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/sci/ - Science & Math

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11275033 No.11275033 [Reply] [Original]

is there a simple and intuitive way of proving this?

>> No.11275041

>>11275033
One way is to use product rule and chain rule.

>> No.11275044

>>11275041
blah blah chain rule that product rule that
show your understanding of this formula and explain it in simple terms in your own words

>> No.11275046

>>11275044
we're not here to do your fucking homework

>> No.11275051

>>11275046
I'm just curious to see whether most people who learned this actually went through to understand it or just memorized it

>> No.11275053

>>11275044
Then fuck off nigger, I'm not holding your hand.
Lemme start you off right: let [math]\epsilon>0[/math] be arbitrary...

>> No.11275056

>>11275053
if you can't explain it in simple terms, you don't understand it

>> No.11275068
File: 130 KB, 840x1120, __nazrin_touhou_drawn_by_mozukuzu_manukedori__cf8d709a1229d37eac51e956104c0604.jpg [View same] [iqdb] [saucenao] [google]
11275068

[math]f'=(g ~ f/g)'=g' ~ fg + g ~ (f/g)' [/math].
Isolate (f/g)' to obtain a solution, and then delete the thread.
Also, ask this stuff in /sqt/.
>>11275041
>chain rule

>> No.11275075

>>11275068
show your understanding of this formula and explain it in simple terms in your own words
do you understand what it means or did you learn how to shuffle letterns around thinking it means anything?

>> No.11275088

>>11275051
Why would anyone need to memorize that hideous formula?
Just derive it when you need it.

>> No.11275093

>>11275088
so you would use it without understanding what it means?
can you not provide an analogy for what this function means? are you just shuffling letters around meaninglessly?

>> No.11275104

>>11275033
trivial

>> No.11275110

How often do you want to write "understanding" in this thread.
It's the product [math](g(x)*h(x))'[/math] rule for [math]h(x):=g(x)^{-1}[/math].
You wouldn't write down [math](g(x)*h(x))'[/math] with [math]h(x):=u(x)^{2}{\mathrm e}^{\sin(v(x))}[/math] and ask about the "understanding" of the resulting expression either.

>> No.11275117

>>11275110
I was asking about intuitive explanation for this formula, not shuffling letters around
can't explain it? dont understand it

>> No.11275119

>>11275117
what do you mean by intuitive? do you want a physical analogy or something?

>> No.11275122

>>11275119
any analogy, yes

>> No.11275133

>>11275122
Then you should have said so from the very start.

>> No.11275138

>>11275133
that's what 'intuitive' means right?
for something to be intuitive to you, you must be able to explain it in simple terms and/or provide an analogy

>> No.11275159 [DELETED] 

d(f/g)
= (f + df)/(g + dg) - fg
= (fg + g.df - fg - f.dg) / (g^2 +g.dg)
= (g.df - f.dg) / g^2

>> No.11275160

>>11275093
Why would you come up with an "intuitive" explanation for every equation you see?

>> No.11275163

>>11275160
why would you memorize something without understanding it?

>> No.11275165 [DELETED] 

>>11275159

also:

d(fg)
= (f + df)(g + dg) - fg
= fg + f.dg + g.df + df.dg - fg
= f.dg + g.df

>> No.11275171

>>11275033
Prove the product rule
Then prove the derivative of 1/g

>> No.11275175

>>11275163
I don't think you understand math as well as you think you do

>> No.11275180

>>11275175
why are you assuming i'd claim i understand it well?
I'm just trying to see if there's anyone who actually does but it doesn't seem like it

>> No.11275182

d(fg)
= (f + df)(g + dg) - fg
= fg + f.dg + g.df + df.dg - fg
= f.dg + g.df

d(f/g)
= (f + df)/(g + dg) - f/g
= (fg + g.df - fg - f.dg) / (g^2 +g.dg)
= (g.df - f.dg) / g^2

>> No.11275184

>>11275041
>>11275044
He told you everything you need. Rewrite the quotient as a product, then use the product and chain rules.
If you can't do that, you don't understand the earlier material.

>> No.11275186

>>11275184
so do you claim it's impossible to provide a simple explanation of it or an analogy?

>> No.11275193

>>11275033
If you know the product rule you simply flip around the functions.
https://en.wikipedia.org/wiki/Quotient_rule#Proof_using_implicit_differentiation

>> No.11275194

>>11275193
have you ever thought outside the box?

>> No.11275206

>>11275186
https://www.youtube.com/watch?v=8Ow_O1JZTLs

>> No.11275208

>>11275186
it's just a special application of the product rule and chain rule that's literally as simple as it gets.

>> No.11275225

>>11275182

nice

>> No.11275240

>>11275225

d(e^x)
=e^(x + dx) - e^x
=e^x.(e^dx - 1)
=e^x.dx

>> No.11275244

>>11275240

nice

>> No.11275250

>>11275225
>>11275240
>>11275244
>samefags
>gives all posts the same distinct formatting of spacing after replying
It's like you're just handing yourself to the police.

>> No.11275266

>>11275250

brainlet

(f(g(x)))'
= [f(g(x + dx)) - f(g(x))]/dx
= [f(g + dg/dx.dx) - f(g(x))] / dx
= df/dg.dg/dx

>> No.11275271

>>11275206
/thread

>> No.11275274

>>11275206
>>11275271
it does not explain why you can treat functions the same way you can treat rectangles

>> No.11275279

f = f(x, y)

df
= f(x + dx, y + dy) - f(x, y)
= f(x + dx, y + dy) - f(x, y + dy) + f(x, y + dy)- f(x, y)
= (∂f/∂x)dx + (∂f/∂y)dy

>> No.11275285

0
= (1-1).(-1)
= -1 + (-1).(-1)

(-1).(-1) = 1

>> No.11275290

>>11275274
You don't understand that if you have a rectangle with side lengths A and B, then the area of that rectangle is A*B?
Really?
Am I supposed to explain that "intuitively" to you as well?

>> No.11275292

>>11275290
a side length is not a function

>> No.11275310

>>11275292
A side length is a variable which is a linear 1st order function.

>> No.11275313

>>11275250
>distinct formatting of spacing
it's hardly distinct

>>11275206
this only works if f and g are strictly increasing, right?

>> No.11275314

>>11275292
For any x, if you have a rectangle with side lengths f(x) and g(x), then the area of that rectangle is f(x)*g(x).

>> No.11275317

>>11275313
and, obviously, non-negative

>> No.11275319

>>11275313
>this only works if f and g are strictly increasing, right?
no, I don't think so
the drawing would look a little different, but it's still the same thing

>> No.11275330

>>11275313
>hardly distinct
>literally the only person in the entire thread

>> No.11275331

>>11275314
f(x) = x^50
g(x) = x^150
h(x) = f(x)g(x) = x^200
are you some 200 dimensional alien able to see 200 dimensional constructs as 2 dimensional rectangles?

>> No.11275338

>>11275274
Prove product rule

[eqn]\frac{\partial}{\partial x}f(x)g(x) = \lim_{h\rightarrow 0} \frac{f(x+h)g(x+h) - f(x)g(x)}{h}[/eqn]
[eqn] = \lim_{h\rightarrow 0} \frac{f(x+h)g(x+h) - f(x+h)g(x) + f(x+h)g(x)- f(x)g(x)}{h}[/eqn]
[eqn] = \lim_{h\rightarrow 0} \frac{f(x+h)g(x+h) - f(x+h)g(x)}{h} \lim_{h\rightarrow 0} \frac{f(x+h)g(x)- f(x)g(x)}{h}[/eqn]
[eqn] = f(x)\frac{\partial g}{\partial x} + \frac{\partial f}{\partial x}g(x)[/eqn]

Now that you've proved product rule, you can prove quotient rule

[eqn] \frac{\partial}{\partial x}\frac{f}{g} = \frac{\partial f}{\partial x}\frac{1}{g} + f\frac{\partial \frac{1}{g}}{\partial x} = \frac{f'g}{g^2} - \frac{fg'}{g^2}[/eqn]

Or you could just forget about this dumb and inane rule and only learn product rule

>> No.11275343

>>11275338
yes I know how to google
give a simple analogy or you're a math-monkey

>> No.11275366

>>11275331
you're fucking retarded
I said for any x
so if you take x=3 for example, then the side lengths are 3^50 and 3^150
and then the area is 3^200, which is correct
How are you not able to understand this simple concept?

>> No.11275378
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11275378

>>11275056
>>11275075
>>11275093
>>11275117
>>11275163
>>11275186
>>11275343
>why yes, I just fiddle around with the algebra and get the result, how did you know?

>> No.11275380

>>11275366
f(x)=x^x
g(x)=x^x^x^x^x
h(x)=f(x)g(x)
still so full of yourself?

>> No.11275382

>>11275343
>analogy
what is this retardation? I dont need to write an analogy for math it can literally just work that way.

Chain rule makes intuitive sense and a video explaining the product rule was already posted. If you cant understand how negative exponents work youre just a fucking ape.

>> No.11275385

>>11275382
>I dont need to write an analogy for math it can literally just work that way
then you don't understand it

>> No.11275437

>>11275380
yes for any x, if one side length is x^x, and the other side length is x^x^x^x^x, then the area is x^x * x^x^x^x^x
I'm starting to think you're trolling at this point

>> No.11275469

>>11275033
Do a linear approximation of the top and bottom near x.
f~f(x), a~f'(x), g~g(x), b~g'(x).
f(x+t) = f+t*a, g(x+t) = g+t*b.
[f+t*a]/[g+t*b] = [f/g + t*a/g]/[1+ t*b/g].
Expand the denominator as a geometric series.
=[f/g + t*a/g]*[1-t*b/g + (t*b/g)^2 - ...]
=f/g + t*[a/g - (f/g)*(b/g)] + higher order terms
The coefficient of t^1 is the derivative.

>> No.11275513

>Liebniz got stuck for years because he expected that the derivative of u(x)v(x) should be u'(x)v'(x). Unfortunately the real answer is a lot more complicated:
>derivative of u(x)v(x) = u'(x)v(x) + u(x)v'(x)
>Apparently Liebniz got the correct answer but didn’t believe it was right. So he spent years trying to redo the problem before he accepted the truth.

It wasn't intuitive even for the creators of calculus.

>> No.11275516

>>11275513
>got stuck for years
I know this feel.

>> No.11275518

maybe

>> No.11275530 [DELETED] 

>>11275513
That sounds very unlikely, given her certainly knew
[math]\frac{d}{dx}x^2=2x.[/math]
While
[math]\frac{d}{dx}x^2=\frac{d}{dx}x\frac{d}{dx}x=1.[/math]

>> No.11275533

>>11275513
That sounds very unlikely, given he certainly knew

[math] (x\cdot x)' = 2x [/math]

While

[math] (x\cdot x)' = x'\cdot x' = 1 [/math]

>> No.11275540

>>11275513
In fact it would imply that the derivative of any x^n is constant.

>> No.11275542

>>11275540

maybe it is

>> No.11275544

>>11275533
>>11275540
So maybe it was the reason he was stuck for so long, he just couldn't understand it intuitively and what he expected was not what he saw.
I don't understand it either, I just memorized it because I didn't see any easy way of conceptualizing it.

>> No.11275547

>>11275513
>u'=(u1)'=u'1'=u'0=0 for any u
Hoiw come this absolute nonce gets to be more talented than me.
This isn't fair.

>> No.11275558

>>11275437
>I'm starting to think you're trolling at this point
what took you so long?

>> No.11275566
File: 90 KB, 512x694, 1573341367999.jpg [View same] [iqdb] [saucenao] [google]
11275566

>>11275558
>y-yeah! y-u-you're right! i-i w-w-w-w-w-was o-only p-p-pr-pr-p-preten-pre-pr-pretending t-t-to b-be r-r-r-retaded

>> No.11275585

>>11275385
fuck off and give me my big mac and fries already

>> No.11275671

>>11275117
There are 3blue1brown videos on this

>> No.11275687

>>11275547
Kek, pretty good observation

>> No.11275695

>>11275033
>is there a simple and intuitive way of proving this?
yes absolutely. erase it because it is incorrect.

>> No.11275703

>>11275033
Think of the function x maps to (g(x), f(x)), a function of R into R^2. This is a curve. What does f'g - fg' measure? It measures the turning rate about the origin, it is a sort of curl (why? if in the first quadrant, we move cw if f' < 0 and g' > 0, then f'g - g'f < 0, and ccw if f' > 0 and g' < 0, then f'g - g'f > 0. check for other quadrants to see that the sign always matches the direction of rotation, + for ccw, - for cw). Now, we have a measure for the rate of rotation but need to normalize it to get the rate of change of the slope f/g. What is our normalization factor? Think dimensional analysis. If f is in units of length and g units of mass, then f'g and g'f both have length*mass/time units, while what we want, (f/g)', has length/mass/time units. So the only way we can normalize using what we've got is by some constant multiple of 1/g^2, which has units 1/mass^2. Now, trying on a simple example, x/x, we see that the constant is 1.
The normalization business is not easy to intuit but the top expression is just how you look at curl of a 2d vector field, so you should get comfortable with why this is related to division. Always think slope.

>> No.11275728

>>11275703
LOL

>> No.11275729

i found the carathéodory's proof too be the most intuitive

>> No.11275788

>>11275033
>is there a simple and intuitive way of proving this
For most people, including you, the answer is no. Simple and intuitive are subjective words. Our epsilon-delta proofs will look the same, but the difference is that I consider the proof simple and intuitive, whereas you, sir, are a brainlet. And if the proofs by linearity of the derivative operator are not considered "intuitive" to you, then you are a troll.

>> No.11275794

>>11275788
damn this board gets worse all the time

>> No.11275795

>>11275794
OP needs to shut up, sit down, be humble, and hit the books.

>> No.11275796

>>11275795
i'm not talking about OP

>> No.11275798

>>11275796
OP is clearly still at intro calculus level math, and so is half the thread. They are all just trolling each other while trying to be smart. What category do you fit into?

>> No.11275799

>>11275728
what are you on about? my whole thing can be reduced to "just think of it as the normalized curl of the space curve [math] x \mapsto (g(x), f(x)) [/math]" which is extraordinarily simple.

>> No.11275800

>>11275798
>They are all just trolling each other while trying to be smart
if you're not sure how you're hitting this to a T...

>> No.11275803

>>11275800
I have a math degree. I'm secure in my mathematical ability and I'm not here to put out free answers. The rest of the thread, on the other hand, uh-uh.

>> No.11275805

>>11275803
haha ok dude you seem really secure

>> No.11275809

>>11275805
Of course I do. I project mathematical masculinity into every paper I write.

>> No.11275814

>>11275703
That's a pretty cool way to visualize it. Neat. Is there a way to show more concretely that f'g - fg' is related to curvature, using the fact that the map (f,g) to f'g - fg' is an antisymmetric bilinear form? Seems like there should be some explicit relation to the curl/wedge products of differential forms, but I can't see it right now.

>> No.11275816

>>11275809
being afraid of ignorance isn't security

>> No.11275835

>>11275816
why would I be afraid of someone's ignorance? Their ignorance is a weakness that I could exploit.

>> No.11275840

>>11275835
you clearly are afraid of your own ignorance. have you been exploiting it?

>> No.11275844

>>11275840
sorry dude, you're projecting your own incompetency onto this other guy. not everyone is coping when they tell you they're confident in themselves.

>> No.11275850

>>11275840
>afraid of my own ignorance
nope. I know what I know and what that gets me, what I don't know and what I miss out on for it, and how to learn what I want. But why use the word 'afraid'? What could there possibly be about math that would strike fear in my heart? Fear is when angry Mexican rebels held me at gunpoint trying to get their government to do shit. Fear is when I walked into an out house and almost backed into a mother scorpion. Fear is when I went drunk driving on the freeway during morning rush and almost swerved into someone. Fear is not, "math is hard, am I stupid?" You've shown that you are not just a brainlet, but also an autist.

>> No.11275862

>>11275850
>>11275840
Fear prompts action. Fear caused my to take the US govt travel suggestions more seriously. Fear caused me to be even more careful around dark corners in rural areas. Fear made me vow to never drive drunk again. "Fear of my own ignorance," if that's even a real thing, did not cause me to obsessively study analytic number theory, just because I realize I'm not so great at it. Not even fear of the final exam did that for me. It's a good thing I have a math degree and I'm too old to get drafted, otherwise fear of Trump's WW3 would cause me to start studying engineering 24/7.

>> No.11276726

The result isn't true on [math]\mathcal{K
}_\zeta[/math] abelianized manifolds

>> No.11276787

>>11275033
algebraically: write [math]f(x+h) = f(x) + f'(x)h + o(h)[/math] and [math]g(x+h) = g(x) + g'(x)h + o(h)[/math]. now divide [math]\tfrac{f(x+h)}{g(x+h)}[/math] with a remainder, the first order term in [math]h[/math] is the derivative.

the same thing geometrically: watch the video on the product rule and now work backwards. denote [math]f[/math] the whole square and [math]g[/math] one of the sides. then [math]\tfrac{f}{g}[/math] is precisely the other side. now it's just simple geometry to write change in [math]\tfrac{f}{g}[/math] using change in [math]f[/math] and [math]g[/math].

>> No.11276795

>>11275033
Write out the fractions as f(x) * g(x)/g(x)^2. Then product rule

>> No.11278762

>>11275056
it just werks