if y=((sin^(−1) x)/(√(1+x^2 ))) show that (dy/dx)(1+x^2 )+xy=1


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Question Number 45413 by mondodotto@gmail.com last updated on 12/Oct/18

$if y = \frac{\sin^{- 1} x}{\sqrt{1 + x^{2}}} show that$ $\frac{d y}{d x} (1 + x^{2}) + x y = 1$
Commented by maxmathsup by imad last updated on 13/Oct/18

$w e h a v e y (x) = \frac{a r c s i n x}{\sqrt{1 + x^{2}}} \Rightarrow y (x) = {(1 + x^{2})}^{- \frac{1}{2}} a c s i n x \Rightarrow$ $y^{^{'}} (x) = - x {(1 + x^{2})}^{- \frac{3}{2}} a r c s i n x + {(1 + x^{2})}^{- \frac{1}{2}} \frac{1}{\sqrt{1 - x^{2}}} \Rightarrow$ $(1 + x^{2}) y^{^{'}} (x) + x y = - x {(1 + x^{2})}^{- \frac{1}{2}} a r c s i n x + \frac{\sqrt{1 + x^{2}}}{\sqrt{1 - x^{2}}} + x {(1 + x^{2})}^{- \frac{1}{2}} a r c s i n x$ $= \frac{\sqrt{1 + x^{2}}}{\sqrt{1 - x^{2}}} f i n a l l y t h r e i s a e r r o r a t t h e q u e s t i o n . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

	$\frac{d y}{d x} + \frac{x}{1 + x^{2}} y = \frac{1}{1 + x^{2}}$ $i n t r e g a y i n g f a c t o r$ $e^{\int \frac{x}{1 + x^{2}} d x}$ $e^{\frac{1}{2} \int \frac{d (1 + x^{3})}{1 + x^{2}}}$ $e^{\frac{1}{2} l n (1 + x^{2})}$ $e^{l n (\sqrt{1 + x^{2}})} = \sqrt{1 + x^{2}}$ $\sqrt{1 + x^{2}} \frac{d y}{d x} + x . \frac{y}{\sqrt{1 + x^{2}}} = \sqrt{1 + x^{2}}$ $\frac{d}{d x} (\sqrt{1 + x^{2}} \times y) = \sqrt{1 + x^{2}}$ $d (y \times \sqrt{1 + x^{2}}) = \sqrt{1 + x^{2}} d x$ $\int d (y \times \sqrt{1 + x^{2}}) = \int \sqrt{1 + x^{2}} d x$ $y \times \sqrt{1 + x^{2}} = \frac{x}{2} \sqrt{1 + x^{2}} + \frac{1}{2} l n (x + \sqrt{x^{2} + a^{)^{2}}}$
	Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

	$c h e c k t h e q u e s t i o n . . .$
	Commented by mondodotto@gmail.com last updated on 12/Oct/18

	$sir we need to prove$
	Answered by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

	$y = \frac{{s i n}^{- 1} x}{\sqrt{1 - x^{2}}}$ $\frac{d y}{d x} = \frac{\sqrt{1 - x^{2}} \times \frac{1}{\sqrt{1 - x^{2}}} - {s i n}^{- 1} x \times \frac{- 2 x}{2 \sqrt{1 - x^{2}}}}{(1 - x^{2})}$ $(1 - x^{2}) \frac{d y}{d x} = 1 + x y$
	Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

	$p l s c h e c k . . .$
	Commented by mondodotto@gmail.com last updated on 12/Oct/18

	$oky sir$
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