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Miscellaneous Exercise · Q8

Q.Find the derivative of ax+bpx2+qx+r\dfrac{ax + b}{px^2 + qx + r}.

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Apply the Quotient Rule to ax+bpx2+qx+r\dfrac{ax+b}{px^2+qx+r}. The derivative is −apx2−2bpx+(ar−bq)(px2+qx+r)2\dfrac{-apx^2-2bpx+(ar-bq)}{(px^2+qx+r)^2}.

ddx(uv)=u′v−uv′v2\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u'v-uv'}{v^2}

Set u=ax+bu=ax+b and v=px2+qx+rv=px^2+qx+r.

Step 1 — Differentiate the parts.

u′=a,v′=2px+q.u'=a,\qquad v'=2px+q.

Step 2 — Apply the Quotient Rule.

a(px2+qx+r)−(ax+b)(2px+q)(px2+qx+r)2\frac{a(px^2+qx+r)-(ax+b)(2px+q)}{(px^2+qx+r)^2}

Step 3 — Expand the numerator.

  • a(px2+qx+r)=apx2+aqx+ara(px^2+qx+r)=apx^2+aqx+ar
  • (ax+b)(2px+q)=2apx2+aqx+2bpx+bq(ax+b)(2px+q)=2apx^2+aqx+2bpx+bq

Step 4 — Subtract.

(apx2+aqx+ar)−(2apx2+aqx+2bpx+bq)(apx^2+aqx+ar)-(2apx^2+aqx+2bpx+bq)

Collecting like terms: apx2−2apx2=−apx2apx^2-2apx^2=-apx^2; the aqxaqx terms cancel; arar remains; then −2bpx-2bpx and −bq-bq:

−apx2−2bpx+(ar−bq)-apx^2-2bpx+(ar-bq) …

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