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

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.