|
|
now is the question
y = sinh ^(2x) /1 +cosh^(2x)
or
y= sinh^2 x/(1+cosh^2 x)
in first case take log on both side
log y =2x log sinh -log(1+cosh^2x)
now differentiate
dy /dx= y( 2logsinh - {[1/1+cosh^2x] cosh^2x log cosh *2
correct me if i m wrong
but surely we have to take log
rate me if i m right
|
Moderation Team
![[Post New]](/templates/default/images/icon_minipost_new.gif){kind=icon}
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
68
