Author Topic: Equating gradients (Read 6591 times) Tweet Share

Mao · « **Reply #15 on:** November 08, 2008, 08:21:51 pm »

Quote from: shinjitsuzx on November 08, 2008, 08:17:01 pm

$\frac{d}{do}(gook)=\frac{d}{do}(gko^{2})$
$\implies \frac{d}{do}(gook)=2gko$

Like that droodles?

provided neither "g" and "k" are functions of "o"

if they both are:

$\frac{d}{do}(gook) = \frac{d}{do}(o^2)\cdot gk + \frac{d}{do}(gk)\cdot o^2 = 2ogk + \left(\frac{dg}{do}k+\frac{dk}{do}g\right)\cdot o^2$

$\implies \frac{d}{do}(gook) = o\cdot \left( 2gk + ok\frac{dg}{do} + og\frac{dk}{do}\right)$

wait... too far?

shinny · « **Reply #16 on:** November 08, 2008, 08:23:11 pm »

Yes Mao, you clearly took it too far.

Mao · « **Reply #17 on:** November 08, 2008, 08:24:31 pm »

droodles · « **Reply #18 on:** November 08, 2008, 08:25:05 pm »

ARE YOU A VICTIM OF RACIAL SLUR?

DO YOU ENJOY MATHS?

WELL HERE'S A SOLUTION...DERIVE THEM AWAY!

Quote from: Mao on November 08, 2008, 08:21:51 pm

provided neither "g" and "k" are functions of "o"

if they both are:

$\frac{d}{do}(gook) = \frac{d}{do}(o^2)\cdot gk + \frac{d}{do}(gk)\cdot o^2 = 2ogk + \left(\frac{dg}{do}k+\frac{dk}{do}g\right)\cdot o^2$

$\implies \frac{d}{do}(gook) = o\cdot \left( 2gk + ok\frac{dg}{do} + og\frac{dk}{do}\right)$

NOW YOUR PROBLEMS ARE 'SOLVED'

bec · « **Reply #19 on:** November 08, 2008, 08:26:08 pm »

oh no he went there

Mao · « **Reply #20 on:** November 08, 2008, 08:29:45 pm »

completely random, but here's an insight I had when I was deriving 'gook':

the multiple products rule:

$\mbox{Let }f(x)=f_1 (x) \cdot f_2 (x) \cdot f_3(x) \cdots$

$\frac{df}{dx} = \frac{df_1}{dx}\cdot \frac{f(x)}{f_1(x)} + \frac{df_2}{dx}\cdot \frac{f(x)}{f_2(x)} + \frac{df_3}{dx}\cdot \frac{f(x)}{f_3(x)} + \cdots$

thank you for your time.

which means, the derivative of gook can be re-expressed as

$\frac{d}{do}(gook) = (gook) \cdot \left( 2o^{-1} + g^{-1} \cdot \frac{dg}{do} + k^{-1}\cdot \frac{dk}{do}\right)$

it appears the racial slur is invincible.....

shinny · « **Reply #21 on:** November 08, 2008, 08:32:22 pm »

I PREFER INTEGRATING RACIAL TENSION INTO THIS FORUM THANK YOU VERY MUCH

$\int (gook) do=\int (gko^{2}) do$

$=[\frac{1}{3}gko^{3}]$

$\int \frac{1}{3}gko^{3} do=[\frac{1}{12}gko^{4}]$

$...$

$=[\frac{1}{9000}gko^{9000}]$

EQUALS
GOOOOOOOOOOOK
Well maybe not quite.

Quote from: bec on November 08, 2008, 08:07:58 pm

So um...disregard this whole thread...? haha

And yes, this is what bec gets for derailing her own thread.

droodles · « **Reply #22 on:** November 08, 2008, 08:36:43 pm »

apparently at mhs the students get erections after they see 50/50 for any of their sacs

shinsjuxusijxxxznarutofannerd confirm?

shinny · « **Reply #23 on:** November 08, 2008, 08:43:12 pm »

Quote from: droodles on November 08, 2008, 08:36:43 pm

apparently at mhs the students get erections after they see 50/50 for any of their sacs

You should some students' reaction when they LOSE a mark.

Quote from: droodles on November 08, 2008, 08:36:43 pm

shinsjuxusijxxxznarutofannerd confirm?

Where'd the Naruto come from? o_O Clearly that description fits ed_saifa better.

Toothpaste · « **Reply #24 on:** November 08, 2008, 08:45:27 pm »

真実
translate: jp -> en
then find romaji.

vce08 · « **Reply #25 on:** November 08, 2008, 08:52:20 pm »

50/50 is clearly standard at MHS

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