**Derivative of absolute value of x at 0 :**

In this section, we are going to see, how to find the derivative of absolute value of (x) at x = 0.

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|**

In the above result |x|' = x/|x|, if we plug x = 0, the denominator becomes zero.

Since the denominator becomes zero, |x|' becomes undefined at x = 0

**Hence, absolute of x is not differentiable**** at x = 0 **

**Problem 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|

**Problem 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|

**Problem 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|.

|x³|' = {3|x|²} . [x/|x|] . (x)'

|x³|' = {3|x|²} . [x/|x|] . (1)

|x³|' = 3x|x|

**Problem 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|

**Problem 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|

**Problem 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|

**Problem 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|

**Problem 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|

**Problem 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|

**Problem 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 "Derivative of absolute value of x at 0".

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