Question

If y= sin-1 x/(1-x2)^1/2,then dy/dx=

Answer 1

Step-by-step explanation:

$y = \frac{ { \sin }^{ - 1}x }{ \sqrt{1 - {x}^{2} } } = \frac{ { \sin }^{ - 1} x}{ {(1 - {x}^{2} )}^{ \frac{1}{2} } } \\ \blue{ \boxed{y = \frac{u}{v} }} \\ \blue{ \boxed{ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx} }{ {v}^{2} } }} \\ \frac{dy}{dx} = \frac{ {({1 - {x}^{2} })}^{ \frac{1}{2} } \times \frac{1}{ {({1 - {x}^{2} })}^{ \frac{1}{2} } } -{ \sin }^{ - 1}x \times \frac{1}{2}{(1 - {x}^{2} )}^{ - \frac{1}{2} } \times ( - 2x) }{ {( {({1 - {x}^{2} })}^{ \frac{1}{2} } )}^{2} } \\ \\ {(1 - {x}^{2} )} \frac{dy}{dx} = 1 +x {\sin }^{ - 1}x \times {(1 - {x}^{2} )}^{ - \frac{1}{2} } \\ \\ \blue{ \rightarrow{let \: \:{\sin }^{ - 1}x \times {(1 - {x}^{2} )}^{ - \frac{1}{2} } = y}} \\ \\ {(1 - {x}^{2} )} \frac{dy}{dx} = 1+xy \\ \\ \blue{ \boxed{correct \: answer \rightarrow \: (ii)}} \\ \\\red{\mathfrak{\underline{\large{Hope \: It \: Helps \: You }}}} \\ \green{\mathfrak{\underline{\large{Mark \: Me \: Brainliest}}}}$

Answer 2

Step-by-step explanation:

$y = \frac{ {sin}^{ - 1}x }{ \sqrt{1 - {x}^{2} } } \\ \\ y \sqrt{1 - {x}^{2} } = {sin}^{ - 1} x \\ \\ differentiate \: \: \: w.r.t \: \: \: \: x \\ \\ y \frac{1}{2 \sqrt{1 - {x}^{2} } } ( - 2x) + \sqrt{1 - {x}^{2} } \frac{dy}{dx} = \frac{1}{ \sqrt{1 - {x}^{2} } } \\ \\ multiplying \: \sqrt{1 - {x}^{2} } both \: sides \\ \\ - xy + (1 - {x}^{2} ) \frac{dy}{dx} = 1 \\ \\ (1 - {x}^{2} ) \frac{dy}{dx} = 1 + xy$