Question

find dy/dx , y=sin(1+x²/1-x²)​

Answer 1

$\boxed{\textbf{\large{Step-by-step explanation:}}}$

y = sin ( (1 + x²)/(1 - x²) )

▶Differentiate the function wrt x

dy / dx = d/dx ( sin ( ( 1 + x² ) / ( 1 - x² )))

dy / dx = ( cos ( 1+ x² ) / ( 1 - x² ) )

x( d/dx ( (1 + x² ) / ( 1 - x² ))

here use the rule ,

if y = u / v

then ,

dy / dx = [ v x ( du / dx) - u x ( dv / dx) ] / v² to differentiate ( (1 + x² ) / ( 1 - x² ))

= [cos ((1 + x² ) / ( 1 - x² ))x(( 1 - x² )x( 0 + 2x))-((1 + x²)x(0 - 2x))] / (1 - x² )²

= [( cos ((1 + x² ) / ( 1 - x² )))x( 2x (1 - x² ) ) - ( - 2x) ( 1 + x² ) ] / ( 1 - x² )²

= [ ( cos ((1 + x² ) / ( 1 - x² )))x( 2x (1 - x² ) + 2x ( 1 + x² )]/(( 1 - x² )( 1 - x² )

= [(cos ((1 + x² ) / ( 1 - x² ))) x ( 2x ((1 - x² ) + ( 1 + x² ))]/( 1 - x² )( 1 - x²)

= [( cos ((1 + x² )/(1 - x² ))) x ( 2x ( 1-x² + 1 + x²))]/( 1 - x² )²

= [(cos ((1 + x² )/( 1 -x²))x

(2x(1+1 ) ] / (1-x²)²

= [( cos (( 1 + x² ) / ( 1 - x² )) x (4x) ]/( 1 - x² )²

= [4x (cos (( 1 + x² ) / (1 - x² ))) ]/ (1 - x²)²

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

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Answer 2

sᴛᴇᴘ ʙʏ sᴛᴇᴘ sᴏʟᴜᴛɪᴏɴ:

concepts used:

quotient rule and chain rule .

sᴏʟᴜᴛɪᴏɴ:

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

use chain rule

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

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

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

i hope it helps you ❤️

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