DERIVATIVE OF ABSOLUTE VALUE OF TRIG FUNCTION

In this section, you will learn how to find derivative of absolute value of trigonometric functions.

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 absolute value of trigonometric functions

Derivative of |sinx| :

|sinx|' = [sinx/|sinx|] ⋅ (sinx)'

|sinx|' = [sinx/|sinx|] ⋅ cosx

|sinx|' = (sinx ⋅ cosx)/|sinx|

Derivative of |cosx| :

|cosx|' = [cosx/|cosx|] ⋅ (cosx)'

|cosx|' = [cosx/|cosx|] ⋅ (-sinx)

|cosx|' = -(sinx ⋅ cosx)/|cosx|

Derivative of |tanx| :

|tanx|' = [tanx/|tanx|] ⋅ (tanx)'

|tanx|' = [tanx/|tanx|] ⋅ sec²x

|tanx|' = sec²x ⋅ tanx/|tanx|

Derivative of |cscx| :

|cscx|' = [cscx/|cscx|] ⋅ (cscx)'

|cscx|' = [cscx/|cscx|] ⋅ (-cscx ⋅ cotx)

|cscx|' = -(csc²x ⋅ cotx)/|cscx|

Derivative of |secx| :

|secx|' = [secx/|secx|] ⋅ (secx)'

|secx|' = [secx/|secx|] ⋅ (secx ⋅ tanx)

|secx|' = -(sec²x ⋅ tanx)/|secx|

Derivative of |cotx| :

|cotx|' = [cotx/|cotx|] ⋅ (cotx)'

|cotx|' = [cot/|cotx|] ⋅ (-csc²x)

|cotx|' = -(csc²x ⋅ cotx)/|cotx|

Solved Problems

Problem 1 :

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|

Problem 2 : 

Differentiate |cosx - sinx| with respect to x

Solution :

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

|cosx - sinx|'  =  [(cosx-sinx) / |cosx-sinx|] ⋅ (cosx-sinx)'

|cosx - sinx|'  =  [(cosx-sinx) / |cosx-sinx|] ⋅ (-sinx-cosx)

|cosx - sinx|'  =  [(cosx-sinx) / |cosx-sinx|] ⋅ (-sinx-cosx)

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

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