Question

derivative of x²+1/x-1 using quotient rule​

Answer 1

Answer:

Mathdude35

Step-by-step explanation:

Answer 2

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

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

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

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

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

Let's solve the problem now!!

$\rm :\longmapsto\:Let \: y \: = \: \dfrac{ {x}^{2} + 1 }{x - 1}$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx}\bigg(\dfrac{ {x}^{2} + 1}{x - 1} \bigg)$

We know,

$\rm :\longmapsto\:\dfrac{d}{dx} \dfrac{u}{v} = \dfrac{v \: \dfrac{d}{dx}u \: - u \: \dfrac{d}{dx}v }{ {v}^{2} }$

Here,

$\: \: \: \: \: \: \: \bull \: \: \sf \: \: u = {x}^{2} + 1$

$\: \: \: \: \: \: \: \bull \: \: \sf \: \: v = x - 1$

So,

On substituting the values, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{(x - 1)\dfrac{d}{dx}( {x}^{2} + 1) - ( {x}^{2} + 1)\dfrac{d}{dx}(x - 1) }{ {(x - 1)}^{2} }$

$\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{(x - 1)(2x + 0) - ({x}^{2} + 1)(1 - 0) }{ {(x - 1)}^{2} }$

$\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{ {2x}^{2} - 2x - {x}^{2} - 1}{ {(x - 1)}^{2} }$

$\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{ {x}^{2} - 2x - 1 }{ {(x - 1)}^{2} }$

$\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{ {x}^{2} - 2x + 1 - 1 - 1 }{ {(x - 1)}^{2} }$

$\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{ {(x - 1)}^{2} - 2}{ {(x - 1)}^{2} }$

$\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{ {(x - 1)}^{2} }{ {(x - 1)}^{2} } - \dfrac{2}{ {(x - 1)}^{2} }$

$\bf\implies \: \boxed{ \bf \: \dfrac{dy}{dx} = 1 - \dfrac{2}{ {(x - 1)}^{2}}}$

Additional Information :-

$\boxed{ \bf \: \dfrac{d}{dx} sinx = cosx}$

$\boxed{ \bf \: \dfrac{d}{dx} cosx = - sinx}$

$\boxed{ \bf \: \dfrac{d}{dx} tanx = {sec}^{2}x}$

$\boxed{ \bf \: \dfrac{d}{dx} secx \: = secx \: tanx}$

$\boxed{ \bf \: \dfrac{d}{dx} cotx = - \: cosec^{2}x}$

$\boxed{ \bf \: \dfrac{d}{dx} cosecx = - cosecx \: cotx}$

$\boxed{ \bf \: \dfrac{d}{dx} {e}^{x} = {e}^{x}}$

$\boxed{ \bf \: \dfrac{d}{dx} log(x) = \dfrac{1}{x} }$