Question

Its, urgent please solve it

Answer 1

$\bold{\huge{\underline{Solution-}}}$

हल-

$\mathbf{y = \frac{ {sin}^{ - 1} x}{ \sqrt{1 - {x}^{2} } } }$

दोनों पक्षों का एक्स के सापेक्ष अवकलन करने पर



$\mathbf{ \frac{dy}{dx} = \frac{( \sqrt{1 - {x}^{2} })( \frac{1}{ \sqrt{1 - {x}^{2} } }) - \: {sin}^{ - 1} \: x \{ \frac{1}{ \sqrt[2]{1 - {x}^{2} } } ( - 2x) }{( \sqrt{1 - {x}^{2} }) {}^{2} } } \\ \\ \\ \\ { \implies {\mathbf{ \frac{dy}{dx} }}} = \mathbf{\frac{1 + ( \frac{x {sin}^{ - 1} x}{ \sqrt{1 - {x}^{2} } }) }{( 1- {x}^{2}) } } \\ \\ \\ \\ \implies \: \mathbf{(1 - {x}^{2}) \frac{dy}{dx} } = \mathbf{1 + xy} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \mathbf {\huge [ \because \: \frac{sin {}^{ - 1}x }{ \sqrt{1 - {x}^{2} } } = y ]}\\ \\ \\ \\ \implies \: \mathbf{(1 - {x}^{2}) \frac{dy}{dx} = xy + 1}$

$\huge \boxed{ \boxed{ \mathbf{HENCE \: PROVED }}}$