Question

Find the derivatives w.r.t.x:
$\rm \frac{px^{2}+qx+r}{(ax+b)^{2}}$

Answer 1

HOPE IT HELPS U ✌️✌️✌️

Answer 2

We know that such form of expression can solve by U/V method of differentiation

$\begin{lgathered}\frac{d}{dx} \bigg(\frac{U}{V}\bigg ) = \frac{V \frac{dU}{dx} - U \frac{dV}{dx} }{ {V}^{2} } \\ \\ here \: U = px^{2}+qx+r \\ \\ V = (ax+b)^{2} \\\\\end{lgathered}$

$\frac{d}{dx} \bigg(\frac{px^{2}+qx+r}{(ax+b)^{2}}\bigg) = \frac{(ax+b)^{2} \frac{d(px^{2}+qx+r)}{dx} - (px^{2}+qx+r) \frac{d(ax+b)^{2}}{dx} }{ { ({(ax + b)}^{2} })^{2} } \\ \\ = \frac{(ax+b)^{2}(2xp+q) - (px^{2}+qx+r)2(ax+b)(a)}{ { {(ax + b)}^{4} }} \\ \\ = \frac{(ax+b)(2xp+q) - 2a(px^{2}+qx+r)}{ { {(ax + b)}^{3} }} \\ \\ = \frac{2a {x}^{2}p + axq + 2xpb + bq - 2apx^{2} - 2aqx - 2ar)}{ { {(ax + b)}^{3} }} \\ \\ \frac{d}{dx} \bigg(\frac{px^{2}+qx+r}{(ax+b)^{2}}\bigg) = \frac{( - axq + 2xpb + bq - 2ar)}{ { {(ax + b)}^{3} }}$
Hope it helps you.