How do you differentiate $y=(2e^x-1)/(5e^x+9)$ ?

Question

1594 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-13T04:13:58

$(23e^x)/(5e^x+9)^2$

$y=(2e^x-1)/(5e^x+9)=u/v$

$(dy)/(dx)=(vu'-uv')/v^2$

$u'=2e^x$

$v'=5e^x$

$(dy)/(dx)=(2e^x(5e^x+9)-5e^x(2e^x-1))/(5e^x+9)^2$
$=(10e^(2x)+18e^x-10e^(2x)+5e^x)/(5e^x+9)^2$
$=(23e^x)/(5e^x+9)^2$

How do you differentiate y=(2e^x-1)/(5e^x+9)y=(2e^x-1)/(5e^x+9)?