
09232008, 09:41 PM

#1

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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)/(1cscX)
Answer is:
(2cscXcotX)/((1cscX)^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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09232008, 09:44 PM

#2

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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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09232008, 09:45 PM

#3

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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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09232008, 09:49 PM

#4

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Re: Anyone good in Calculus here? Derivatives?
Umm... By limits I guess... Limit as H aproaches 0... I think...
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09232008, 09:50 PM

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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))(1cscX)
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)



09232008, 09:52 PM

#6

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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= 1cscX
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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09232008, 10:26 PM

#7

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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(1cscX)^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.



09242008, 06:15 PM

#8

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Posts: 5,574

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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