Question

find the derivative f(x)=tan⁡x/(1+cos⁡x )
I need help not sure this is right

y^'=((1+cos⁡〖x)d/dx (Tan x)-(Tan x) d/dx(1+cos⁡〖x)〗 〗)/((1+cos⁡〖x)^2 〗 )

y^'= ((1+cos⁡〖x)(sec^2 )-(Tan x)(0-sin⁡〖x)〗 〗)/((1+cos⁡〖x)^2 〗 )

y’ =(sec^2+Tan x Sin X)/((1+cos⁡〖x)〗 ) (I do not believe this is correct)

Answer 1

this looks ok assuming the bottom reads (1+ cos x ) ^2
y^'= ((1+cos⁡〖x)(sec^2 )-(Tan x)(0-sin⁡〖x)〗 〗)/((1+cos⁡〖x)^2 〗 )

distribute the sec^2 to get
(sec^2 x + sec x + tan x sin x) /( 1+ cos x)^2

I don't see how to get from this line to your last line

Answer 2

\[my second line is : y'=(1+\cos x)(\sec^2 x)-(\tan x)(-\sin x)/(1+\cos x)^2\] do not know if I should take one of the (1+cos x) out of the bottom or not

Answer 3

\[ \frac{(1+\cos x)(\sec^2x) + \tan x \sin x}{(1+\cos x)^2} \]

yes, you could write its as
\[ \sec^2x + \frac{ \tan x \sin x}{(1+\cos x)^2} \]
or in quite a few variations
\[ \frac{\sec^2 x + \sec x + \tan x \sin x}{( 1+ \cos x)^2} \]
\[ \frac{\sec x ( 1+ \sec x + \sin^2 x)}{( 1+ \cos x)^2} \]
and so on...