Question

y√x²+1=log(√x²+1-x) then prove that dy/dx+xy+1=0

Answer 1

Step-by-step explanation:

$y \sqrt{( {x}^{2 } + 1)} = log( \sqrt{( {x}^{2} + 1 } - x ) \\ \\ then \: prove \: that \\ \\ ( {x}^{2} + 1) \frac{dy}{dx} + xy + 1 = 0$

because x and y can't separate simply,so apply implicit Differentiation

$y \frac{d( \sqrt{ {x}^{2} + 1 }) }{dx} + \sqrt{ {x}^{2} + 1 } \frac{dy}{dx} = \frac{d( log( \sqrt{ {x}^{2} + 1} - x)) }{dx} \\ \\ y. \frac{1}{2 \sqrt{ {x}^{2} + 1 } } .(2x + 0) + \sqrt{ {x}^{2} + 1} \frac{dy}{dx} = \frac{1}{( \sqrt{ {x}^{2} + 1} - x))} . (\frac{1}{2 \sqrt{ {x}^{2} + 1 } } .(2x + 0) - 1) \\ \\ \frac{xy}{ \sqrt{ {x}^{2} + 1 } } + \sqrt{ {x}^{2} + 1} \: \frac{dy}{dx} = \frac{x - \sqrt{ {x}^{2} + 1 } }{{( \sqrt{ {x}^{2} + 1} - x))} \sqrt{ {x}^{2} + 1 } } \\ \\ \frac{xy}{ \sqrt{ {x}^{2} + 1 } } + \sqrt{ {x}^{2} + 1} \: \frac{dy}{dx} = \frac{ - 1}{ \sqrt{ {x}^{2} + 1 } } \\ \\ take \: LCM \: and \: cancel \: denominator \: both \: sides \\ \\ xy + ( {x}^{2} + 1) \frac{dy}{dx} = - 1 \\ \\ ( {x}^{2} + 1) \frac{dy}{dx} + xy + 1 = 0 \\ \\ hence \: proved$