/sci/ - Science & Math » Thread #12416337

20KiB, 469x206, quotient-rule.jpg

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Who made the quotient rule

Anonymous Thu 03 Dec 20:28:39 2020 No.12416337 View Reply Original Report

Quoted By: >>12416465 >>12416522 >>12416654 >>12416843 >>12417595 >>12418420

Kinda fucking cringe, who do I bully

Anonymous

Anonymous Thu 03 Dec 2020 20:29:26 No.12416345 Report

Quoted By:

the product rule

Anonymous

Anonymous Thu 03 Dec 2020 20:40:51 No.12416410 Report

Quoted By: >>12416485

$\frac{Po'-oP'}{PP}$

Anonymous

Anonymous Thu 03 Dec 2020 20:51:12 No.12416465 Report

Quoted By: >>12416489 >>12416522

>>12416337
It's literally just the chain rule

Anonymous

Anonymous Thu 03 Dec 2020 20:56:07 No.12416485 Report

Quoted By:

>>12416410
Based and calculuspilled

Anonymous

Anonymous Thu 03 Dec 2020 20:57:09 No.12416489 Report

Quoted By: >>12416526

>>12416465
literally,>>12416465
everything is the chain rule, Fuck off

Anonymous

Anonymous Thu 03 Dec 2020 21:01:23 No.12416508 Report

Quoted By: >>12416532

>People get filtered by this shit
>They need Kahn Academy and 3blue1brown to understand such a basic rule
Daily remainder you are wasting your time and money and you will never goign to make it.

Anonymous

Anonymous Thu 03 Dec 2020 21:05:14 No.12416522 Report

Quoted By: >>12416564 >>12416611

>>12416337
>>12416465
in fact, since i'm bored here's the proof:

First note that
$\frac{f(x)}{g(x)}=f(x) \cdot (g(x))^{-1}$
By the product rule and chain rule,
$\Bigg(\frac{f(x)}{g(x)}\Bigg)'=f'(x) \cdot (g(x))^{-1} -(g(x))^{-2} \cdot f(x) \cdot g'(x)$
This is just
$\frac{f'(x)}{g(x)} - \frac{g'(x) \cdot f(x)}{[g(x)]^{2}}$
Combining into a single fraction, we have
$\frac{f'(x) \cdot (g(x))^{2} - g(x) \cdot g'(x) \cdot f(x)}{[g(x)]^{3}}$
Finally, cancelling out the common $g(x)$ we arrive at the quotient rule:
$\frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{[g(x)]^{2}}$

Anonymous

Anonymous Thu 03 Dec 2020 21:05:45 No.12416526 Report

Quoted By: >>12416631

>>12416489
product/quotient rule (derivative of product) is distinct from chain rule (derivative of functional composition)

Anonymous

View Same Google iqdb SauceNAO what.jpg, 70KiB, 470x470

Anonymous Thu 03 Dec 2020 21:06:47 No.12416532 Report

Quoted By:

>>12416508
The rule makes sense anon it's just cringe, stop projecting and drink some hot chocolate

Anonymous

Anonymous Thu 03 Dec 2020 21:11:09 No.12416564 Report

Quoted By: >>12416571

>>12416522
but who made it

Anonymous

Anonymous Thu 03 Dec 2020 21:12:53 No.12416571 Report

Quoted By:

>>12416564
just blame Newton and Leibniz independently.

Anonymous

Anonymous Thu 03 Dec 2020 21:19:54 No.12416611 Report

Quoted By: >>12416622

>>12416522
The 4th line is utterly redundant since you can go from line 3 to the result in one step (multiple left term by g/g).

I am curious though why we have to take the modulus of g^2?

Anonymous

Anonymous Thu 03 Dec 2020 21:22:08 No.12416622 Report

Quoted By: >>12416632

>>12416611
i know its redundant, its a proof for brainlets.
you dont take modulus, i just formatted it that way to match OP's pic.

Anonymous

Anonymous Thu 03 Dec 2020 21:24:14 No.12416631 Report

Quoted By: >>12417202

>>12416526
>consider g(x,y) = x*y

Anonymous

Anonymous Thu 03 Dec 2020 21:25:48 No.12416632 Report

Quoted By:

>>12416622
Ahh, I blame the pixels.

Anonymous

Anonymous Thu 03 Dec 2020 21:26:12 No.12416634 Report

Quoted By: >>12416661 >>12416686 >>12417618

$h(x) = \frac{f(x)}{g(x)}$

$h'(x) \approx \frac{1}{\delta_x} \left( \frac{f(x+\delta_x)}{g(x+\delta_x)} - \frac{f(x)}{g(x)} \right)$

$h'(x) \approx \frac{1}{\delta_x} \left( \frac{f(x+\delta_x) g(x) - f(x) g(x+\delta_x)}{g(x) g(x+\delta_x)} \right)$

$h'(x) \approx \frac{1}{\delta_x} \left( \frac{f(x) g(x) + \delta_x f'(x) g(x) - f(x) g(x) - \delta_x f(x) g'(x)}{g^2(x) + \delta_x g(x) g'(x)} + o(\delta_x^2) \right)$

$h'(x) \approx \left( \frac{f'(x) g(x) - f(x) g'(x)}{g^2(x) + \delta_x g(x) g'(x)} + o(\delta_x) \right)$

$h'(x) \to \frac{f'(x) g(x) - f(x) g'(x)}{g^2(x)}$

Anonymous

Anonymous Thu 03 Dec 2020 21:30:26 No.12416654 Report

Quoted By:

>>12416337
Let $A$ and $B$ two symetric positive matrix of size $n$ .

Extremum of

$g(x) = \frac{ \langle x, Ax \rangle }{ \langle x, Bx \rangle }$

Anonymous

Anonymous Thu 03 Dec 2020 21:31:41 No.12416661 Report

Quoted By:

>>12416634
clean

Anonymous

Anonymous Thu 03 Dec 2020 21:32:25 No.12416667 Report

Quoted By:

>tfw just do the product rule with f(x)*h(x) where h(x)=1/g(x)

Anonymous

Anonymous Thu 03 Dec 2020 21:35:48 No.12416686 Report

Quoted By:

>>12416634
i like this proof best

Anonymous

Anonymous Thu 03 Dec 2020 22:05:38 No.12416843 Report

Quoted By:

>>12416337
Cash when there a square root in the denominator

Anonymous

Anonymous Thu 03 Dec 2020 23:47:54 No.12417202 Report

Quoted By: >>12417290

>>12416631
that's a multivariate function. OP is sucking dick doing functions of single variables. your example is meaningless

Anonymous

Anonymous Fri 04 Dec 2020 00:06:24 No.12417290 Report

Quoted By:

>>12417202
it shows that Leibniz rule follows from chain rule

El Arcón

El Arcón Fri 04 Dec 2020 00:38:42 No.12417399 Report

Quoted By:

I use the product rule with a negative exponent.

Anonymous

Anonymous Fri 04 Dec 2020 02:11:58 No.12417595 Report

Quoted By:

>>12416337
That form is shit for memorization purposes.

f'g - fg' as the numerator is better because the function names are in simple alphabetical order and you can think of the numerator being "quoted" (new word "quotiented") -- i.e., derivative ticks on the outside of the expression.

Anonymous

Anonymous Fri 04 Dec 2020 02:22:15 No.12417618 Report

Quoted By: >>12418358

>>12416634
math fags
> but it's not rigorous !
physics fags
> your mom is not rigorous

Anonymous

Anonymous Fri 04 Dec 2020 07:49:56 No.12418358 Report

Quoted By:

>>12417618
It is rigorous though

Anonymous

Anonymous Fri 04 Dec 2020 08:25:57 No.12418420 Report

Quoted By:

>>12416337
God, actually.

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