Let f(x) = 1/1-x and g(x) = x/x-1.

Question

Compare f'(x) and g'(x) and explain.

Answer 1

f(x) = 1 / 1 - x

Apply derivative to each side with respective x

f'(x) = -1 / (1 - x)² (o - 1)

f'(x) = 1 / (1 - x)²   

Therefore f'(x) = 1 / (x - 1)²     -(1) 

g(x) = x / x - 1 = (x - 1 + 1) / x - 1 = 1 + (1 / x - 1)

Apply derivative to each side with respective x

g'(x) = 0 + (-1) / (x - 1)²

Simplify

g'(x) = -1 / (x - 1)²    -(2)

From (1) , (2)

f'(x) = -(-1) / (x - 1)² = -g'(x)

Add g'(x) to each side

f'(x) + g'(x) = 0

Therefore f'(x) is a additive inverse of g'(x).