Derivatives Involving Absolute Value

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:

\[ f'(x) = \dfrac{df}{du} \dfrac{du}{dx} = \left(\dfrac{1}{2} \dfrac{2u}{\sqrt{u^2}}\right) \dfrac{du}{dx} = u \cdot \dfrac{u'}{|u|} = \dfrac{x-1}{|x-1|} \]

Note the following:

1. If \( x > 1 \), then \( |x - 1| = x - 1 \) and \( f'(x) = 1 \).
2. If \( x < 1 \), 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 \( f' \) show that it is not possible to draw a unique tangent to the graph of \( f \) at \( x = 1 \), which explains the non-existence of the derivative at \( x = 1 \).

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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 \( x < 2 \), \( |-x + 2| = -x + 2 \) and \( f'(x) = -2 \).
2. If \( x > 2 \), \( |-x + 2| = -(-x + 2) \) 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.

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Example 3

Find the first derivative of \( f \) given by:

\[ f(x) = \dfrac{x+1}{|x^2 - 1|} \]

Solution to Example 3:

Using the quotient rule and absolute value derivative formula:

\[ f'(x) = \dfrac{1 \cdot |x^2 - 1| - (x+1)(2x)\dfrac{x^2 - 1}{|x^2 - 1|}}{|x^2 - 1|^2} \]

Split the fraction into two terms 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. \( i(x) = |-2x^2 + 2x - 1| \)
5. \( j(x) = e^{|2x-1|} \)
6. \( k(x) = |\ln(-3x+1)| \)
7. \( l(x) = \sin |2x| \)

Answers to above exercises:

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. \( i'(x) = \dfrac{(-2x^2+2x-1)(-4x+2)}{|-2x^2+2x-1|} \)
5. \( j'(x) = \dfrac{2e^{|2x-1|}(2x-1)}{|2x-1|} \)
6. \( k'(x) = -\dfrac{3\ln(-3x+1)}{|\ln(-3x+1)|(-3x+1)} \)
7. \( l'(x) = \dfrac{2x\cos(2|x|)}{|x|} \)