Question

$\boxed{ \small \tt \pink{if \: 2x = {y}^{ \frac{1}{m} } + {y}^{ - \frac{1}{m} } \: prove \: that \: ( {x}^{2} - 1)y _{2} + xy _{1} = {m}^{2} y.}}$
Answer with proper explanation.

Don't Dare for spam.

→ Expecting answer from :

★ Moderators
★ Brainly stars and teacher
★ Others best users ​​​​​

Answer 1

Answer:

y1/m + y-1/m = 2x Differentiate wrt 'x'. Squaring both sides, we get Read more on Sarthaks.com - https://www.sarthaks.com/1044905/if-y-1-m-y-1-m-2x-show-that-x-2-1-y2-xy1-m-2y

Step-by-step explanation:

Pls follow and mark me as brainliest

Answer 2

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

Given that,

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

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

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

We know,

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

and

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

So, using these, we get

$\rm :\longmapsto\:2 = \dfrac{1}{m} {y}^{ \frac{1}{m} - 1} y_1 - \dfrac{1}{m} {y}^{ - \frac{1}{m} - 1 }y_1$

$\rm :\longmapsto\:2 = \dfrac{y_1}{m} {y}^{ - 1}({y}^{ \frac{1}{m} } - {y}^{ - \frac{1}{m} })$

$\rm :\longmapsto\:2 = \dfrac{y_1}{my} ({y}^{ \frac{1}{m} } - {y}^{ - \frac{1}{m} })$

$\rm :\longmapsto\:2my = y_1({y}^{ \frac{1}{m} } - {y}^{ - \frac{1}{m} })$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ x - y = \sqrt{ {(x + y)}^{2} - 4xy}}}}$

So, using this identity, we get

$\rm :\longmapsto\:2my = y_1 \sqrt{ {({y}^{ \frac{1}{m} } + {y}^{ - \frac{1}{m}}})^{2} - 4({y}^{ \frac{1}{m} } {y}^{ - \frac{1}{m}})}$

So, using equation (1), we get

$\rm :\longmapsto\:2my = y_1 \sqrt{(2x)^{2} - 4}$

$\rm :\longmapsto\:2my = y_1 \sqrt{4x^{2} - 4}$

$\rm :\longmapsto\:2my = y_1 \sqrt{4(x^{2} - 1)}$

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

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

On squaring both sides, we get

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

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

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

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

$\rm :\longmapsto\: {m}^{2}(2yy_1) = {y_1}^{2}(2x - 0) + ( {x}^{2} - 1)(2y_1y_2)$

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

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

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

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

Hence, Proved

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

LEARN MORE

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