# Derivatives Involving Absolute Value

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

## Derivative of an Absolute Value Function

Let $f(x) = | u(x) |$.
Note that $|u(x)| = \sqrt {u^2(x)}$
Use the
chain rule of differentiation to find the derivative of $f = | u(x) | = \sqrt {u^2(x)}$.
$\dfrac{df}{dx} = \dfrac{df}{du} \dfrac{du}{dx}$

$\dfrac{df}{du} = \dfrac{1}{2} \dfrac{2 u }{\sqrt {u^2}} = \dfrac{u}{|u|}$
Hence
$\large \color{red}{\dfrac{df}{dx} = \dfrac{du}{dx} \dfrac{u}{|u|}}$

## Examples with Solutions

Example 1
Find the first derivative $f \,'(x)$, if $f(x)$ is given by $f(x) = |x - 1|$ Solution to Example 1
Let $u = x - 1$ so that $f(x)$ may be written as
$f(x) = |u| = \sqrt{u^2}$

Use the chain rule
$f \, '(x) = \dfrac{df}{du} \dfrac{du}{dx}$

$f \, '(x) = (1/2) \dfrac{2u}{\sqrt{u^2}} \dfrac{du}{dx}$

$f \, '(x) = u \cdot \dfrac{u \, '}{|u|}$

$f \, '(x) = u \cdot \dfrac{1}{\sqrt{u^2}} = \dfrac{x-1}{|x-1|}$

Note the following:
1) if $x \gt 1$, then $|x - 1| = x - 1$ and $f \, '(x) = 1$.
2) If $x \lt 1$, then $|x - 1| = -(x - 1)$ and $f \, '(x) = -1$.
3) $f \,'(x)$ does not exist at $x = 1$.
The graphs of $f$ and its derivative $f'$ are shown below and we see that it is not possible to have a tangent to the graph of $f$ at $x = 1$ which explains the non existence of the derivative at $x = 1$.

Example 2

Find the first derivative of $f$ given by $f(x) = - x + 2 + |- x + 2|$ Solution to Example 2
$f(x)$ is made up of the sum of two functions. Let $u = - x + 2$ so that
$f\,'(x) = -1 + u \, ' \dfrac {u}{|u|} = -1 + \dfrac{-1(-x+2)}{|-x+2|}$
Simplify
$f\,'(x) = - 1 - \dfrac{-x+2}{|-x+2|}$
Note the following:
1) If $x \lt 2$, $|- x + 2 | = - x + 2$ and $f \, '(x) = -2$.
2) If $x \gt 2$, $|- x + 2 | = -(- x + 2)$ and $f \, '(x) = 0$.
3) $f \, '(x)$ does not exist at $x = 2$.

As an exercise, plot the graph of $f$ and explain the results concerning $f'(x)$ obtained above.

Example 3

Find the first derivative of $f$ given by $f(x) = \dfrac{x+1}{ |x^2 - 1| }$ Solution to Example 3
$f\,'(x) = \dfrac{1.|x^2 - 1|-(x+1)(2x)\dfrac{x^2 - 1}{|x^2 - 1|}}{|x^2 - 1|^2}$

Split the fraction into two fractions and simplify the fraction on the left side
$f\,'(x) = \dfrac{1}{|x^2-1|} - \dfrac{2x(x+1)(x^2-1)}{(x^2-1)^2|x^2-1|}$

Simplify the fraction on the right side
$f\,'(x) = \dfrac{1}{|x^2-1|} - \dfrac{2x}{(x-1)|x^2-1|}$

Set the two fractions to the same denominator
$f\,'(x) = \dfrac{x-1}{(x-1)|x^2-1|} - \dfrac{2x}{(x-1)|x^2-1|}$

Add the two fractions and simplify
$f\,'(x) = - \dfrac{x+1}{(x-1)|x^2-1|}$

Find the first derivatives of these functions
Hint: In some of the questions below you might have to apply the chain rule more than once.
1. $f(x) = |2x - 5|$
2. $g(x) = (x - 2)^2 + |x - 2|$
3. $h(x) = \left |\dfrac{x+1}{x-3} \right|$
4. $i(x) = \left | -2x^2 + 2x -1 \right|$
5. $j(x) = e^{|2x-1|}$
6. $k(x) = | \ln(-3x+1)|$
7. $l(x) = \sin |2x|$

1. $f \, '(x) = 2 \dfrac{2x-5}{|2x-5|}$

2. $g \, '(x) = 2 (x - 2) + \dfrac{x-2}{|x-2|}$

3. $h \, '(x) = -4 \left|\dfrac{x-3}{x+1}\right| \dfrac{x+1}{(x-3)^3}$

4. $i\,'(x) = \dfrac{\left(-2x^2+2x-1\right)\left(-4x+2\right)}{\left|-2x^2+2x-1\right|}$

5. $j\,'(x) = \dfrac{2e^{\left|2x-1\right|}\left(2x-1\right)}{\left|2x-1\right|}$

6. $k\,'(x) = -\dfrac{3\ln \left(-3x+1\right)}{\left|\ln \left(-3x+1\right)\right|\left(-3x+1\right)}$

7. $l\,'(x) = \dfrac{2x\cos \left(2\left|x\right|\right)}{\left|x\right|}$