Question

if y=x-√1+x^2, show that (1+x^2)(dy/dx)^2 = y^2​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

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

TO PROVE

$\displaystyle \: \: (1 + {x}^{2} ) {( \frac{dy}{dx})}^{2} = {y}^{2}$

CALCULATION

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

Differentiating both sides with respect to x we get

$\displaystyle \: \frac{dy}{dx} = 1 - \frac{1}{2} \times \frac{2x}{ \sqrt{1 + {x}^{2} } }$

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

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

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

$\implies \: \displaystyle \: \frac{dy}{dx} = \frac{ - y}{ \sqrt{1 + {x}^{2} } }$

Squaring both sides

$\implies \: \displaystyle \: {( \frac{dy}{dx})}^{2} = \frac{ {y}^{2} }{{1 + {x}^{2} } }$

$\therefore \: \displaystyle \: \: (1 + {x}^{2} ) {( \frac{dy}{dx})}^{2} = {y}^{2}$