Question:
Precalculus show that f'(x) = 2/4x-x^2 when g(f(x)) = x-2
Last updated: 8/6/2022
Precalculus show that f'(x) = 2/4x-x^2 when g(f(x)) = x-2 and g'(x) = 2 -[f(x)]^2
Last updated: 8/6/2022
Precalculus show that f'(x) = 2/4x-x^2 when g(f(x)) = x-2 and g'(x) = 2 -[f(x)]^2