Question

differenriate x+3 / x²+4x+2​

Answer 1

Answer:

x

$x { }^{2} + 4x + 2 \\ - x + 3 \\ans \: 3x - 1$

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Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:\dfrac{x + 3}{ {x}^{2} + 4x + 2}$

Let we assume that,

$\rm :\longmapsto\:y = \dfrac{x + 3}{ {x}^{2} + 4x + 2}$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx} \dfrac{x + 3}{ {x}^{2} + 4x + 2}$

We know,

$\boxed{\tt{ \dfrac{d}{dx} \frac{u}{v} \: = \: \frac{v\dfrac{d}{dx}u \: - \: u\dfrac{d}{dx}v}{ {v}^{2} } \: }}$

So, using this Quotient Rule, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{( {x}^{2} + 4x + 2)\dfrac{d}{dx}(x + 3) - (x + 3)\dfrac{d}{dx}( {x}^{2} + 4x + 2)}{ {( {x}^{2} + 4x + 2)}^{2} }$

We know,

$\boxed{\tt{ \dfrac{d}{dx}x = 1}} \: \: \: \: \: \: \boxed{\tt{ \dfrac{d}{dx}k = 0}} \: \: \: \: \: \: \boxed{\tt{ \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \: }}$

So, using these, we get

$\rm \:  =  \: \dfrac{( {x}^{2} + 4x + 2)(1 + 0) - (x + 3)(2x + 4)}{ {( {x}^{2} + 4x + 2)}^{2} }$

$\rm \:  =  \: \dfrac{{x}^{2} + 4x + 2- (x + 3)(2x + 4)}{ {( {x}^{2} + 4x + 2)}^{2} }$

$\rm \:  =  \: \dfrac{{x}^{2} + 4x + 2- (2 {x}^{2} + 4x + 6x + 12)}{ {( {x}^{2} + 4x + 2)}^{2} }$

$\rm \:  =  \: \dfrac{{x}^{2} + 4x + 2- (2 {x}^{2} + 10x + 12)}{ {( {x}^{2} + 4x + 2)}^{2} }$

$\rm \:  =  \: \dfrac{{x}^{2} + 4x + 2- 2 {x}^{2} - 10x - 12}{ {( {x}^{2} + 4x + 2)}^{2} }$

$\rm \:  =  \: \dfrac{- {x}^{2} - 6x - 10}{ {( {x}^{2} + 4x + 2)}^{2} }$

$\rm \:  =  \: - \: \dfrac{{x}^{2} + 6x + 10}{ {( {x}^{2} + 4x + 2)}^{2} }$

Hence,

$\boxed{\tt{ \bf \: \dfrac{dy}{dx} =  \: - \: \dfrac{{x}^{2} + 6x + 10}{ {( {x}^{2} + 4x + 2)}^{2} } \: }}$

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More to Learn

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$