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Posted by AiSH on 04-30-2002 03:56 PM:

calc

help!

if g(t) = csc 2t, find g''' (-pi/8)

and can tell me wha you get for each?
g'
g''
g'''

thx! whoever can do it..

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Posted by annabanana on 05-01-2002 04:20 PM:

hmmm, derivatives..... let's see.... deng, you are screwed. i got out of calculus a long time ago. i just took it for one semester, and that was too much even though i ended up failing the class and now i have to take two math classes like math models and algebra III to make up for the dmn credit.

Posted by Crazydeb8ter on 05-01-2002 06:37 PM:

i hate derivitives

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Posted by AiSH on 05-01-2002 07:27 PM:

nvm. too late now. thx neway.

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Posted by Babu2002 on 05-02-2002 07:18 AM:

Answer in Calc.

Math is so gay... datz why I chose pink. Neway, all you have to do is get the answer to the first problem, rite? Then, repeat the problem ON to the first answer to the first problem,~~ and so forth and so forth. That's why they call them derivatives of the last "derivative".

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Posted by Alchemist on 05-02-2002 12:12 PM:

Re: Answer in Calc.

quote:
Originally posted by Babu2002
Math is so gay...

i don't think that math has a sexuality

Posted by Babu2002 on 05-03-2002 05:46 AM:

True dat! haha... How's Red for a color instead, then???

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Posted by Crazydeb8ter on 05-03-2002 01:45 PM:

Re: Answer in Calc.

quote:
Originally posted by Babu2002
Math is so gay... datz why I chose pink. Neway, all you have to do is get the answer to the first problem, rite? Then, repeat the problem ON to the first answer to the first problem,~~ and so forth and so forth. That's why they call them derivatives of the last "derivative".

oh man are you a genius

Posted by Drunken Master on 05-03-2002 08:46 PM:

These are higher derivatives and can get damn tedious...
Like mentioned before.. just figure out the derivative of the first function. After you have that, you figure out the derivative of THAT function and so on and so on...

I just figured these ones out but I'm not guaranteeing that this is correct... It's late at night and i really don't feel like thinking(not to mention that higher dervatives are amazingly annoying).
I didn't bother to figure out for g"'(-pi/8), you can do that yourself.

ok here goes...

g(t)= csc 2t
g'(t)= -2cot2tcsc2t
g"(t)= 4[(cot2t)^2(csc2t) + (csc2t)^3]
g"'(t)= 8[(cot2t)(csc2t)^3 - (cot2t)^3(csc2t)]

Well I hope these are correct and I hope that they helped...

Posted by jakez0r on 05-14-2002 08:13 AM:

quote:
Originally posted by Drunken Master
These are higher derivatives and can get damn tedious...
Like mentioned before.. just figure out the derivative of the first function. After you have that, you figure out the derivative of THAT function and so on and so on...

I just figured these ones out but I'm not guaranteeing that this is correct... It's late at night and i really don't feel like thinking(not to mention that higher dervatives are amazingly annoying).
I didn't bother to figure out for g"'(-pi/8), you can do that yourself.

ok here goes...

g(t)= csc 2t
g'(t)= -2cot2tcsc2t
g"(t)= 4[(cot2t)^2(csc2t) + (csc2t)^3]
g"'(t)= 8[(cot2t)(csc2t)^3 - (cot2t)^3(csc2t)]

Well I hope these are correct and I hope that they helped...

well i don't think you got that first derivative right. For this kind of problem you have to use the chain rule which states that you take the derivative of the inside function and multiply it by the derivative of the whole function. Now i don't know the derivative of cosecant but g'(t) would look something like this

g'(t) = 2*csc'(2t)

basically 2 times the derivative of cosecant with (2t)

i hope that kind of helped :)

Posted by Drunken Master on 05-14-2002 08:18 PM:

quote:
Originally posted by jakez0r

well i don't think you got that first derivative right. For this kind of problem you have to use the chain rule which states that you take the derivative of the inside function and multiply it by the derivative of the whole function. Now i don't know the derivative of cosecant but g'(t) would look something like this

g'(t) = 2*csc'(2t)

basically 2 times the derivative of cosecant with (2t)

i hope that kind of helped :)

No, you don't do it like that sorry.
just think of 2t as B lets say...
they it would be g(t)=csc B
the derivative of that would be g'(t)= -cot Bcsc B * dB
we sub B=2t then it would be g'(t) = -cot 2t csc2t *2.
The chain rule is used but you also forgot to derive the csc value.

Posted by jakez0r on 05-16-2002 07:54 AM:

quote:
Originally posted by Drunken Master

No, you don't do it like that sorry.
just think of 2t as B lets say...
they it would be g(t)=csc B
the derivative of that would be g'(t)= -cot Bcsc B * dB
we sub B=2t then it would be g'(t) = -cot 2t csc2t *2.
The chain rule is used but you also forgot to derive the csc value.

i know i did because i forgot what the derivative of csc is :) if you look i put a little " ' " there to mean csc prime or derivative of csc. But whatever :)

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