Question

whats the derivative of f(x)= 1+2cosx/sinx

Answer 1

number five

Answer 2

cosx/sinx is cotx
so you can differentiate that
if you do not know how
then you could differentiate by doing
vdu-udv/v²

this means
the function V multiplied by the derivative of U - function U multiplied by derivative of V
all divided by function V squared

Answer 3

Is it?\[f(x)=\frac{ 1+2\cos(x) }{ \sin(x)}\]

Answer 4

@sandy524

Answer 5

im not sure but im scared to put that answer in

Answer 6

@Zekarias im pretty sure that is his question so you should talk him through solving the one you wrote

Answer 7

that is what im trying to find the derivative of yahoo

Answer 8

i mean zekarias

Answer 9

please help :(

Answer 10

Can you apply the quotient rule?

Answer 11

idk

Answer 12

im sure the answer previously said was right but it has to be in rational form

Answer 13

in order to get one of the answer choices

Answer 14

I did not post any answer yet.

Answer 15

ok

Answer 16

help please

Answer 17

are u still there

Answer 18

amorfide can u help

Answer 19

\[y=\frac{ 1+2\cos(x) }{ \sin(x) }\]
\[y'=\frac{ (1+2\cos(x))^{'}\sin(x)-(\sin(x))^{'}(1+2\cos(x)) }{ \sin^{2}(x) }\]
\[y^{'}=\frac{ (-2\sin(x))\sin(x)-(\cos(x))(1+2\cos(x)) }{ \sin^{2}(x) }\]
\[y^{'}=\frac{ -2\sin^{2}(x))-\cos(x)-2\cos^{2}(x)) }{ \sin^{2}(x) }\]
\[y^{'}=\frac{ -2(\sin^{2}(x)+\cos^{2}(x))-\cos(x) }{ \sin^{2}(x) }\]
\[y^{'}=\frac{ -2-\cos(x) }{ \sin^{2}(x) }\]

Answer 20

yes that is the question what is the answer

Answer 21

nvm thanks

Answer 22

that again not one of the choices

Answer 23

\[y^{'}=\frac{ -2-\cos(x) }{ \sin^{2}(x) }=-\frac{ 2+\cos(x) }{ \sin^{2}(x) }\]

Answer 24

Thanks!