Question

if y= log{x+√(x²+1)} prove that (x²+1)y+xy=0

Answer 1

Appropriate Question :-

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

Prove that

$\bf\longmapsto \:( {x}^{2} + 1)y_2 + xy_1 = 0$

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

Given that

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

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

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

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

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

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

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

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

$\bf\implies \:y_1 = \dfrac{1}{ \sqrt{ {x}^{2} + 1} }$

$\rm :\longmapsto\: \sqrt{ {x}^{2} + 1} \: y_1 = 1$

On squaring both sides, we get

$\rm :\longmapsto\: ({x}^{2} + 1) \: {y_1}^{2} = 1$

Now, Differentiate both sides w. r. t. x, we get

$\rm :\longmapsto\: \dfrac{d}{dx}({x}^{2} + 1) \: {y_1}^{2} = \dfrac{d}{dx}1$

$\rm :\longmapsto\: {y_1}^{2} \dfrac{d}{dx}({x}^{2} + 1) \: + ( {x}^{2} + 1) \dfrac{d}{dx} {y_1}^{2} = 0$

$\rm :\longmapsto\: {y_1}^{2}(0 + 2x) + ( {x}^{2} +1 )2y_1y_2 = 0$

$\rm :\longmapsto\: {y_1}^{2}(2x) + ( {x}^{2} +1 )2y_1y_2 = 0$

$\bf\implies \:( {x}^{2} + 1)y_2 + xy_1 = 0$

Hence, Proved

Formula Used :-

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

$\rm :\longmapsto\:\dfrac{d}{dx}k = 0$

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

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

$\rm :\longmapsto\:\dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}$

$\rm :\longmapsto\:\dfrac{d}{dx}y = y_1$

$\rm :\longmapsto\:\dfrac{d}{dx}y_1 = y_2$

$\rm :\longmapsto\:\dfrac{d}{dx}u.v \: = \: u\dfrac{d}{dx}v \: \: + \: v\dfrac{d}{dx}u$