Question

Differentiate the trigonometric function

1-sinx/cosx

Answer 1

(1-sin(x))/cos(x) ?

Answer 2

yep thats right

Answer 3

you can avoid using quotient rule

Answer 4

i used the quotient rule but my answer didnt come out right

Answer 5

1/cos(x) - sin(x)/cos(x)
D'(sec(x) - tan(x))
sec(x)tan(x) - sec(x)^2

Answer 6

\[\frac{1}{cos(x)+1}\] yes?

Answer 7

yeah thats right can u explain

Answer 8

so many forms this can take

Answer 9

let me write it out. may take a minute

Answer 10

k thanks

Answer 11

wow i had it and i lost it. let me try this

Answer 12

just use the quotient rule.. the denominator is
\[cos^2(x)\] the numerator is
\[-cos^2(x) + sin(x)(1-sin(x))\]

Answer 13

the numerator simplifies to \[-cos^2(x) + sin(x)-sin^2(x)=-cos^2(x)-sin^2(x)+sin(x)\]

Answer 14

=\[sin(x)-1\]

Answer 15

so we get
\[\frac{sin(x)-1}{cos^2(x)}\]

Answer 16

oh i see i got the same except my sin^2(x) was positive aswell

Answer 17

now i think we can rewrite the denominator

Answer 18

no it should be negative i think.

Answer 19

ok im gona have to check my working

Answer 20

put \[f(x)=1-sin(x)\] so
\[f'(x)=-cos(x)\]
\[g(x)=cos(x)\]
\[g'(x)=-sin(x)\]
numerator is
\[gf'-fg'=cos(x)(-cos(x))-(1-sin(x))(-sin(x))\]

Answer 21

=\[-cos^2(x)+sin(x)(1-sin(x))\]
\[=-cos^2(x)+sin(x)-sin^2(x)=sin(x)-1\]

Answer 22

so at least we got that far. deminator is \[cos^2(x)=1-sin^2(x)=(1-sin(x))(1+sin(x))\]

Answer 23

damn now i am getting
\[\frac{-1}{1+sin(x)}\]

Answer 24

maybe i screwed up somewhere

Answer 25

no thats right

Answer 26

i looked at the wrong answer before lol

Answer 27

oh lord. what is the right answer in the book?

Answer 28

yes it was lol

Answer 29

what is it?

Answer 30

-1/1+sin(x)

Answer 31

whew!

Answer 32

done finito halas stick a fork in it

Answer 33

you still got it

Answer 34

:)

Answer 35

:)