Question

find the derivative of y=ln|secx + tanx|

Answer 1

http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4e7f8f020b8b9dc9cf29a637

Answer 2

\[y'=\frac{(\sec(x)+\tan(x))'}{\sec(x)+\tan(x)}\]

Answer 3

\[(\sec x \tan x+\sec ^{2}x)/(\sec x+\tan x)\]

Answer 4

\[=\sec(x) , \sec(x) \neq -\tan(x)\]

Answer 5

should mine be more simplified?

Answer 6

i was just saying we can simplify if sec(x)
if sec(x) does not equal -tan(x)

Answer 7

(tanx+secx)/(1+sinx)?

Answer 8

\[\frac{\sec(x)(\tan(x)+\sec(x))}{\sec(x)+\tan(x)}=\sec(x) if \sec(x) \neq -\tan(x)\]

Answer 9

oh, I see