IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: SGVaibhav on November 10, 2009, 02:57:51 pm

Title: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
Post by: SGVaibhav on November 10, 2009, 02:57:51 pm
[WARNING: VERY ANNOYING STUFF INSIDE ---> ENTER AT UR OWN RISK]



need to explain to full class, that how do we derive
the derivatives of arcsin, arccos, arctan, arcsec, (need to explain to full class, that how do we derive and get this values).
need to submit with 4-5 pages of handouts.
if anyone can help in any way, please do,
right now, i am reading it in wiki, (but am pretty confused ???) (lol :D still maths is fun :D)
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
Post by: slvri on November 10, 2009, 03:02:55 pm
y dont u post this in the a level thread instead?
and inverse trig derivatives...dont know how to do them.......sory.....i bet astar knows
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
Post by: Zain-Xa on November 10, 2009, 03:03:11 pm
Yeah ill move it there :)
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time
Post by: Tyserius on November 10, 2009, 03:07:07 pm
Gimme a moment, I'll give you an example and you can try proving it on other inverse trigo functions :) Will post it after this.
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
Post by: SGVaibhav on November 10, 2009, 03:10:28 pm
well, i was reading on wiki, how to derive the inverse sine function.
and i came across this
they have reached till here

http://upload.wikimedia.org/math/a/b/3/ab3d5da927e35f4370bb0b9c0b483189.png

and now they wrote this thing
http://upload.wikimedia.org/math/8/8/a/88aedc93821b18cfc22c28cf970cd133.png

from where does this come ???
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time
Post by: Tyserius on November 10, 2009, 03:16:52 pm
Is this what you need?
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
Post by: SGVaibhav on November 10, 2009, 03:20:01 pm
reading
looks effective so far
excited
will try that on cos and will tell if it works
Be right back reading :D
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
Post by: SGVaibhav on November 10, 2009, 03:27:34 pm
how do we know that we have to find the adjacent side ???
and how do we get (root)1-x^2, that is confusing me
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
Post by: SGVaibhav on November 10, 2009, 03:28:47 pm
im not getting it  :'(
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time
Post by: Tyserius on November 10, 2009, 03:31:04 pm
'Cause sin y = x which from that you know the opposite side is x and the hypotenuse is 1. What about the adjacent? Imagine a right-angled triangle with one of its acute angles as y.
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
Post by: SGVaibhav on November 10, 2009, 03:32:24 pm
how is the hyp 1
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time
Post by: Tyserius on November 10, 2009, 03:33:16 pm
sin y = x meaning sin y = x/1 haha. And sin = opposite/hyp
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
Post by: SGVaibhav on November 10, 2009, 03:34:12 pm
lol thanks,
let me try to continue
so stupid  >:(
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time
Post by: Tyserius on November 10, 2009, 03:39:25 pm
Haha, no prob :D You might have a problem with sec and cosec though, but do try it xD
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
Post by: SGVaibhav on November 10, 2009, 03:43:52 pm
lool

i was typing, BRAINS r like parachutes, plz open
and was going to type a doubt.
but got the answer

im convinced how do i get the answer, but im not sure if this method will be enough to convince others. im practicing how to convince others :D  ;D ;)
thanks for help :D
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time
Post by: Tyserius on November 10, 2009, 03:45:39 pm
It will be, trust me :D By the way, you wanna show how you showed your working? Maybe you can try include a figure of a right-angled triangle to explain the specific lengths :)
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
Post by: SGVaibhav on November 10, 2009, 03:48:15 pm
found out, how to explain

the only confusing part to explain is the adjacent part, so i will find out adj in the beginning, when i draw a triangle.
now going to COS
hope its easy :D
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time
Post by: Tyserius on November 10, 2009, 03:50:27 pm
found out, how to explain

the only confusing part to explain is the adjacent part, so i will find out adj in the beginning, when i draw a triangle.
now going to COS
hope its easy :D

Haha, I guess you can do it :D you can do it like let's say you draw a triangle ABC. Angle ABC = 90. Put angle CAB as y then you can illustrate :)

And be careful on the cosine part. Remember that its derivative is a negative sine. :)
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
Post by: SGVaibhav on November 10, 2009, 03:52:51 pm
DONE xD
first try, did not use the net also, no helps
wow :D,

now i got 2 more functions remaining
tan inverse and sec inverse
tan should not be a big prob

sec looks scary  :o ::) O0
going with Tan now
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time
Post by: Tyserius on November 10, 2009, 03:54:25 pm
Haha! You can do it! :D Once you get the hang of it you can do it :) You wanna try the following?

Find the derivative of y = sin^-1 3x^2 :D
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
Post by: SGVaibhav on November 10, 2009, 03:58:19 pm
crap crap crap
im stuck with sec, i know how to do it.
but i hope other classmates understand, because im not good at EXPLAINING algebra, because i do it mentally, specially stuff such as sec y = 1 / cos y
so that is = 1 / (adj / hyp)
crap boring to explain  :-\

let me try to finish it now  :)
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
Post by: SGVaibhav on November 10, 2009, 04:00:38 pm
got the answer

is the answer 1 / (1 + x^2) ???
lol i dont know.
w8 checking book

right :D :D :D

now the scariest part ever
the don of DONS   sec filled with modulus

doing sec inverse now
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time
Post by: Tyserius on November 10, 2009, 04:01:30 pm
crap crap crap
im stuck with sec, i know how to do it.
but i hope other classmates understand, because im not good at EXPLAINING algebra, because i do it mentally, specially stuff such as sec y = 1 / cos y
so that is = 1 / (adj / hyp)
crap boring to explain  :-\

let me try to finish it now  :)

Haha, you want me to try do it for you? :)
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
Post by: SGVaibhav on November 10, 2009, 04:03:14 pm
let me try
practice
trying
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time
Post by: Tyserius on November 10, 2009, 04:04:04 pm
let me try
practice
trying

Haha! Okay, lemme know if you're stuck xD
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
Post by: SGVaibhav on November 10, 2009, 04:08:00 pm
awww

reached

dy/dx = 1 / (x .  root (x^2 - 1) )

lol hope u got that part
now thinking how to proceed.
if i multiply the whole thing by x^2 - 1 , then the root still stays.
if i multiply the whole thing by the root of x^2 ....., then the root comes above
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
Post by: SGVaibhav on November 10, 2009, 04:12:45 pm
w8

i checked my book now, and this is the final answer,

except the modulus part exists for x.
how does the modulus come ???
need help now
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time
Post by: Tyserius on November 10, 2009, 04:16:18 pm
Hmm I saw that too. Lemme think about the modulus part :) Gimme a while.
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time
Post by: Tyserius on November 10, 2009, 04:20:12 pm
w8

i checked my book now, and this is the final answer,

except the modulus part exists for x.
how does the modulus come ???
need help now

Hmmm... I think this site can help :)

http://archives.math.utk.edu/visual.calculus/3/inverse_trig.1/5.html
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time
Post by: Tyserius on November 10, 2009, 04:48:40 pm
So, do you understand it now? xD
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
Post by: SGVaibhav on November 10, 2009, 04:54:00 pm
sorry i was away,
trying it now
this method is little different and making it little confusing
trying now

is there any such rule that sec x tan x should be > 0 or something
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
Post by: SGVaibhav on November 10, 2009, 05:11:45 pm
ok got that part

confusion confusion confusion
in that site....

when they find the derivative of sec inverse for the range (pi/2) till pi.
they get a negative sign in the denomintor before the square root
how is that now
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time
Post by: Tyserius on November 10, 2009, 05:39:00 pm
ok got that part

confusion confusion confusion
in that site....

when they find the derivative of sec inverse for the range (pi/2) till pi.
they get a negative sign in the denomintor before the square root
how is that now

'Cause for cos x where pi/2 < x < pi, it is negative haha.
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
Post by: SGVaibhav on November 10, 2009, 06:12:38 pm
this was truly confusing  >:( :(
it will be very hard to explain.

but why do we need to care to put modulus
if there is no modulus, then the answer comes negatives (LET IT STAY NEGATIVE IF IT IS NEGATIVE  >:()

annoying  :-\
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
Post by: SGVaibhav on November 10, 2009, 06:18:58 pm
yep one more thing
when sec goes negative, tan also goes negative
so negative x negative = positive

im going off pc now probably
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time
Post by: astarmathsandphysics on November 10, 2009, 06:21:24 pm
I differenthated them all but i posted in the ib board
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
Post by: SGVaibhav on November 10, 2009, 06:25:22 pm
nice
lol i edited the main page
haha
warning

thanks for help :D

will never forget this now.
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time
Post by: Tyserius on November 10, 2009, 06:35:51 pm
yep one more thing
when sec goes negative, tan also goes negative
so negative x negative = positive

im going off pc now probably

Yep yep! Haha that explains the modulus :) Anyway, all the best for the presentation xD
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
Post by: SGVaibhav on November 10, 2009, 06:49:32 pm
loooooooool
haha, that -ve x -ve = +ve was a doubt
but when u said it explains the modulus, i realised that the doubt is the solution ???
cya tom
thanks for help
bye
Take care
goodnight
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time
Post by: Tyserius on November 10, 2009, 07:18:10 pm
loooooooool
haha, that -ve x -ve = +ve was a doubt
but when u said it explains the modulus, i realised that the doubt is the solution ???
cya tom
thanks for help
bye
Take care
goodnight

Haha! Take care and goodnight :)
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
Post by: astarmathsandphysics on November 10, 2009, 10:21:22 pm
https://studentforums.biz/index.php/topic,4796.msg150350.html#msg150350

inverse trig functions are differentiated here.
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
Post by: SGVaibhav on November 11, 2009, 09:20:47 am
 :(

sir wanted it written in a paper, while i thought it had to be explained to the full class.
now i will open word 2007, write down steps, print it and give it to him
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time
Post by: Tyserius on November 11, 2009, 09:25:29 am
:(

sir wanted it written in a paper, while i thought it had to be explained to the full class.
now i will open word 2007, write down steps, print it and give it to him

Lol. But did you get everything now? :) I guess you've done it all :D
Title: Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
Post by: SGVaibhav on November 11, 2009, 01:37:50 pm
yeah i got it