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Please help. ;-;

I think a mathematics and science board would be cool.

A general school board pls

i voat for the monkees they make good songs.

Quote from: JMV290 on February 26, 2008, 08:14:53 PM
I think a science board would be cool.

srsly

shut up felt

Quote from: Commodore Guff on February 27, 2008, 12:34:47 PM
shut up felt

.

Just had a test on this today...

Quote from: Your Posting Rival on February 27, 2008, 01:31:29 PM
Just had a test on this today...

Really?

Quote from: su??? n????o? on February 27, 2008, 01:32:33 PM
Really?

If this is what I think it is, then yes... or at least the product rule.

Quote from: Your Posting Rival on February 27, 2008, 01:33:24 PM
If this is what I think it is, then yes... or at least the product rule.

Is it something like this:
f(x)g(x)=h(x)
h'(x)=f(x)g'(x)+g(x)f'(x)

Quote from: su??? n????o? on February 27, 2008, 01:35:47 PM
Is it something like this:
f(x)g(x)=h(x)
h'(x)=f(x)g'(x)+g(x)f'(x)

Yep, that's it

Quote from: Commodore Guff on February 27, 2008, 12:34:47 PM
shut up felt

felt requests that you be friends with him regardless of his face

get out, Felt

Please help. ;-;

What part do you need help with?

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.

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.

Quote from: suÃ,,±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...

Oh shit.  Why did I add them?

fixed so it multiplies.

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).

Quote from: suÃ,,±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

Quote from: JMV290 on February 29, 2008, 02:16:37 PM
dats wut i said except i used u

And used addition originally caterpie;

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.

You did not get C right. Try again.

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

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

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.

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

sin u = 5/13

find sin (u/2) plz

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:
(http://home.alltel.net/okrebs/Ch10-15.gif)

Ok guys, I need help again.
(http://www.codecogs.com/eq.latex?%20\sqrt{xy}=x-2y)
I need to find dy/dx of this.
I got: (http://www.codecogs.com/eq.latex?%20\frac{dy}{dx}=\frac{2\sqrt{x}-\sqrt{y}(\sqrt{y})}{\sqrt{x}(\sqrt{x}+4\sqrt{y})})

hay mib why arent you in geo h?

Quote from: Tomboh on March 07, 2008, 04:18:29 PM
hay mib why arent you in geo h?

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.

Quote from: Original_MIB on March 07, 2008, 05:46:32 PM
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.

DON'T USE A CALCULATOR.
Have they taught you how to factor radicals?
Could you scan the problem? You may have wrote the wrong notation, thus you are having problems.
a=6.
b=6.
This is why it is causing you problems.
a²+b²=c² by the way of the Pythagorean theorem.
So we get 72=c²