Question

Take the derivative of
y=(1-cosx)/(sin^2x)

Answer 1

So far I got this, and I'm not sure if the first is correct or not
\[y=\frac{ -(-sinx)(\sin^2x)-(1-cosx)2sinx(cosx)}{ (sinx)^4 }\]
???

Answer 2

I used the quotient rule and the chain rule

Answer 3

That is looks good

Answer 4

I think I can make your problem simpler though
\[y=\frac{1-\cos(x)}{1-\cos^2(x)}=\frac{1-\cos(x)}{(1-\cos(x))(1+\cos(x))}=\frac{1}{1+\cos(x)}=(1+\cos(x))^{-1}\]
if you think that is simpler

Answer 5

and then differentiate but what you have is good so far

Answer 6

http://prntscr.com/73jtkp sorry to interrupt
hmm

Answer 7

My teacher wants us to use the original question... I got stuck in simplfying it more. I tried common factoring the sinx out and then I got stuck...

Answer 8

\[y'=\frac{\sin^3(x)-(1-\cos(x))2\sin(x)\cos(x)}{\sin(x)(\sin(x))^3} \\ y'=\frac{\sin(x)[\sin^2(x)-(1-\cos(x))2\cos(x)]}{\sin(x)(\sin(x))^3}\]
so if I write this
does this lose you?

Answer 9

Okay, I got that

Answer 10

then we cancel out the common factor sin(x) on top and bottom
(and yes trig everywhere tonight @Nnesha )

Answer 11

Yes

Answer 12

\[y'=\frac{\sin^2(x)+(\cos(x)-1)2 \cos(x)}{(\sin(x))^3} \\ y'=\frac{\sin^2(x)+2\cos^2(x)-2\cos(x)}{\sin^3(x)} \\ y'=\frac{\sin^2(x)+\cos^2(x)+\cos^2(x)-2\cos(x)}{\sin^3(x)}\]
can you think of any identity you can apply to the numerator right now?

Answer 13

sin^2x+cos^2x=1 ?

Answer 14

yea

Answer 15

\[y'=\frac{\cos^2(x)-2\cos(x)+1}{\sin^3(x)}\]
you can factor the numerator

Answer 16

Oh I got it now! Thank you so much! :D
I forgot to use my identities.

Answer 17

oh is that the way their final answer looks ?

Answer 18

It's more simplify:
\[y'=\frac{ (cosx-1)^2 }{ \sin^3x }\]

Answer 19

ok cool we can actually "simplify it further "

\[y'=\frac{(\cos(x)-1)^2}{\sin^2(x) \sin(x)} =\frac{(\cos(x)-1)^2}{(1-\cos^2(x))\sin(x)} \\ y'=\frac{(\cos(x)-1)(\cos(x)-1)}{(1-\cos(x))(1+\cos(x))\sin(x)} \\ y'=\frac{-(\cos(x)-1)}{(1+\cos(x) )\sin(x)} \\ y'=\frac{1-\cos(x)}{(1+\cos(x))\sin(x)} \cdot \frac{1+\cos(x)}{1+\cos(x)} \\ y'=\frac{1-\cos^2(x)}{(1+\cos(x))^2(\sin(x)}=\frac{\sin^2(x)}{(1+\cos(x))^2\sin(x)} \\ y'=\frac{\sin(x)}{(1+\cos(x))^2}\]
depending on what "more simplified" means to you :p

Answer 20

the only thing you have have to be careful about is the domain thing

Answer 21

my y' shouldn't exist at x=npi
where n is an integer

Answer 22

Wow! I didn't realized that you could actually simplified even more! Thank you for showing me that! :)

Answer 23

no problem
that is also what you get when you differentiate the simplified form of your function (though when you simplify functions you may be messing with the domain of the original function\[y=(1+\cos(x))^{-1} \\ y=(-1)(1+\cos(x))^{-2}(-\sin(x))=\frac{\sin(x)}{(1+\cos(x))^2}\])

Answer 24

:)

Answer 25

anyways peace
and have fun