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Author Topic: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)  (Read 4059 times)

Offline Tyserius

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Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time
« Reply #15 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 :)
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Offline SGVaibhav

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Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
« Reply #16 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
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Offline Tyserius

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Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time
« Reply #17 on: November 10, 2009, 03:50:27 pm »
Quote from: 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

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. :)
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Offline SGVaibhav

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Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
« Reply #18 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
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Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time
« Reply #19 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
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Offline SGVaibhav

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Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
« Reply #20 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  :)
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Offline SGVaibhav

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Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
« Reply #21 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
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Offline Tyserius

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Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time
« Reply #22 on: November 10, 2009, 04:01:30 pm »
Quote from: 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  :)

Haha, you want me to try do it for you? :)
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Offline SGVaibhav

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Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
« Reply #23 on: November 10, 2009, 04:03:14 pm »
let me try
practice
trying
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Offline Tyserius

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Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time
« Reply #24 on: November 10, 2009, 04:04:04 pm »
Quote from: sgvaibhav on November 10, 2009, 04:03:14 pm
let me try
practice
trying

Haha! Okay, lemme know if you're stuck xD
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Offline SGVaibhav

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Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
« Reply #25 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
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Offline SGVaibhav

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Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time :D)
« Reply #26 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
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Offline Tyserius

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Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time
« Reply #27 on: November 10, 2009, 04:16:18 pm »
Hmm I saw that too. Lemme think about the modulus part :) Gimme a while.
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Offline Tyserius

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Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time
« Reply #28 on: November 10, 2009, 04:20:12 pm »
Quote from: 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

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

http://archives.math.utk.edu/visual.calculus/3/inverse_trig.1/5.html
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Offline Tyserius

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Re: help needed in Deriving Inverse Trigonmetric Functions (Not urgent this time
« Reply #29 on: November 10, 2009, 04:48:40 pm »
So, do you understand it now? xD
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