Question

differentiate√1-x^2/√1+x^2​

Answer 1

Step-by-step explanation:

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

Differentiate the function wrt x

$\frac{dy}{dx} = \frac{d}{dx} ( \frac{ \sqrt{1 - {x}^{2} } }{ \sqrt{1 + {x}^{2} } } )$

$\frac{dy}{dx} = ( \frac{ (\sqrt{1 + {x}^{2} } \times \frac{1}{ \sqrt{1 - {x}^{2} } } \times (0 - 2x)) - {} {} {( \sqrt{1 -{x}^{2}} ) \frac{1}{( \sqrt{1 + {x}^{2} ) } } \times (0 + 2x)} )} {( { \sqrt{1 + {x}^{2} }) }^{2} }$

$= \frac{ \frac{ - 2x \sqrt{1 + {x}^{2} } }{2 \sqrt{1 - {x}^{2} } } - \frac{2x \sqrt{1 - {x}^{2} } }{2 \sqrt{1 + {x}^{2} } } }{ ({ \sqrt{(1 + {x}^{2} } )}^{2} }$

$= \frac{( \frac{1}{( \sqrt{1 + {x}^{2} } ) \times \sqrt{1 - {x}^{2} } )}) \times ( - x \sqrt{1 + {x}^{2} } ( \sqrt{1 + {x}^{2}) } - x \sqrt{1 - {x}^{2} } ( \sqrt{1 - {x}^{2}))}) }{( { \sqrt{(1 + {x}^{2} }) }^{2} }$

$= \frac{ - {( \sqrt{1 + {x}^{2} } ) }^{2} - ( { \sqrt{1 - {x}^{2} )} }^{2} }{ \sqrt{(1 - {x}^{4}) } ( { \sqrt{1 + {x}^{2} )} }^{2} }$

$= \frac{ - 1 - {x}^{2} - 1 + {x}^{2} }{ \sqrt{(1 - {x}^{4} )} (1 + {x}^{2} )}$

$= \frac{ - 2}{ \sqrt{(1 - {x}^{4} )} (1 + {x}^{2} )}$

$\frac{dy}{dx} = \frac{ - 2}{ \sqrt{(1 - {x}^{4} )} (1 + {x}^{2} )}$

Answer 2

Answer:-

Given in the attachment