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Question 24 of 43

Q.If y=2x2+3x+4x2+5y = \dfrac{2x^2 + 3x + 4}{x^2 + 5} the find dydx\dfrac{dy}{dx}.

Gujarat GsebGujarat Board (GSEB) HSC Commerce Board 2022Subjective· 3mImportance★★★★★
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Quotient rule with u=2x2+3x+4u=2x^2+3x+4, v=x2+5v=x^2+5 gives dydx=−3x2+12x+15(x2+5)2\dfrac{dy}{dx}=\dfrac{-3x^2+12x+15}{(x^2+5)^2}.

Given: y=2x2+3x+4x2+5y=\dfrac{2x^2+3x+4}{x^2+5}.

Let u=2x2+3x+4⇒u′=4x+3u=2x^2+3x+4\Rightarrow u'=4x+3, and v=x2+5⇒v′=2xv=x^2+5\Rightarrow v'=2x.

Quotient rule:

dydx=v u′−u v′v2=(x2+5)(4x+3)−(2x2+3x+4)(2x)(x2+5)2.\frac{dy}{dx}=\frac{v\,u'-u\,v'}{v^2}=\frac{(x^2+5)(4x+3)-(2x^2+3x+4)(2x)}{(x^2+5)^2}.

Expand the numerator:

(x2+5)(4x+3)=4x3+3x2+20x+15,(x^2+5)(4x+3)=4x^3+3x^2+20x+15,

(2x2+3x+4)(2x)=4x3+6x2+8x.(2x^2+3x+4)(2x)=4x^3+6x^2+8x. …

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