Question

y=(x + √x² - 1)^M, prove that (M²-1)(dy/dx)²= M²y²​

Answer 1

Answer:

Hope this helps.

N.B. It appears the "M²-1" should be "x²-1".

$\displaystyle y = (x+\sqrt{x^2-1})^M\\ \\\frac{dy}{dx} = M(x+\sqrt{x^2-1})^{M-1}\times\bigl(1 + \tfrac12(x^2-1)^{-1/2}\times 2x\bigr)\\ \\= M(x+\sqrt{x^2-1})^{M-1}\times\left(1+\frac{x}{\sqrt{x^2-1}}\right)\\ \\= M(x+\sqrt{x^2-1})^{M-1}\times\left(\frac{x+\sqrt{x^2-1}}{\sqrt{x^2-1}}\right)\\ \\= \frac{My}{\sqrt{x^2-1}}\\ \\\Rightarrow (x^2-1)\left(\frac{dy}{dx}\right)^2 = M^2y^2$