Question

find the derivative of function 1÷ax^2+bx+c​

Answer 1

$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

$\boxed{ \purple{ \rm \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} }}$

$\large\underline\purple{\bold{Solution :- }}$

$: \implies \tt \: Let \: y \: = \dfrac{1}{ {ax}^{2} + bx + c}$

$: \implies \tt \: y \: = \: {( {ax}^{2} + bx + c) }^{ - 1}$

★ On differentiating both sides w r. t. x, we get

$: \implies \tt \: \dfrac{dy}{dx} = \dfrac{d}{dx} {( {ax}^{2} + bx + c) }^{ - 1}$

$\therefore \tt \: \dfrac{dy}{dx} = - \: {( {ax}^{2} + bx + c) }^{ - 2} \dfrac{d}{dx} ( {ax}^{2} + bx + c)$

$\therefore \tt \: \dfrac{dy}{dx} = - \: {( {ax}^{2} + bx + c) }^{ - 2}(2ax + b)$

$: \implies \tt \: \dfrac{dy}{dx} = \dfrac{ - (2ax + b)}{ {( {ax}^{2} + bx + c)}^{2} }$