Question

$Find\ the\ derivative\ of\ :$
$\displaystyle \frac{1}{ax^2+bx+c}$​

Answer 1

Step-by-step explanation:

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Answer 2

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$Find\ the\ derivative\ of\ :$

$\displaystyle \frac{1}{ax^2+bx+c}$

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$\boxed{\red{\bold{To \ find:}}}$

$\purple{\implies \displaystyle \frac{d}{dx} \frac{1}{ax^2+bx+c}}$

$\boxed{\red{\bold{By \ product \ rule:}}}$

$\red{\implies}\displaystyle \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{\frac{d}{dx} u\; \: . \: v - \frac{d}{dx} v\: \: . \: u}{v^2}$

$\red{\implies}\displaystyle{\frac{d}{dx}} \frac{1}{ax^2+bx+c} = \\ \frac{\frac{d}{dx} 1 \: \: . \: (ax^2 +bx+c) - \frac{d}{dx}(ax^2+bx+c) \: \; . \: 1}{{(ax^2+bx+c)}^2}$

$\red{\implies}\displaystyle \frac{0.(ax^2+bx+c) - (2ax+b). 1}{{(ax^2+bx+c)}^2}$

$\blue{\left( \because \displaystyle \frac{d}{dx} c = 0 \: and \: \frac{d}{dx} ax^n = nax^{n-1} \right)}$

$\green{\boxed{\implies{\boxed{\displaystyle \frac{-(2ax+b)}{{(ax^2+bx+c)}^2}}}}}$

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