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Old 09-23-2008, 10:41 PM   #1
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Default Anyone good in Calculus here? Derivatives?

So I am totally lost... I need some help. Thought I might ask here...
Problem:
Find the derivative of the following:

(1+cscX)/(1-cscX)

Answer is:

(-2cscXcotX)/((1-cscX)^2)

Can someone explain how I get to that answer? I am lost... That's just one problem. But I don't understand how to get there.

The d/dX of cscX = -cscXcotX
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Old 09-23-2008, 10:44 PM   #2
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Default Re: Anyone good in Calculus here? Derivatives?

Might be a little hard to explain through the internet, but I'll give it a shot. Are you solving by definition of derivative (limits) or using rules?
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Old 09-23-2008, 10:45 PM   #3
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Default Re: Anyone good in Calculus here? Derivatives?

haha, i was about to post saying that im pretty sure JogaBonito1502 would be able to help you out.
but it looks like he beat me to the punch.
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Old 09-23-2008, 10:49 PM   #4
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Default Re: Anyone good in Calculus here? Derivatives?

Umm... By limits I guess... Limit as H aproaches 0... I think...
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Old 09-23-2008, 10:50 PM   #5
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Default Re: Anyone good in Calculus here? Derivatives?

Oh geez. This is going to be a long one. Here I go:

There's 2 ways to do by limits. As x -> c and as delta x (for lack of sign I'll use #) -> x. So with signs as #x -> x. I'll do the #x method because I find it easier to simplify. It can be even easier once you start using Power, Product, Quotient, and Chain rules.

So:

lim ........................... 1 + csc(#X + X) ................... 1 + cscX
#X->0 ...................... ---------------- ..... - ........ -----------
................................1 - csc(#X + X) ...................... 1 - cscX
...............................-------------------------------------------
.................................................. .........#X

lim.................................... (1 - cscX)(1 + csc(#X+X)) - (1 - csc(#X+X))(1 + cscX)....................... All that I've done here is common denominators
#X->0.............................. -------------------------------------------------------................................... followed by simplification.
.................................................. .......#X(1 - csc(#X+X))(1 - cscX)


lim....................................1 + csc(#X+X) - cscX - (cscX)(csc(#X+X)) - 1 - cscX + csc(#X+X) + (cscX)(csc(#X+X))
#X->0..............................------------------------------------------------------------------------------------------................Distribute.
.................................................. .......................#X(1 - csc(#X+X))(1 - cscX)............................................. .........I've already taken care of distributing the -1.

lim............................................... ...........2csc(#X+X) - 2cscX
#X->0.............................................. ..----------------------------.............Simplify
.................................................. .......#X(1 - csc(#X+X))(1 - cscX)

lim............................................... .....2(csc(#X+X) - cscX)
#X->0............................................-----------------------.............Simplify more (it's always good in a hard problem like this; I just use rules)
.................................................. .....#X(1 - csc(#X+X))(1-cscX)


Taken out of context to simplify:

...................................1.............. .................................1................ ...........
...................------------------------......-........--------------------------........ Rule for simplifying sin(A+B)
...................sin#X cosX + cos#X sinX..............................sinX............ ............

............................sinx - sin#XcosX - cos#XsinX
...................--------------------------------------
......................sinXsin#XcosX + cos#X(sinX)^2

Alright dude, screw this we're doing rules, as proving trig stuff is really nasty. Even my calc teacher said so. After tons of simplification you should get:

lim............................................... .....................2cscX
#X->................................................. ............-----------.........................The #X on the side needs to be cancelled off so you get:
.................................................. .......#X(1 - csc(#X+X))(1 - cscX)

lim............................................... .....................2cscX
#X->................................................. ............-----------.........................Plug 0 for #X
.................................................. ...... (1 - csc(#X+X))(1 - cscX)

lim............................................... .....................2cscX
#X->................................................. ............-----------.........................And there's your answer.
.................................................. ............(1 - cscX)(1 - cscX)
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Old 09-23-2008, 10:52 PM   #6
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Default Re: Anyone good in Calculus here? Derivatives?

Ok, might be rules.

Thinking about it again. The formal definition is F(X+H)-F(X)
-------------
H
Correct? But with the qoutent rule, I would take u1v - uv1
-----------
v^2
u= 1+cscX
v= 1-cscX
1= take derivative

maybe that's it... Or is there simplifying I can do? I know there are rules I can apply to csc sec and cot, but I cant remember them... secx = 1/sinx correct? cotx = 1/tanx? and cscx = 1/cosx
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Old 09-23-2008, 11:26 PM   #7
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Default Re: Anyone good in Calculus here? Derivatives?

K man, I started with definition I'll keep going. I'll show rule after k?


Through rules:

(1 + cscX)
----------
(1 - cscX)


(1 + cscX)(1 - cscX)^-1

-cscXcotX(1-cscX)^-1 + (1+cscX)(-1)(-cscXcotX)^-2

-cscXcotX.........1 + cscX
-----------..+..-----------
1 - cscX...........cscXcotX

Do your simplification, etc. I used product rule because I hate quotient rule. Here's quotient:

d.......................................'(1 + cscX)(1 - cscX) - (1 + cscX)'(1 - cscX)
--.... (1 + cscX)/(1 - cscX) = -------------------------------------------
dx................................................ ...........(1 - cscX)^2

General Form:

d...............('a)(b) - (a)('b)
-- (a/b) = ----------------
dx....................(b)^2

edit: Remind me to never again attempt to help with calculus over the internet. Also, go to calcchat.com and they give you answers to odds questions along with explanations.

edit2: Sorry I missed your explanation of quotient rule. I was too busy trying to get this. You didn't need my help after all =o. Had it in you from the beginning.
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Old 09-24-2008, 07:15 PM   #8
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Default Re: Anyone good in Calculus here? Derivatives?

I'm not bad at derivatives (even though we haven't actually done them in this class we did the derivative of quadratics last year) and I wouldn't ahve a clue as to what to do.


So I give you mathway.
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