Yet another calculus problem..last one I promise!

[TheDude]

I'm so glad the members of SF know calculus better than I do lol. Here is the function to derive: y = ( (1+sin x) / (1-cos x) )^2 This is what I get, but I cannot figure out how to simplify it (hopefully my calculus is correct) 2*( (1+sin x)/(1-cos x) ) * [(cos x)(1-cos x) - {(sin x)(1+sin x)} / (1-cos x)^2 ] For clarification, in the big trig term, (cos x)(1-cos x) - {(sin x)(1+sin x)} is all over (1-cos x)^2. (If anyone needs this to be clarified, I'll scan a picture of the problem). My teacher remarked that this problem simplifies really nicely, which leads me to believe that some trig identity must come into play, but I cannot see any that would help, or at least I can't figure out how to rearrange this to allow the identity to become evident. Thanks! I can't believe how helpful this forum has been in more ways than one!!

 

[LA Guy]

I'm gonna smack you upside the head. How about separating first, simplifying, and then doing the derivative. Okay... go.

Edit: To clarify, separate out the terms in y using the laws of multiplication, figure out the identities, then derive by parts.

 

[TheDude]

umm yeah, that is the problem, what exactly do you mean?

 

[LA Guy]

Originally Posted by TheDude

umm yeah, that is the problem, what exactly do you mean?

I mean, don't blindly do the derivative first. It is often easier to figure out the identities first.

 

[rach2jlc]

Styleforum: come for the clothes, stay for the Calculus.

(This has to be one of the few places on planet earth where members can, in one stop, receive criticism on the fit of their bespoke shirts, use a word like "cordwainer" without feeling guity, read hilarious political debates, see shoes made from Bullfrog hide, and get answers to complicated math questions.)

 

[TheDude]

damn, the only thing I have come up with is chaning it to this:

[ (1+sin x)/ 1 * (1) / (1-cos x) ] ^2 but I'm not seeing how that helps...

 

[Go Surface]

Shouldn't you be using this as an excuse to find a hot tutor?

 

[tiecollector]

Originally Posted by GoSurface

Shouldn't you be using this as an excuse to find a hot tutor?

he'll be looking for quite a while.

 

[TheDude]

so yeah, anyone want to actually help me out? I realize how obnoxious it is to be asking for calculus help on a style forum, but I'm just trying to figure out how to do some good ol' math

 

[LA Guy]

*******. Just do it the long way that you've started already, use your cotan and cosec as well as your tan identities, and simplify slowly but surely.

 

[raley]

There are multiple calculators online that can do derivatives for you step by step, FYI.

 

[Homme]

After a ton of gathering like terms, and simplifying (mainly using the results (sinx)^2 + (cosx)^2 = 1 and (cosx - sinx)^2 = 1 - 2sinxcosx); i got

2((cosx - sinx)^3) - 3((cosx-sinx)^2) - 1
--------------------------------------------------
(1 - cosx)^4

for the top line, if you let (cosx - sinx) = y; the function 2y^3 - 3y^2 - 1 does have a root; but it isn't 1/2/3/something else trivial; so i didn't bother trying to multiply out the denominator and then finding a comon factor to remove etc etc.

 

[yace]

I don't know if you know about the following two equations:

sinx = 2*(tan(x/2))/(1+(tan(x/2))^2)
cosx = (1-(tan(x/2))^2)/(1+(tan(x/2))^2)

Then 4y=(1+ctan(x/2))^4

The remaining is quite simple because (ctan(x))'=-csc(x)^2. I believe this is the simplest way to get the derivative.

 

[LA Guy]

^^^ For the win. Wow, I should tell all profs teaching first year Calculus that putting questions on a Style-forum will get people doing homework for no credit

Styleforum=teaching tool of 2007.

 

[ms244]

Originally Posted by LA Guy

^^^ For the win. Wow, I should tell all profs teaching first year Calculus that putting questions on a Style-forum will get people doing homework for no credit

Styleforum=teaching tool of 2007.
Any experts in diff (icult) equations (regular and partial) and electron microscopy? Maybe we should have a post your hw and get an answer thread?