Question

if y=(x+√x^2-1)^m then (x^2-1)(dy/dx)^2-m^2y^2 is proved to be​

Answer 1

Answer:

$(x^2-1)(\frac{dy}{dx})^2-m^2y^2=0$

Step-by-step explanation:

$y=(x+\sqrt{x^2-1})^m$

Differentiate with respect to x

$\frac{dy}{dx}=m(x+\sqrt{x^2-1})^{m-1}*[1+\frac{2x}{2\sqrt{x^2-1}}]$

$\frac{dy}{dx}=m(x+\sqrt{x^2-1})^{m-1}*[1+\frac{x}{\sqrt{x^2-1}}]$

$\frac{dy}{dx}=m(x+\sqrt{x^2-1})^{m-1}*[\frac{\sqrt{x^2-1}+x}{\sqrt{x^2-1}}]$

squaring on both sides we get

$(\frac{dy}{dx})^2=m^2(x+\sqrt{x^2-1})^{2m-2}*[\frac{(\sqrt{x^2-1}+x)^2}{x^2-1}]$

$\implies\:(x^2-1)(\frac{dy}{dx})^2=m^2(x+\sqrt{x^2-1})^{2m-2}*[\(x+\sqrt{x^2-1})^2}]$

$\implies\:(x^2-1)(\frac{dy}{dx})^2=m^2(x+\sqrt{x^2-1})^{2m}$

$\implies\:(x^2-1)(\frac{dy}{dx})^2=m^2[(x+\sqrt{x^2-1})^m]^2$

$\implies\:(x^2-1)(\frac{dy}{dx})^2=m^2[y]^2$

$\implies\:(x^2-1)(\frac{dy}{dx})^2-m^2y^2=0$