BEHOLD THE MATH ORGY

"TheHondaBro" (wwaveform)
10/26/2015 at 21:43 • Filed to: None

I’m bored.

DISCUSSION (21)

bob and john > TheHondaBro
10/26/2015 at 21:47

NO.
NO. GO FUCK YOURSELF.
NO.

(a very tired peter who has been studying dynamics, thermal and industrial for well over 50 hours now

Nibby > bob and john
10/26/2015 at 21:53

hot

ttyymmnn > TheHondaBro
10/26/2015 at 21:53

Kinja'd!!!

Dr. Zoidberg - RIP Oppo > TheHondaBro
10/26/2015 at 21:59

Uhh... Goddammit I don’t remember any of this stuff... I hate these because you’re like, “well of COURSE it’s true!”
= f’gh + fg’h + fgh’
= ...

No.

DrJohannVegas > TheHondaBro
10/26/2015 at 22:01

I keep trying to type LaTeX into the editor, but nothing happens.

Eh, screw it:

\begin{align}
(fgh)‘ & = f’(gh) + f(gh)‘ \\

(fgh)‘ & = f’(gh) + f(g’h + gh’) \\

(fgh)‘ & = f’gh + fg’h + fgh’

\end{align}

Edited for sloppy parentheses.

traderQAMobileTestAutomationMobileBoostOn > TheHondaBro
10/26/2015 at 22:02

That brought back horrible memories of last semester’s Pre-Cal...and foreshadows to the even horribler Pre-Cal II next semester.

Sportwägen, Driver Of The Red Sportwagen > TheHondaBro
10/26/2015 at 22:03

Calculus is so much fun!

*says the guy who is contemplating changing majors so he won’t have to take calc 2 or 3*

Dr. Zoidberg - RIP Oppo > TheHondaBro
10/26/2015 at 22:05

WAIT! Can’t you just basically cheat with the product rule? I think I’m missing a major concept here but...

LOOSELY...

(fgh)‘
(fgh)?(fgh)
=ff+fg+fh+gf+gg+gh+hf+hg+hh
=f(fgh)+g(fgh)+h(fgh)
=f’gh+g’fh+h’fg
which is f’gh+fg’h+fgh’

... Am i warm?

TractorPillow > TheHondaBro
10/26/2015 at 22:05

I’ve taken pre cal, calc 1 and 2 and got a’s in all of them. That was 8 years ago. I have no clue what I am looking at.

TheHondaBro > Dr. Zoidberg - RIP Oppo
10/26/2015 at 22:06

Just stahp pls.

Dr. Zoidberg - RIP Oppo > TheHondaBro
10/26/2015 at 22:06

But... :(

TheHondaBro > Dr. Zoidberg - RIP Oppo
10/26/2015 at 22:07

I don’t see f, g, and h. I see gdejslghjdskgjklsgjklgbjfgbjfkdlghjfkldgh.

Dr. Zoidberg - RIP Oppo > TheHondaBro
10/26/2015 at 22:16

Oh... well basically the question actually gives you the answer. A differentiation *here* is really an application of the product rule. So you turn (whatever) prime into (whatever) “times” (whatever).

So in this case you end up with fgh quantity prime rewritten as the identical fgh quantity prime squared, and remember that (ab)^2 is NOT a^2+b^2, but rather (ab)(ab) which give a^2+b^2+2ab

Thus, turning (fgh)‘ into “(fgh)(fgh)“ (quotes because we’re multiplying in application, except ff become f’ here), so (fgh)‘ is just f times f plus f times g plus f times h plus g times f plus g times g plus g times h plus h times f plus h times g plus h times h. Then you group like terms so you end up with f(fgh) + g(fgh) + h(fgh). And here, we have aa=a^2=”a’ “ here so using the distribution method on all three terms, you get f’gh + fg’h + fgh’.

Maybe.