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Q.If y=px2+qx+rax+by = \dfrac{px^2 + qx + r}{ax + b}, find dydx\dfrac{dy}{dx}.

Jammu Kashmir JkboseJammu and Kashmir Board of School Education (Class 11) 2025Subjective· 4mImportance★★★★★
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Quotient rule on y=px2+qx+rax+by=\frac{px^2+qx+r}{ax+b} gives dydx=pax2+2pbx+qb−ar(ax+b)2\frac{dy}{dx}=\frac{pax^2+2pbx+qb-ar}{(ax+b)^2} after simplifying.

Let u=px2+qx+ru = px^2+qx+r so u′=2px+qu' = 2px+q, and v=ax+bv = ax+b so v′=av'=a.

By the quotient rule:

dydx=u′v−uv′v2=(2px+q)(ax+b)−(px2+qx+r)(a)(ax+b)2.\dfrac{dy}{dx} = \dfrac{u'v - uv'}{v^2} = \dfrac{(2px+q)(ax+b) - (px^2+qx+r)(a)}{(ax+b)^2}.

Expand the numerator:

(2px+q)(ax+b)=2pax2+(2pb+qa)x+qb.(2px+q)(ax+b) = 2pax^2 + (2pb+qa)x + qb.

a(px2+qx+r)=apx2+aqx+ar.a(px^2+qx+r) = apx^2 + aqx + ar.

Subtract:

2pax2+(2pb+qa)x+qb−apx2−aqx−ar2pax^2 + (2pb+qa)x + qb - apx^2 - aqx - ar …

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