### Here Is How You Can Calculate The Derivative Of Absolute Value

Derivatives are defined as the differentiation of the independent variable with respect to the dependent variable. We can solve this by expressing the function for an absolute value of a variable x, and then dividing its value by x. Thus, let’s let x equal absolute value, y = |x|

**Here is a complete solution step-by-step:**

In general, we can observe that the absolute value of a given variable is the non-negative value of the variable. A common term for this is the modulus of x. If we consider a function y in terms of

⇒y=|x| ……………. Equation (i)

Keeping that as a background, we know the derivative of a function is the differentiation of an independent variable by its dependent variable. We will therefore need to differentiate y with respect to x in order to find the derivative for the absolute value. Consequently, we should simplify the expression for modulus of x.

We can always calculate the square root and absolute value of any number without any difficulty. It follows that x’s modulus shall equal its square root. So, the equation (i) can also be written as:

⇒ y = …….. Equation (ii)

⇒ y = (x^{2})^{1/2}

Differentiating both sides of the equation with respect to x, then we will get.

⇒ =

Now we apply chain rule for differentiation.

⇒ = (x^{2})^{1/2 -1 }

⇒ = (x^{2})^{-1/2 }

⇒ =

From equation (ii) we can write the denominator from the right hand side (RHS)

⇒ =

⇒ =

Replacing the equation (i) in the above written equation, we will get

⇒ =

Hence, we find out that the absolute value of x is equal to

**Note:** To find the derivative of the absolute value of x will take the value equals to or greater than 1 for x > 0, and −1 for x < 0. By solving the equation we find out that for the absolute value of x, the value of x cannot be equal to 0 as it will return us which cannot defined.

Let’s now solve some examples:

**Example 1:** Find out the derivative of |2x+1| with respect to x.

**Solution: **

Using the formula of derivative.

|2x+1|’ = [(2x+1)/|2x+1|] * (2x+1)’

|2x+1|’ = [(2x+1)/|2x+1|] * 2

|2x+1|’ = 2 * (2x+1)/|2x+1|

**Example 2:** Solve and find the derivative of |x^{3}+1| with respect to x.

**Solution:**

|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:**

Solve and find the 3ifferentiate and find the value of |x|^{3} with respect to x

**Solution:**

(|x|^{3})’ = {3|x|^{2}} ⋅ [x/|x|] ⋅ (x)’

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

(|x|^{3})’ = 3x|x|

**Example 4:**

Solve and differentiate the value of (x-2)^{2} + |x-2| with respect to x

**Solution :**

{(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 5:**

Solve and the value of differentiate 3|5x+7| with respect to x

**Solution:**

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 6:**

Solve and the value of differentiate |sinx| with respect to x

**Solution :**

|sinx|’ = [sinx/|sinx|] ⋅ (sinx)’

|sinx|’ = [sinx/|sinx|] ⋅ cosx

|sinx|’ = (sinx ⋅ cosx) / |sinx|

**Example 7:**

Solve and the value of differentiate |cosx| with respect to x

**Solution:**

|cosx|’ = [cosx/|cosx|] ⋅ (cosx)’

|cosx|’ = [cosx/|cosx|] ⋅ (-sinx)

|cosx|’ = – (sinx ⋅ cosx) / |cosx|

**Example 8:**

Solve and the value of differentiate |tanx| with respect to x

**Solution:**

|tanx|’ = [tanx/|tanx|] ⋅ (tanx)’

|tanx|’ = [tanx/|tanx|] ⋅ sec^{2}x

|tanx|’ = sec²x ⋅ tanx / |tanx|

**Example 9:**

Solve and the value of differentiate |sinx + cosx| with respect to x

**Solution :**

|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|