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Exercises · Q15

Q.Differentiate y=2x+3x2+1y = \dfrac{2x + 3}{x^{2} + 1} using the quotient rule.

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Let u=2x+3u=2x+3 (u′=2u'=2) and v=x2+1v=x^2+1 (v′=2xv'=2x). By the quotient rule:

dydx=v u′−u v′v2=(x2+1)(2)−(2x+3)(2x)(x2+1)2.\frac{dy}{dx}=\frac{v\,u'-u\,v'}{v^2}=\frac{(x^2+1)(2)-(2x+3)(2x)}{(x^2+1)^2}.

Expand the numerator: (x2+1)(2)=2x2+2(x^2+1)(2)=2x^2+2 and (2x+3)(2x)=4x2+6x(2x+3)(2x)=4x^2+6x, so

2x2+2−(4x2+6x)=2x2+2−4x2−6x=−2x2−6x+2.2x^2+2-(4x^2+6x)=2x^2+2-4x^2-6x=-2x^2-6x+2.

Hence dydx=−2x2−6x+2(x2+1)2\dfrac{dy}{dx}=\dfrac{-2x^2-6x+2}{(x^2+1)^2}, which factors as −2(x2+3x−1)(x2+1)2\dfrac{-2(x^2+3x-1)}{(x^2+1)^2}. …

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