Question

Q25. If $\sf{y=[x+\sqrt{x^2+1}]^m}$, show that (x² + 1) y₂ + xy₁ - m²y = 0.​

Answer 1

Formula Used :-

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

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

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

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

$\boxed{ \red{ \sf \: \dfrac{dy}{dx} = y_1}}$

$\boxed{ \red{ \sf \: \dfrac{d}{dx}y_1 = y_2}}$

$\boxed{ \red{ \sf \:log {x}^{y} = y \: logx}}$

$\boxed{ \red{ \sf \: \dfrac{d}{dx}u.v = u\dfrac{d}{dx}v \: + \: v\dfrac{d}{dx}u}}$

Step by step calculation :-

↝ Consider,

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

↝ On taking log on both sides, we get

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

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

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

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

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

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

$\rm :\longmapsto\:\dfrac{1}{y}y_1 = \dfrac{m}{x + \sqrt{ {x}^{2} + 1}} \bigg(1 + \dfrac{1}{2\sqrt{ {x}^{2} + 1}} \times 2x \bigg)$

$\rm :\longmapsto\:\dfrac{1}{y}y_1 = \dfrac{m}{x + \sqrt{ {x}^{2} + 1}} \bigg(1 + \dfrac{x}{\sqrt{ {x}^{2} + 1}} \bigg)$

$\rm :\longmapsto\:\dfrac{1}{y}y_1 = \dfrac{m}{ \cancel{x + \sqrt{ {x}^{2} + 1}}} \bigg(\dfrac{ \cancel{\sqrt{ {x}^{2} + 1} + x}}{\sqrt{ {x}^{2} + 1}} \bigg)$

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

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

↝ On squaring both sides, we get

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

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

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

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

$\rm :\longmapsto\:( {x}^{2} + 1)2y_1y_2 + {y_1}^{2}(2x) = {m}^{2}2yy_1$

$\rm :\longmapsto\:2y_1(( {x}^{2} + 1)y_2 + {y_1}x) = {m}^{2}2yy_1$

$\rm :\longmapsto\:( {x}^{2} + 1)y_2 + {y_1}x = {m}^{2}y$

$\rm :\longmapsto\:( {x}^{2} + 1)y_2 + {y_1}x - {m}^{2}y = 0$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

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

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

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

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

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

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