Question

find derivative of ax+b/px²+qx+r
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Answer 1

check the attachment.

here we use the formula

$\frac{du}{dv}$

Answer 2

Correct question:

$\sf Find\;Derivative\;of\;\dfrac{px^{2}+qx+r}{ax+b}.$

Solution:

$\sf Now,\;let\;y=\dfrac{px^{2}+qx+r}{ax+b}\\ \\ \\ \sf We\;will\;solve\;this\;question\;by\;using\;following\;formula,\\ \\ \\ \implies \sf \dfrac{d}{dx}\Bigg(\dfrac{u}{v}\Bigg) =\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^{2}} \\ \\ \\ \implies \sf \dfrac{dy}{dx}=\dfrac{(ax+b)\dfrac{d}{dx}(px^{2}+qx+r)-(px^{2}+qx+r)\dfrac{d}{dx}(ax+b)}{(ax+b)^{2}}\\ \\ \\ \implies \sf \dfrac{dy}{dx}=\dfrac{(ax+b)(2px+q+0)-(px^{2}+qx+r)(a+0)}{(ax+b)^{2}}$

$\implies \sf \dfrac{dy}{dx}=\dfrac{2pax^{2}+axq+2pbx+bq-pax^{2}-qax-ax}{(ax-b)^{2}}\\ \\ \\ \implies {\boxed{\sf \dfrac{dy}{dx}= \dfrac{pax^{2}+2pbx+bq-ax}{(ax+b)^{2}}}}$