Computer Forums Anyone good in Calculus here? Derivatives?

 09-23-2008, 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)/(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 __________________ __________________ “Adults are just obsolete children and the hell with them.” - Dr. Suess
 09-23-2008, 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? __________________ __________________
 09-23-2008, 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 880G-E45 Motherboard/2 x 2GB DDR3 G.Skill Ripjaws X/Sapphire HD 5870 1GB GDDR5/CORSAIR CMPSU-550VX/Seagate 7200.12 500GB (OS), Samsung F3 1TB (Storage)/Sony DVD Burner/Windows 7 Professional 64-bit
 09-23-2008, 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
 09-23-2008, 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))(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) __________________
 09-23-2008, 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= 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 __________________ “Adults are just obsolete children and the hell with them.” - Dr. Suess
 09-23-2008, 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(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. __________________
 09-24-2008, 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