Question

differentiate the following w.r.t. x, y=x2+3x+5/x2+2

Answer 1

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

Given function is

$\rm :\longmapsto\:y = \dfrac{ {x}^{2} + 3x + 5 }{ {x}^{2} + 2}$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx}\dfrac{ {x}^{2} + 3x + 5 }{ {x}^{2} + 2}$

We know,

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

So, on substituting the values, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{( {x}^{2} + 2)\dfrac{d}{dx}({x}^{2} + 3x + 5) - ( {x}^{2} + 3x + 5)\dfrac{d}{dx}( {x}^{2} + 2) }{ {( {x}^{2} + 2)}^{2} }$

We know,

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

and

$\red{ \boxed{ \sf{ \:\dfrac{d}{dx}x = 1}}}$

and

$\red{ \boxed{ \sf{ \:\dfrac{d}{dx}k = 0}}}$

So, using this identity, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{( {x}^{2} + 2)(2x + 3)- ( {x}^{2} + 3x + 5)(2x) }{ {( {x}^{2} + 2)}^{2} }$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{( {x}^{2} + 2)(2x + 3)- ( {x}^{2} + 3x + 5)(2x) }{ {( {x}^{2} + 2)}^{2} }$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{ {2x}^{3} + {3x}^{2} + 4x + 6 - {2x}^{3} - {6x}^{2} - 10x}{ {( {x}^{2} + 2)}^{2} }$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{ - {3x}^{2} - 6x + 6 }{ {( {x}^{2} + 2)}^{2} }$

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

Additional Information :-

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