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Q.Differentiate (px² + qx + r)/(ax + b) w.r.t. x. OR Find the derivative of x³ - 3x from first principles w.r.t. x.

Haryana BsehBoard of School Education Haryana (Senior Secondary Part-I / Class 11) 2026Subjective· 3mImportance★★★★★
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Apply the quotient rule d/dx(u/v) = (u′v − uv′)/v² with u = px²+qx+r and v = ax+b, then simplify.

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

By the quotient rule:

ddx(uv)=u′v−uv′v2=(2px+q)(ax+b)−(px2+qx+r)(a)(ax+b)2\frac{d}{dx}\left(\frac uv\right) = \frac{u'v-uv'}{v^2} = \frac{(2px+q)(ax+b) - (px^2+qx+r)(a)}{(ax+b)^2}

Expand the numerator:

(2px+q)(ax+b)=2apx2+2bpx+aqx+bq=2apx2+(2bp+aq)x+bq(2px+q)(ax+b) = 2apx^2 + 2bpx + aqx + bq = 2apx^2 + (2bp+aq)x + bq

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

Subtract: …

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