How To Find The Derivative Of An Absolute Value Function

Derivatives Involving Accented Value

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

Derivative of an Absolute Value Office

Allow \( f(x) = | u(10) | \).
Note that \( |u(x)| = \sqrt {u^ii(10)} \)
Use the chain rule of differentiation to find the derivative of \( f = | u(10) | = \sqrt {u^ii(x)}\).
\( \dfrac{df}{dx} = \dfrac{df}{du} \dfrac{du}{dx} \)

\( \dfrac{df}{du} = \dfrac{1}{2} \dfrac{2 u }{\sqrt {u^ii}} = \dfrac{u}{|u|} \)
Hence
\[ \big \color{scarlet}{\dfrac{df}{dx} = \dfrac{du}{dx} \dfrac{u}{|u|}} \]

Examples with Solutions

Example 1
Find the first derivative \( f \,'(x) \), if \( f(x) \) is given by \[ f(ten) = |x - 1| \]

Solution to Example 1

Let \( u = x - ane\) so that \( f(ten) \) may be written equally
\( f(x) = |u| = \sqrt{u^2} \)

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

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

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

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

Note the following:
1) if \( 10 \gt 1 \), so \( |x - one| = ten - 1 \) and \( f \, '(x) = 1 \).
two) If \( x \lt one \), and so \( |x - ane| = -(x - 1) \) and \( f \, '(x) = -ane \).
3) \( f \,'(10) \) does not exist at \( x = i \).
The graphs of \( f \) and its derivative \( f' \) are shown beneath and nosotros run into 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 \( x = 1 \).

Case two

Observe the first derivative of \( f \) given by \[ f(x) = - x + 2 + |- ten + two| \]

Solution to Instance ii

\( f(x) \) is made up of the sum of two functions. Allow \( u = - x + 2 \) and then that
\( f\,'(x) = -one + u \, ' \dfrac {u}{|u|} = -i + \dfrac{-i(-10+ii)}{|-ten+2|} \)
Simplify
\( f\,'(ten) = - 1 - \dfrac{-x+ii}{|-x+2|} \)
Note the following:
one) If \( ten \lt two \), \( |- x + 2 | = - x + 2 \) and \( f \, '(ten) = -2 \).
ii) If \( ten \gt ii \), \( |- ten + 2 | = -(- x + 2) \) and \( f \, '(x) = 0 \).
3) \( f \, '(10) \) does non exist at \( x = 2 \).

As an practise, plot the graph of \( f \) and explain the results concerning \( f'(10) \) obtained above.

Example three

Notice the first derivative of \( f \) given past \[ f(x) = \dfrac{x+1}{ |10^2 - 1| } \]

Solution to Example 3

\( f\,'(x) = \dfrac{ane.|x^2 - 1|-(x+one)(2x)\dfrac{x^2 - one}{|10^2 - 1|}}{|10^2 - 1|^2} \)

Split the fraction into two fractions and simplify the fraction on the left side
\( f\,'(x) = \dfrac{1}{|ten^2-ane|} - \dfrac{2x(x+1)(10^2-1)}{(x^2-one)^2|10^2-one|} \)

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

Set the two fractions to the same denominator
\( f\,'(10) = \dfrac{x-one}{(x-1)|ten^2-1|} - \dfrac{2x}{(10-1)|x^2-1|} \)

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

Exercises with Answers

Discover the first derivatives of these functions
Hint: In some of the questions below you might have to utilise the chain rule more one time.
1. \( f(x) = |2x - v| \)
ii. \( g(10) = (x - 2)^two + |x - two| \)
iii. \( h(x) = \left |\dfrac{x+1}{x-iii} \right| \)
4. \( i(x) = \left | -2x^2 + 2x -1 \right| \)
5. \( j(ten) = e^{|2x-one|} \)
6. \( k(x) = | \ln(-3x+i)| \)
7. \( l(x) = \sin |2x| \)

Answers to above exercises:

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

2. \( g \, '(ten) = ii (ten - 2) + \dfrac{10-2}{|x-2|} \)

iii. \( h \, '(10) = -iv \left|\dfrac{x-iii}{x+1}\right| \dfrac{x+1}{(x-3)^3} \)

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

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

6. \( thou\,'(x) = -\dfrac{iii\ln \left(-3x+1\correct)}{\left|\ln \left(-3x+1\right)\right|\left(-3x+ane\correct)} \)

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

More Links and References

1. Chain Dominion
2. differentiation and derivatives

Source: https://www.analyzemath.com/calculus/Differentiation/absolute_value.html

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