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..

__________________
wha

Report this post to a moderator | IP: Logged

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.

Report this post to a moderator | IP: Logged

i hate derivitives

__________________
ni pour ni contre; ça m'est égal

"The weight of this sad time we must obey,/ Speak what we feel, not what we ought to say./ The oldest hath borne most; we that are young/ Shall never see so much, nor live so long."
King Lear (V.3.300-304)

Report this post to a moderator | IP: Logged

nvm. too late now. thx neway.

__________________
wha

Report this post to a moderator | IP: Logged

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".

__________________
"Please call me... please kiss me..." FinKL [Now]

Report this post to a moderator | IP: Logged

quote:

---

Originally posted by Babu2002
Math is so gay...

---

Report this post to a moderator | IP: Logged

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

__________________
"Please call me... please kiss me..." FinKL [Now]

Report this post to a moderator | IP: Logged

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

__________________
ni pour ni contre; ça m'est égal

"The weight of this sad time we must obey,/ Speak what we feel, not what we ought to say./ The oldest hath borne most; we that are young/ Shall never see so much, nor live so long."
King Lear (V.3.300-304)

Report this post to a moderator | IP: Logged

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...

Last edited by Drunken Master on 05-03-2002 at 08:49 PM

Report this post to a moderator | IP: Logged

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...

---

Report this post to a moderator | IP: Logged

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 :)

---

Report this post to a moderator | IP: Logged

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.

---

__________________
Thread Bumper Extraordinaire

Bumping Threads Since 2001

Report this post to a moderator | IP: Logged