Question

IS ANYONE ON HERE??

Answer 1

@johnweldon1993 please helpp

Answer 2

So you have a choice here, we can either do the quotient rule...or the product rule...which do you prefer?

Answer 3

product rule

Answer 4

Alright so since that's the case...we need to write this now as

\[\large (1 + sin(x))(1 - cos(x))^{-1}\]

now we just use the product rule...that states that

\[\large f'g + fg'\]

let \(\large (1 + sin(x))\) = f
and let
\(\large (1 - cos(x))\) = g

Answer 5

g=(1-cos(x))^(-1)?

Answer 6

so i was right(:

Answer 7

^thank you for catching that mistake @freckles :)

Answer 8

I'm not sure about the answer 2csc(x)sec(x)
did you apply the product rule to the rewrite of the problem that @johnweldon1993 has provided

Answer 9

oh wait one sec

Answer 10

would it be 1?

Answer 11

\[\frac{d}{dx}(1+\sin(x))=?\]

Answer 12

\[\frac{d}{dx}(1-\cos(x))^{-1}=?\]

Answer 13

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

Answer 14

i keep getting the same answer

Answer 15

-2csc(x)sec(x)

Answer 16

@Loser66 am i right?

Answer 17

its been 35 minutes lol..

Answer 18

@satellite73 HELPPPPP

Answer 19

http://www.wolframalpha.com/input/?i=%28%281%2Bsin+x%29%2F%281-cosx%29%29%27