find dy/dx
y=log((1-x2)/(1+x2))
y = log( ( 1 – x2) / ( 1 + x2) )
Or y = log( 1 – x2) – log( 1 + x2) [log(a/b) = loga – logb ]
Now differentiating , dy/dx = ( 1/ 1 - x2) ( d/dx( 1 - x2) ) - ( 1/ 1 + x2)( d/dx( 1 + x2)) [Chain rule]
or dy/dx = (1/1-x2)( -2x) - ( 1/1 + x2)(2x)
= -2x( 1 + x2 + 1 – x2 ) / ( 1 – x2) ( 1 + x2 )
= -4x/( 1 – x4 )
Hope that helps.
Or y = log( 1 – x2) – log( 1 + x2) [log(a/b) = loga – logb ]
Now differentiating , dy/dx = ( 1/ 1 - x2) ( d/dx( 1 - x2) ) - ( 1/ 1 + x2)( d/dx( 1 + x2)) [Chain rule]
or dy/dx = (1/1-x2)( -2x) - ( 1/1 + x2)(2x)
= -2x( 1 + x2 + 1 – x2 ) / ( 1 – x2) ( 1 + x2 )
= -4x/( 1 – x4 )
Hope that helps.