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Obligatory Math Help Thread

Started by [REDACTED], February 26, 2008, 08:13:17 PM

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YPrrrr


[REDACTED]

Quote from: Your Posting Rival on February 29, 2008, 01:34:45 PM
What part do you need help with?
What is the exact rule?
I know you take a function and make it into two composite functions, but I don't know how you differentiate from there.
I do not have HIV/AIDS.

Daddy

February 29, 2008, 01:38:54 PM #17 Last Edit: February 29, 2008, 01:55:08 PM by JMV290
Say that you had


(5x2+2x)2 and you wanted derivative this is what you'd do

Substitute 5x2+2x with U so you'd have U2 and then multiply it by (5x2+2x)'

so U2' * (5x2+2x)' =

2U(10x +2)  =

2(5x2+2x) (10x+2) =

(10x2 + 4x) (10x + 2) =

100x3+20x2 + 40x2+8x =

100x3+60x2+8x

I suck at explaining stuff, but that's right.

YPrrrr

February 29, 2008, 01:50:24 PM #18 Last Edit: February 29, 2008, 01:55:11 PM by Your Posting Rival
Quote from: su,,±ÊÊ‡ nɐǝʇɔoÉ” on February 29, 2008, 01:37:19 PM
What is the exact rule?
I know you take a function and make it into two composite functions, but I don't know how you differentiate from there.
It's all about derivitives...

Using JMV's example: (5x2+2x)2

so you find the derivitive of f(x), which is 2, and multiply it by g(x) which equals (5x2+2x) * g'(x), which is (10x+2)

So in the end, it would be 2(5x2+2x)(10x+2)

In summary... chain rule = f '(g(x))g'(x)

f '(x) = 2
g(x) = (5x2+2x)
g'(x) = (10x+2)

that is, if what i'm thinking of is correct...

Daddy

February 29, 2008, 01:52:00 PM #19 Last Edit: February 29, 2008, 01:55:48 PM by JMV290
Oh shit.  Why did I add them?

fixed so it multiplies.

[REDACTED]

Ok, I now understand it.
I just need to make a composite function of the function I want to differentiate, h(x)=f(g(x)).
Then find f' and g'. Then, multiply f'(g(x) by g'(x).
I do not have HIV/AIDS.

Daddy

Quote from: su,,±ÊÊ‡ nɐǝʇɔoÉ” on February 29, 2008, 02:13:15 PM
Ok, I now understand it.
I just need to make a composite function of the function I want to differentiate, h(x)=f(g(x)).
Then find f' and g'. Then, multiply f'(g(x) by g'(x).
dats wut i said except i used u

YPrrrr


6M69I69B9

February 29, 2008, 05:28:49 PM #23 Last Edit: March 02, 2008, 10:32:11 PM by suıʍʇ nɐǝʇɔoɔ
my head is broken, please help the freshman with algebra.



Right triangle

A : 6^2
B : 6^2
C : Square root of 3.

Find C.


I do 6^2 to both A and B, which are 36. Next, I add them together to get 72.

Then I do 3x72, which equals 216.

Then I have to find the square root of that.



Did I do it right? I know I have to get the square root, but I don't have a calculator right now.
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[REDACTED]

You did not get C right. Try again.
I do not have HIV/AIDS.

UnagiPower

At least you don't have to do proofs.  :(

Which are hard if you don't pay attention in class/read the textbook.

Gladjaframpf

I'm confused. Are A, B, and C the side lengths of the triangle? If so, you can't possibly have a right angle triangle with those side lengths, so something is wrong. Also, you're trying to find C, but you already said it was the square root  of 3. I'm just not sure what you're trying to do here.

Squirtlejazz

A2+B2=C2
But that isn't possible on your triangle.
iSnake

C.Mongler

sin u = 5/13

find sin (u/2) plz

[REDACTED]

March 05, 2008, 02:16:55 PM #29 Last Edit: March 05, 2008, 02:20:41 PM by suıʍʇ nɐǝʇɔoɔ
Use the half angle formula, silly.
Also, does it give you what quadrant or region the angle is located in?
Use the pythagorean trig identities to find cos u.
Here's the half angle formula:

I do not have HIV/AIDS.

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