Question

derivative of (x2/x+1) - (1/1-x)​

Answer 1

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

Given that .

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

Let reduced this to simplest form first,

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

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

$\rm :\longmapsto\:y = \dfrac{(x + 1)(x - 1)}{x + 1} + \dfrac{1}{x + 1} + \dfrac{1}{x - 1}$

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

$\rm :\longmapsto\:y = x - 1 + {(x + 1)}^{ - 1} + {(x - 1)}^{ - 1}$

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

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

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

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

$\bigg(\sf\because\:\dfrac{d}{dx}x = 1\bigg) \: \: \: \: \: \: \: \: \: \\ \bigg(\sf\because\:\dfrac{d}{dx} {x}^{n} = n {x}^{n - 1} \bigg) \\ \bigg(\sf\because\:\dfrac{d}{dx}x = 1\bigg) \: \: \: \: \: \: \: \:$

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

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

Additional information :-

$\rm :\longmapsto\:\dfrac{d}{dx} {e}^{x} = {e}^{x}$

$\rm :\longmapsto\:\dfrac{d}{dx} {a}^{x} = {a}^{x} log(a)$

$\rm :\longmapsto\:\dfrac{d}{dx}logx = \dfrac{1}{x}$