
09232008, 09:41 PM

#1

Golden Master
Join Date: May 2006
Posts: 7,904

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
__________________
__________________
“Adults are just obsolete children and the hell with them.”  Dr. Suess



09232008, 09:44 PM

#2

Guru
Join Date: Jan 2008
Location: U.S.
Posts: 7,841

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?
__________________
__________________



09232008, 09:45 PM

#3

(╯°□°）╯︵ ┻━┻
Join Date: Dec 2006
Location: United States
Posts: 5,957

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.
__________________
Desktop Cooler Master NV690/AMD Phenom II X4 965 BE/MSI 880GE45 Motherboard/2 x 2GB DDR3 G.Skill Ripjaws X/Sapphire HD 5870 1GB GDDR5/CORSAIR CMPSU550VX/Seagate 7200.12 500GB (OS), Samsung F3 1TB (Storage)/Sony DVD Burner/Windows 7 Professional 64bit



09232008, 09:49 PM

#4

Golden Master
Join Date: May 2006
Posts: 7,904

Re: Anyone good in Calculus here? Derivatives?
Umm... By limits I guess... Limit as H aproaches 0... I think...
__________________
“Adults are just obsolete children and the hell with them.”  Dr. Suess



09232008, 09:50 PM

#5

Guru
Join Date: Jan 2008
Location: U.S.
Posts: 7,841

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

Golden Master
Join Date: May 2006
Posts: 7,904

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
__________________
“Adults are just obsolete children and the hell with them.”  Dr. Suess



09232008, 10:26 PM

#7

Guru
Join Date: Jan 2008
Location: U.S.
Posts: 7,841

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

Golden Master
Join Date: Mar 2006
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.
__________________
__________________
AMD Athlon 64X2 4200+, 2 GB ram, 2 TB HDD, Ubuntu+SAMBA ♫ Lenovo T410, i5@ 2.4 GHz, 4 GB ram, 7




Thread Tools 
Search this Thread 


Display Modes 
Linear Mode

Posting Rules

You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts
HTML code is Off




» Recent Threads 
















