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Worked Examples · Example 6

Q.Differentiate y=x2+1x−1y = \dfrac{x^{2} + 1}{x - 1} using the quotient rule.

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Let u=x2+1u=x^2+1 and v=x−1v=x-1. Then u′=2xu'=2x and v′=1v'=1.

By the quotient rule ddx(uv)=v u′−u v′v2\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\,u'-u\,v'}{v^2} (denominator ≠0\neq0, i.e. x≠1x\neq1):

dydx=(x−1)(2x)−(x2+1)(1)(x−1)2.\frac{dy}{dx}=\frac{(x-1)(2x)-(x^2+1)(1)}{(x-1)^2}.

Expand the numerator: (x−1)(2x)=2x2−2x(x-1)(2x)=2x^2-2x, so

2x2−2x−(x2+1)=2x2−2x−x2−1=x2−2x−1.2x^2-2x-(x^2+1)=2x^2-2x-x^2-1=x^2-2x-1.

Hence dydx=x2−2x−1(x−1)2\dfrac{dy}{dx}=\dfrac{x^2-2x-1}{(x-1)^2}. …

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