Question

find dy/dx
y=(cosx)/(1+sinx)
I have it figured out I think, but I just want to get it checked to see if I am approaching the problem the correct way. I will post my answer and how I got it in a minute.

Answer 1

Did you use quotient rule?

Answer 2

my answer is
\[y'=-1/(1+\sin x)\]

Answer 3

is that where it is
\[(u*dv/dx -v*du/dx)v ^{2}\]?
where the numerator is u and denominator is v?

Answer 4

[(1+sinx)(-sinx) - (cosx)(cosx)]/(1+sinx)^2
[sinx/1+sinx] - [(cosx)^2] / [(1+sinx)^2]
is what I got

Answer 5

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

Answer 6

@VeritasVosLiberabit
when I factored out, I got
(-(sin x +sin^2 x +cos^2 x))/(1+sin x)^2

Answer 7

we have different signs I could be wrong have to check again

Answer 8

timmtamm that's it. didn't see you factor out the negative

Answer 9

the numerator was originally
-sin x -sin^2 x -cos ^2 x
but I took out the negative to get factor it out to
-(sin x +sin^2 x +cos^2 x)
to then simplify it out to
-(sin x +1) as the numerator

Answer 10

and then
-(sin x +1)/(1+sin x)^2
can then be simplified to
-1/(1+sin x)
right?