check the result[arc tan (3x-1) + arc cot (3x-1)]'=0 why? explain

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sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate derivatives of composed functions `arctan (3x-1)` and `arc cot (3x-1)` , using chain rule, such that:

`(arctan (3x-1))' = 3/(1 + (3x - 1)^2)`

`(arc cot (3x-1))' = -3/(1 + (3x - 1)^2)`

Adding the relations above, yields:

`(arctan (3x-1) + arc cot (3x-1))' = (arctan (3x-1))' + (arc cot (3x-1))' `

`(arctan (3x-1) + arc cot (3x-1))' = 3/(1 + (3x - 1)^2) - 3/(1 + (3x - 1)^2)`

`(arctan (3x-1) + arc cot (3x-1))' = 0`

Hence, evaluating the derivative of the summation `arctan (3x-1) + arc cot (3x-1)` yields `(arctan (3x-1) + arc cot (3x-1))' = 0.`

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

The result is correct! I'll prove to you why:

We'll differentiate each term of the sum with respect to x:

d[arc tan (3x-1)]/dx = (3x-1)'/[1 + (3x-1)^2]

d[arc tan (3x-1)]/dx = 3/[1 + (3x-1)^2] (1)

d[arc cot (3x-1)]/dx = -(3x-1)'/[1 + (3x-1)^2]

d[arc cot (3x-1)]/dx = -3/[1 + (3x-1)^2] (2)

We'll add (1) + (2):

[arc tan (3x-1) + arc cot (3x-1)]' = (3-3)/[1 + (3x-1)^2]

Since the terms of numerator are cancelling out each other, we'll get:

[arc tan (3x-1) + arc cot (3x-1)]' = 0

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