Question

find derivative of 1/(root x+1)-(rootx-1)​

Answer 1

Formula Used :-

$\boxed{ \red{ \bf \: \dfrac{d}{dx} \sqrt{x} = \dfrac{1}{2 \sqrt{x} }}}$

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

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

Solution :-

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

On rationalizing the denominator, we get

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

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

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

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

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

$\rm :\longmapsto\:\dfrac{d}{dx} y =\dfrac{d}{dx} \: \dfrac{\sqrt{x + 1} + \sqrt{x - 1}}{2}$

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

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

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{4}\bigg(\dfrac{1}{\sqrt{x + 1} } - \dfrac{1}{\sqrt{x - 1} } \bigg)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{4}\bigg(\dfrac{ \sqrt{x - 1} - \sqrt{x + 1} }{\sqrt{ {x}^{2} - 1} } \bigg)$

Additional Information :-

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

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

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

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

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

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