Question

Find dy/DX when:
Y=√x+1 +√x-1 /√x+1 -√x-1

Answer 1

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

Given function is

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

On rationalizing the denominator, we get

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

We know,

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

So, using this, we get

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

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

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

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

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

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

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

We know, .

$\boxed{\tt{ \dfrac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} }}} \: \: and \: \: \boxed{\tt{ \dfrac{d}{dx}x = 1}}$

So, using this, we get

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

We know,

$\boxed{\tt{ \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}} \\ \\ and \\ \\ \boxed{\tt{ \dfrac{d}{dx}k \: = \: 0 \: }} \: \: \: \:$

So, using this, we get

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

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

Hence,

$\red{\rm :\longmapsto\:\boxed{\tt{ \dfrac{dy}{dx} = 1 + \dfrac{x}{\sqrt{ {x}^{2} - 1} }}}}$

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Additional Information

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

Answer 2

Answer:

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

Given function is

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

On rationalizing the denominator, we get

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

We know,

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

So, using this, we get

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

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

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

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

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

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

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

We know, .

$\boxed{\tt{ \dfrac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} }}} \: \: and \: \: \boxed{\tt{ \dfrac{d}{dx}x = 1}}$

So, using this, we get

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

We know,

$\boxed{\tt{ \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}} \\ \\ and \\ \\ \boxed{\tt{ \dfrac{d}{dx}k \: = \: 0 \: }} \: \: \: \:$

So, using this, we get

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

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

Hence,

$\red{\rm :\longmapsto\:\boxed{\tt{ \dfrac{dy}{dx} = 1 + \dfrac{x}{\sqrt{ {x}^{2} - 1} }}}}$

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Additional Information

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