Question

y = sin^{-1} (1-x^{2} / 1+x^{2}), 0 < x < 1 dy/dx ज्ञात कीजिए

Answer 1

★彡 Question ★彡

$\rm if \: y = { \sin}^{ - 1} \big( \frac{1 - {x}^{2} }{1 + {x}^{2} } \big) \: , \: 0 < x < 1 \: find \: \frac{dy}{dx}$

★彡 Answer 彡★

$\to \boxed{ \rm\frac{dy}{dx} = \frac{ - 2}{1 + {x}^{2} } } \:$

★彡 Solution 彡★

We have ,

$\rm \: y = { \sin}^{ - 1} \big( \frac{1 - {x}^{2} }{1 + {x}^{2} } \big) \: \: \\ \\ \sf \red{ differentiate \: with \: respect \: to \: x} \\ \\ \to \rm \frac{dy}{dx} = \frac{d}{dx} { \sin}^{ - 1} \big( \frac{1 - {x}^{2} }{1 + {x}^{2} } \big) \: \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \because \rm \frac{d}{dx} { \sin}^{ - 1} x = \frac{1}{ \sqrt{1 - {x}^{2} } } } \\ \\ \to \rm \frac{dy}{dx} = \frac{1}{ \sqrt{1 - { \big(\frac{1 - {x}^{2} }{1 + {x}^{2} } \big) }^{2} } } \frac{d}{dx} \frac{1 - {x}^{2} }{1 + {x}^{2} } \\ \\ \to \rm \frac{dy}{dx} = \frac{1 + {x}^{2} }{ \sqrt{ {(1 + {x}^{2}) }^{2} - { {(1 - {x}^{2} )}}^{2} } } \frac{d}{dx} \frac{1 - {x}^{2} }{1 + {x}^{2} } \\ \\ \to \rm \frac{dy}{dx} = \frac{1 + {x}^{2} }{ \sqrt{2 {x}^{2} + 2 {x}^{2} } } \frac{d}{dx} \frac{1 - {x}^{2} }{1 + {x}^{2} } \:....eq.(1) \\ \\ \sf \: for \: find \\ \\ \to \rm \frac{d}{dx} \frac{1 - {x}^{2} }{1 + {x}^{2} } \\ \\ \sf we \: use \: quotient \: rule \\ \\ \to \rm\frac{(1 + {x}^{2}) \frac{d}{dx} (1 - {x}^{2} ) - (1 - {x}^{2}) \frac{d}{dx} (1 + {x}^{2} ) }{ {(1 + {x}^{2}) }^{2} } \\ \\ \to \rm \frac{(1 + {x}^{2}) ( - 2x) - (1 - {x}^{2})( 2x)}{ {(1 + {x}^{2}) }^{2} } \\ \\ \to \rm \frac{ - 2 {x} - 2 {x}^{3} - 2x + 2 {x}^{3} }{ {(1 + {x}^{2} )}^{2} } \\ \\ \to \rm \: \frac{ - 4x}{ {(1 + {x}^{2} )}^{2} } \\ \\ \sf so \: eq.(1) \: wil \: be \:$

$\rm \frac{dy}{dx} = \frac{1 + {x}^{2} }{ \sqrt{2 {x}^{2} + 2 {x}^{2} } } \bigg(\: \frac{ - 4x}{ {(1 + {x}^{2} )}^{2} } \bigg) \\ \\ \to \rm \frac{dy}{dx} = \frac{ - 4x}{ \sqrt{4 {x}^{2}} (1 + {x}^{2} )} \\ \\ \to \rm \: \frac{dy}{dx} = \frac{ - 4x}{2x(1 + {x}^{2}) } \\ \\ \to \boxed{ \rm\frac{dy}{dx} = \frac{ - 2}{1 + {x}^{2} } }$

Answer 2

dy/dx  = - 2/(1 + x²) यदि y   = Sin⁻¹ ( 1 - x²)/(1 + x²)

Step-by-step explanation:

y   = Sin⁻¹ ( 1 - x²)/(1 + x²)

=> Siny =  ( 1 - x²)/(1 + x²)

Cosy dy/dx  =  ((1 - x²) (-1)/(1 + x²)² )2x    + (-2x)/(1 + x²)

=> Cosy dy/dx  =  (2x³ - 2x)/(1 + x²)²     - 2x /(1 + x²)

=> Cosy dy/dx  =  (2x³ - 2x  - 2x³ - 2x)/(1 + x²)²

=> Cosy dy/dx  = -4x/(1 + x²)²

Cosy = √(1 - Sin²y)   =  √(1 -  (( 1 - x²)/(1 + x²))²

=> Cosy = 2x/(1 + x²)

=>(2x/ (1 + x²) )dy/dx  = -4x/(1 + x²)²  

=> dy/dx  = - 2/(1 + x²)

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