## DERIVATIVE OF ABSOLUTE VALUE FUNCTION

Derivative of Absolute Value Function :

In this section, we will learn, how to find the derivative of absolute value function.

Let |f(x)| be the absolute value function.

Then the formula to find the derivative of |f(x)| is given below. Based on the formula given, let us find the derivative of |x|

|x|'  =  [x/|x|] . (x)'

|x|'  =  [x/|x|] . (1)

|x|'  =  x/|x|

Therefore, the derivative of |x| is

x / |x|

Let y = |x|'

Then, we have y = x/|x|

In y = x/|x|, if we plug x = 0, the denominator becomes zero.

Since the denominator becomes zero, y becomes undefined at x = 0

Let us plug some random values for "x" in y

When x = -3,   y  =  -3/|-3|  =  -3/3  =  -1

When x = -2,   y  =  -2/|-2|  =  -2/2  =  -1

When x = -1,   y  =  -1/|-1|  =  -1/1  =  -1

When x = 0,   y  =  0/|0|  =  0/0  =  undefined

When x = 1,   y  =  1/|1|  =  1/1  =  1

When x = 2,   y  =  2/|2|  =  2/2  =  1

When x = 3,   y  =  3/|3|  =  3/3  =  1

Let us summarize the above calculation in table. Now, based on the table given above, we can get the graph of derivative of |x| ## Derivative of Absolute Value Functions - Examples

Example 1 :

Differentiate |2x+1| with respect to x.

Solution :

Using the formula of derivative of absolute value function, we have

|2x+1|'  =  [(2x+1)/|2x+1|]  (2x+1)'

|2x+1|'  =  [(2x+1)/|2x+1|]  2

|2x+1|'  =  2(2x+1) / |2x+1|

Example 2 :

Differentiate |x3+1| with respect to x

Solution :

Using the formula of derivative of absolute value function, we have

|x3+1|'  =  [(x3+1)/|x3+1|]  (x3+1)'

|x3+1|'  =  [(x3+1)/|x3+1|]  3x2

|x3+1|'  =  3x2(x3+1) / |x3+1|

Example 3 :

Differentiate |x|3 with respect to x

Solution :

In the given function |x|3, using chain rule, first we have to find derivative for the exponent 3 and then for |x|.

(|x|3)'  =  {3|x|2 [x/|x|]  (x)'

(|x|3)'  =  {3|x|2 [x/|x|]  (1)

(|x|3)'  =  3x|x|

Example 4 :

Differentiate |2x-5| with respect to x

Solution :

Using the formula of derivative of absolute value function, we have

|2x-5|'  =  [(2x-5)/|2x-5|]  (2x-5)'

|2x-5|'  =  [(2x-5)/|2x-5|]  2

|2x-5|'  =  2(2x-5) / |2x-5|

Example 5 :

Differentiate (x-2)2 + |x-2| with respect to x

Solution :

Using the formula of derivative of absolute value function, we have

{(x-2)2 + |x-2|}'  =  [(x-2)2]' + |x-2|'

{(x-2)2 + |x-2|}'  =  2(x-2) + [(x-2)/|x-2|] ⋅ (x-2)'

{(x-2)2 + |x-2|}'  =  2(x-2) + [(x-2)/|x-2|] ⋅ (1)

{(x-2)2 + |x-2|}'  =  2(x-2) + (x-2) / |x-2|

Example 6 :

Differentiate 3|5x+7| with respect to x

Solution :

Using the formula of derivative of absolute value function, we have

3|5x+7|'  =  3  [(5x+7)/|5x+7|]  (5x+7)'

3|5x+7|'  = 3  [(5x+7)/|5x+7|]  5

3|5x+7|'  =  15(5x+1) / |5x+7|

Example 7 :

Differentiate |sinx| with respect to x

Solution :

Using the formula of derivative of absolute value function, we have

|sinx|'  =  [sinx/|sinx|]  (sinx)'

|sinx|'  =  [sinx/|sinx|] ⋅ cosx

|sinx|'  =  (sinx  cosx) / |sinx|

Example 8 :

Differentiate |cosx| with respect to x

Solution :

Using the formula of derivative of absolute value function, we have

|cosx|'  =  [cosx/|cosx|]  (cosx)'

|cosx|'  =  [cosx/|cosx|]  (-sinx)

|cosx|'  =  - (sinx  cosx) / |cosx|

Example 9 :

Differentiate |tanx| with respect to x

Solution :

Using the formula of derivative of absolute value function, we have

|tanx|'  =  [tanx/|tanx|]  (tanx)'

|tanx|'  =  [tanx/|tanx|]  sec2x

|tanx|'  =  sec²x  tanx / |tanx|

Example 10 :

Differentiate |sinx + cosx| with respect to x

Solution :

Using the formula of derivative of absolute value function, we have

|sinx + cosx|'  =  [(sinx+cosx) |sinx+cosx|]  (sinx+cosx)'

|sinx + cosx|'  =  [(cosx+sinx) |sinx+cosx|]  (cosx-sinx)

|sinx + cosx|'  =  (cos2x - sin2x) |sinx+cosx|

|sinx + cosx|'  =  cos2x / |sinx+cosx| After having gone through the stuff given above, we hope that the students would have understood, how to find the derivative of ab absolute value function.

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