Twice-differentiable real functions f,g satisfy f(x)g(x)=1 for every real x, with f(x), f'(x) and g'(x) never zero. What is f''(x)/f'(x)-g''(x)/g'(x)?

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From fg=1, g=1/f and g'=-f'/f². Since all displayed denominators are nonzero, differentiation gives g''/g'=f''/f'-2f'/f. Thus f''/f'-g''/g'=2f'/f=-2g'/g. Options B and D are correct.

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