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Derivative of a rational function via the quotient rule

Let g(x)=x2+1x2−1g(x) = \dfrac{x^2+1}{x^2-1}. Find g′(x)g'(x).

Hint

Use the quotient rule (fh)′=f′h−fh′h2\left(\frac{f}{h}\right)' = \frac{f'h - fh'}{h^2} with f(x)=x2+1f(x) = x^2+1 and h(x)=x2−1h(x) = x^2-1. Compute f′(x)f'(x) and h′(x)h'(x) first, then substitute carefully, keeping track of the order of subtraction in the numerator.

Step-by-step solution

Identify f(x)=x2+1f(x) = x^2+1 and h(x)=x2−1h(x) = x^2-1, so f′(x)=2xf'(x) = 2x and h′(x)=2xh'(x) = 2x. By the quotient rule, g′(x)=f′(x)h(x)−f(x)h′(x)[h(x)]2g'(x) = \dfrac{f'(x)h(x) - f(x)h'(x)}{[h(x)]^2}. Substituting gives g′(x)=2x(x2−1)−(x2+1)(2x)(x2−1)2g'(x) = \dfrac{2x(x^2-1) - (x^2+1)(2x)}{(x^2-1)^2}.

Expand the numerator: 2x(x2−1)=2x3−2x2x(x^2-1) = 2x^3 - 2x and (x2+1)(2x)=2x3+2x(x^2+1)(2x) = 2x^3+2x. Subtracting, 2x3−2x−(2x3+2x)=−4x2x^3 - 2x - (2x^3+2x) = -4x. Thus g′(x)=−4x(x2−1)2g'(x) = \dfrac{-4x}{(x^2-1)^2}.

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