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

Precalculus show that f'(x) = 2/4x-x^2 when g(f(x)) = x-2 and g'(x)

Precalculus show that f'(x) = 2/4x-x^2 when g(f(x)) = x-2 and g'(x) = 2-[f(x)]^2

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