Prove that the result of differentiating arc tan (1-x^2) + arc cot (1-x^2) is 0.

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to prove that the derivative of arc tan (1-x^2) + arc cot (1-x^2) is 0.

 

We know that the derivative of arc tan x = 1/[1 + (1-x^2)^2] and the derivative of arc cot x = -1 / [1 + (1-x^2)^2]

Let f(x) = arc tan (1-x^2) + arc cot (1-x^2)

f'(x) = -2x * 1/[1 + (1-x^2)^2] + (-2x)* -1/[1 + (1-x^2)^2]

=> -2x/[1 + (1-x^2)^2] + -(-2x)/[1 + (1-x^2)^2]

=> -2x/[1 + (1-x^2)^2] + 2x/[1 + (1-x^2)^2]

=> 0

The derivative of arc tan (1-x^2) + arc cot (1-x^2) = 0.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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

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

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

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

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

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

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

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

d[arc tan (1-x^2) + arc cot (1-x^2)]/dx = 0 q.e.d.

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