Question

Find the derative of (px + q) (r/x + s)

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Answer 1

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Answer 2

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

Given function is

$\rm :\longmapsto\:(px + q)\bigg(\dfrac{r}{x} + s \bigg)$

Let assume that

$\rm :\longmapsto\:y \: = \: (px + q)\bigg(\dfrac{r}{x} + s \bigg)$

$\rm :\longmapsto\:y \: = \: pr + psx + \dfrac{qr}{x} + qs$

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

$\rm :\longmapsto\:\dfrac{d}{dx} y \: =\dfrac{d}{dx}\bigg[ \: pr + psx + \dfrac{qr}{x} + qs\bigg]$

$\rm :\longmapsto\:\dfrac{dy}{dx} \: = \: \dfrac{d}{dx}pr + ps\dfrac{d}{dx}x + qr\dfrac{d}{dx} {x}^{ - 1} + \dfrac{d}{dx}qs$

We know,

$\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx}k = 0}}$

and

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

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = 0 + ps(1) - qr {x}^{ - 1 - 1} + 0$

$\rm :\longmapsto\:\dfrac{dy}{dx} = ps - qr {x}^{ - 2}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = ps - \dfrac{qr}{ {x}^{2} }$

Hence,

$\sf\implies\:\boxed{\tt{ \: \dfrac{dy}{dx} = ps - \dfrac{qr}{ {x}^{2} } \: }}$

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