In this section, you 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|

**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 |x^{3}+1| with respect to x

**Solution :**

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

|x^{3}+1|' = [(x^{3}+1)/|x^{3}+1|] ⋅ (x^{3}+1)'

|x^{3}+1|' = [(x^{3}+1)/|x^{3}+1|] ⋅ 3x^{2}

|x^{3}+1|' = 3x^{2}(x^{3}+1) / |x^{3}+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|] ⋅ sec^{2}x

|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|' = (cos^{2}x - sin^{2}x) / |sinx+cosx|

|sinx + cosx|' = cos2x / |sinx+cosx|

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