1) / |-x + 2| We note that if x < 2, |- x + 2 | = - x + 2 and f'(x) = -2 . If x > 2, f'(x) = 0. f'(x) does not exist at x = 2. As an exercise, plot the graph of f and explain the results concerning f'(x) obtained above.
## ExercisesFind the first derivatives of these functions1. f(x) = |2x - 5| 2. f(x) = (x - 2) ^{2} + |x - 2|
## Answers to Above Exercises1. f'(x) = 2 (2x - 5) / |2x - 5| 2. f(x) = 2 (x - 2) + (x - 2) / |x - 2| ## More Links and Referencesdifferentiation and derivatives |