# Derivatives Involving Absolute Value

 Tutorial on how to find derivatives of functions in calculus (Differentiation) involving the absolute value. Example 1: Find the first derivative $f \,'(x)$, if $f$ is given by $f(x) = |x - 1|$ Solution to Example 1 Noting that $| u | = \sqrt {u^2}$, let $u = x - 1$ so that $f(x)$ may be written as $f(x) = y = \sqrt{u^2}$ We now use the chain rule $f \, '(x) = \dfrac{dy}{du} \dfrac{du}{dx}$ $= (1/2) \dfrac{2u}{\sqrt{u^2}} \dfrac{du}{dx}$ $= u \cdot \dfrac{u \, '}{|u|}$ $= u \cdot \dfrac{1}{\sqrt{u^2}} = \dfrac{x-1}{|x-1|}$ 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 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 $f'(x) = -1 + u \, ' \dfrac {u}{|u|} = -1 + \dfrac{-1(-x+2)}{|-x+2|}$ 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. Exercises: Find the first derivatives of these functions 1. $f(x) = |2x - 5|$ 2. $f(x) = (x - 2)^2 + |x - 2|$ Answers to above exercises: 1. $f \, '(x) = 2 \dfrac{2x-5}{|2x-5|}$ 2. $f(x) = 2 (x - 2) + \dfrac{x-2}{|x-2|}$ More on differentiation and derivatives

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Updated: 2 April 2013