$\displaystyle Cos x= e^y $

0<x<pi/2

What is dy/dx in terms of x?

Thanks!

Printable View

- Jan 17th 2008, 03:43 PMLinnusFinding the derivative
$\displaystyle Cos x= e^y $

0<x<pi/2

What is dy/dx in terms of x?

Thanks! - Jan 17th 2008, 04:06 PMJhevon
- Jan 17th 2008, 04:17 PMLinnus
ah, thanks, i can't believe I overlooked that