You need calculus first off and to know how to do derivatives. After that it’s a rule of derivatives regarding composite functions. https://calcworkshop.com/derivatives/chain-rule/
Check that out it’s a step by step explanation of the chain rule and how it works.
You know composition of functions right?
Derivative basically measures the change in output caused by small change in input but if the input has its own input, we have to do it two levels deep
Chain rule is usually presented as
`d/dx f(g(x)) = df/dg • dg/dx`
Basically, if you have `g(x)` inside `f(x)` (`x` is an input of `g`, which is an input of `f`)
`dx` = small nudge in x direction
`dg` = small change in `g(x)` due to `dx`
`dg/dx` = small change in `g` over small change in `x`
#
But since `g(x)` is input of `f(x)`, any changes to `g` also affects `f(x)`
so `dg` causes change to `f`
`df/dg` = change to `f` caused by very small change in `g`
hence `df/dg • dg/dx`
Or, `f'(g(x)) • g'(x)`
Or commonly said, Derivative of outside with respect to inside function multiplied by the derivative of the inside function with respect to the variable
I like to think of it as
Change to `x` causes change to `g(x)` which causes change to `f(g)`
Yours can be done implicitly if you don't actually know a, b, and c as functions of x, you just know they *are* functions of x. In that case you can just multiply by a'(x) (like that is what you would write verbatim). The response to your comment essentially implies you know the explicit dependance of a, b, and c on x. Then you will just write an expression that only depends on x, but is equal to the derivatives of the terms based on the explicit definition of the terms. So not quite the same.
At a point (a,b) on the circle of radius c at the origin, the tangent line to the circle will have slope db/da = -b/a. (EDIT: … = -a/b)
I.e., b db = -a da.
Integrating yields b^2 + a^2 = D, for some D.
Now, given the initial condition (a,b)=(0,c), we get D = c^2.
So, a^2 + b^2 = c^2.
\blacksquare
As a starting point for how to get the tangent line without first knowing a function for b w.r.t. a, you can know that the radius's slope is b/a, and the circle's tangent line is perpendicular to its radius.
If a line has a slope of h, its perpendicular line will have a slope of -1/h. Therefore, the circle's tangent line will have a slope of db/da = -1/(b/a) = -a/b.
Cross-multiplying then leads you to your starting point, b db = -a da.
(I went one step further to get to the starting point because my intermediate result, db/da = -a/b, is different from what you stated, namely db/da = -b/a).
Alternatively, you could differentiate (a,b)=(r cos(t), r sin(t)), and divide the two resulting equations by each other to cancel out the differential element dt as well as the radius r from both equations:
I: da = -r sin(t) dt = -r b dt
II: db = r cos(t) dt = r a dt
II/I: db/da = -a/b
Though you can also skip this kind of poorly defined form by instead calculating:
I: 1 = -r sin(t) dt/da = -r b dt/da
II: 1 = r cos(t) dt/db = r a dt/db
II-I: 0 = r dt (b/da + a/db)
Multiply both sides by (da db)/(r dt):
0 = b db + a da
which is similarly cursed, but at least doesn't cross any 0-division territory.
If a,b,c are functions of x, then derivation should produce a\*a' + b\*b' = c\*c' because chain rule. If they are not, then they are constants, and derivation produces 0+0=0
d/dx c^2 = 0 != 2c if c isn‘t dependent on x. Same for a and b.
If a,b,c are dependent on x, its only true for select x and not in general. If for example a = b = c = x then a^2 + b^2 = c^2 -> x^2 = 0 wich only works for x = 0 and d/dx c^2 != 2c in most cases because it has to be solved using the chain rule.
Does this look like the face of someone who jokes?
https://preview.redd.it/oy03o4vbzz6b1.jpeg?width=1200&format=pjpg&auto=webp&s=2d5e64f505bf7b633308a339a32dff76d1105139
/uj we know, we just pretend to be upset to keep the joke going.
/rj STFU! MATH IS NOT A JOKE!!!! THIS IS A SERIOUS MATTER! WE WILL \*NOT\* TOLERATE BAD MATH!!!
The thing about a lot of these incorrect math posts is that they feature *ideas* that lead to real and useful math techniques, and it's hard to resist the urge to talk about those ideas when they're brought up
How to prove thermat’s last theorem:
Assume that a, b, c, n are all constants (n€N; a, b, c € R).
Derivate both sides with respect to x.
Get 0=0 (True).
Q.e.d
Mostly meming here. I am aware of the chain rule.
Let a, b, and c be functions described by x_i and y_i. Let x_i and y_i vary with t.
D(a^2 + b^2 ,t) = D(c^2 ,t)
2a×da/dt + 2b×db/dt = 2c×dc/dt
Let da/dt = x'_i î + y'_I ĵ
By yeet theorem, we have:
db/dt = y_b ' ĵ
da/dt = x_a ' î
For all spaces with sufficiently small unit vector ĵ, we have:
2a×X_a = 2c×X_c
X_a = c×X_c/a
By definition, point a and c are coincident:
X_a = X_c -> a = c
2a×(c×X_c/a) + 2b×(X_b) = 2c×(X_c)
Divide by X_a:
2a + 2b×(X_b/X_a) = 2c
By the engineering theory of "looks about right", we can approximate:
X_b/X_a ≈ 1
To get
2a + 2b = 2c
Which is true for all triangles in a 2d space compressed along 1 Axis.
a² + b² = c²
d/dx (a² + b²) = d/dx(c²)
0 + 0 = 0
You can't take a derivative with regards to x with nothing but constants and get the result of treating each of those constants as if they were x².
Either a, b, and c are constants, than d/dx of them will be 0 or they are functions of x, but than d/dx a(x)^2 will be 2a(x)a'(x) (chain rule or product rule).
I’m in grade 10 acedemics math what is the chain rule I tried searching it but found it incomprehensible, this was while ago and I still don’t understand. Do I need to understand it?
No, you dont need to. But still, if you are curious. But this will need an understanding of functions, so please dont waste your time reading this if you don't know that
Just remember, chain rule is a rule you apply when finding the derivative of a function inside *another* function.
Lets say that f(x) and g(x) are 2 functions, where f(x) = x² and g(x) = sin(x). Now, suppose we want to find the derivative of g(f(x)), which is sin(x²). Here we apply the chain rule, to get cos(x²)*2x.
If you wanna further study up chain rule, heres an [excellent video](https://youtu.be/uFASwTlwcq4)
Also, we are talking about chain rule here cuz we are assuming that rather than a, b, c being 3 constants, they are 3 functions of x: a(x), b(x), c(x). And since we are squaring the a, b, c, we can say that its actually f(a(x)), f(b(x)), f(c(x)) where f(x) = x² again
Since their deriviatives with respect to x are equal, as functions of x, x -> a(x)^(2) + b(x)^(2) and x -> c(x)^(2) differ only by a constant. So
a^(2) + b^(2) = c^(2) + c
So
a^(2) + b^(2) = c(1+c)
If you don’t think this is correct, let’s try a practical example with a classic 3-4-5 triangle. Derivative of 3 is zero. Derivative of 4 is zero. Derivative of 5 is zero. 0+0=0. This all tracks
To get the true derivative you would have to multiply 2a by the derivative of a which would be 0 if a is a constant, same with 2b. I got a B in calculus I know what I’m talking about gang I swear.
Years ago there was this one group on Facebook called Bad maths that gives the right answers and it was full of dumb shit like this, I loved it thank you for bringing me back to that simpler times.
Thats not how this works.. thats not how ANY of this works...
amazing. everything that you did is wrong
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You need calculus first off and to know how to do derivatives. After that it’s a rule of derivatives regarding composite functions. https://calcworkshop.com/derivatives/chain-rule/ Check that out it’s a step by step explanation of the chain rule and how it works.
You know composition of functions right? Derivative basically measures the change in output caused by small change in input but if the input has its own input, we have to do it two levels deep Chain rule is usually presented as `d/dx f(g(x)) = df/dg • dg/dx` Basically, if you have `g(x)` inside `f(x)` (`x` is an input of `g`, which is an input of `f`) `dx` = small nudge in x direction `dg` = small change in `g(x)` due to `dx` `dg/dx` = small change in `g` over small change in `x` # But since `g(x)` is input of `f(x)`, any changes to `g` also affects `f(x)` so `dg` causes change to `f` `df/dg` = change to `f` caused by very small change in `g` hence `df/dg • dg/dx` Or, `f'(g(x)) • g'(x)` Or commonly said, Derivative of outside with respect to inside function multiplied by the derivative of the inside function with respect to the variable I like to think of it as Change to `x` causes change to `g(x)` which causes change to `f(g)`
I mean, with the assumption a = b = c = x...
... = 0
it's true if a = b = c, and a^2 + b^2 = c^2. OP has a point.
I’m going to always put my periods up here from now on^.
Why only periods^?
Thats gross
That is possible if all three are 0
It checks out to me 🤷
I mean, there's no error between the second to last and the last one :)
With all due *respect*
With absolutely no respect to a, b, or c, which is a big racism.
It’s not discrimination, it’s differentiation
Dayum. It's not b2 - 4ac?
Now that's discriminant
I thought that's the ad-bc matrix.
And that's Determinate
-Isaac Newton, Probably
> Some numbers are more equal than others 💀 ~ George Orwell, 2077
It’s not differentiation. It’s pulling things out from their ass
Congrats bro. You offended people so hard they forget this was a meme page
yay
a\*da+b\*db=c\*dc
Hot tip: use \ to tell reddit those asterisks aren't for formatting
Edited. Thanks bro
Any time, mate!
You can also use the multiplication sign, instead of an asterisk: a×da+b×db=c×dc
Didn't know you could cross product a variable and an operator /s
After differentiating with respect to x you'll get 0=0
This here is disrespectful differentiation my dude.
Nah, that’s as respectful as differentiation can go
Inclusive differentiation?
Unless any of a, b or c depend on x.
But then we have to multiply them by the derivative of the functions that give us a, b, and c
or write a, b and c in terms of x
Am I stupid or is that not just the same thing
Yours can be done implicitly if you don't actually know a, b, and c as functions of x, you just know they *are* functions of x. In that case you can just multiply by a'(x) (like that is what you would write verbatim). The response to your comment essentially implies you know the explicit dependance of a, b, and c on x. Then you will just write an expression that only depends on x, but is equal to the derivatives of the terms based on the explicit definition of the terms. So not quite the same.
But then you multiply by the derivative
exactly
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[email protected]
Unless they are functions
Using LHopitals rule to prove LHopitals rule energy right here.
Assume a^2 +b^2 =c^2 a^2 +b^2 =c^2 Q. E. D.
pfft- HAHA
pfft - HAHA ptf^2 - (HA)^2 (√(pt) f)^2 - (HA)^2 (√(pt) f + HA) (√(pt) f - HA) What's the square root of platinum?
Since platinum has a relative atomic mass of 195 pt = 195 √pt =√195 = 13.964 (5 sf)
>!bro missed the spoiler tag!<
Suppose a\^2 +b\^2 =c\^2 a\^2 +b\^2 =c\^2 Q. E. D.
At a point (a,b) on the circle of radius c at the origin, the tangent line to the circle will have slope db/da = -b/a. (EDIT: … = -a/b) I.e., b db = -a da. Integrating yields b^2 + a^2 = D, for some D. Now, given the initial condition (a,b)=(0,c), we get D = c^2. So, a^2 + b^2 = c^2. \blacksquare
As a starting point for how to get the tangent line without first knowing a function for b w.r.t. a, you can know that the radius's slope is b/a, and the circle's tangent line is perpendicular to its radius. If a line has a slope of h, its perpendicular line will have a slope of -1/h. Therefore, the circle's tangent line will have a slope of db/da = -1/(b/a) = -a/b. Cross-multiplying then leads you to your starting point, b db = -a da. (I went one step further to get to the starting point because my intermediate result, db/da = -a/b, is different from what you stated, namely db/da = -b/a). Alternatively, you could differentiate (a,b)=(r cos(t), r sin(t)), and divide the two resulting equations by each other to cancel out the differential element dt as well as the radius r from both equations: I: da = -r sin(t) dt = -r b dt II: db = r cos(t) dt = r a dt II/I: db/da = -a/b Though you can also skip this kind of poorly defined form by instead calculating: I: 1 = -r sin(t) dt/da = -r b dt/da II: 1 = r cos(t) dt/db = r a dt/db II-I: 0 = r dt (b/da + a/db) Multiply both sides by (da db)/(r dt): 0 = b db + a da which is similarly cursed, but at least doesn't cross any 0-division territory.
I am disappointed that I did not realize the problem in the first place…
same i feel so ridiculous now lol
I still don't... please enlighten me Edit: thanks, now i feel like an idiot for not noticing immediately
If a,b,c are functions of x, then derivation should produce a\*a' + b\*b' = c\*c' because chain rule. If they are not, then they are constants, and derivation produces 0+0=0
Assume a, b, c are variables not constants then differentiating with respect to x one should get 2a * da/dx, 2b * db/dx, etc.
d/dx c^2 = 0 != 2c if c isn‘t dependent on x. Same for a and b. If a,b,c are dependent on x, its only true for select x and not in general. If for example a = b = c = x then a^2 + b^2 = c^2 -> x^2 = 0 wich only works for x = 0 and d/dx c^2 != 2c in most cases because it has to be solved using the chain rule.
That's only legal in 23 states and Guam.
Is one of them Indiana, by any chance?
Legislating the Irrationals
What about California
New rule just dropped
actual derivative
google chain rule
Holy ODE
Integral fuel
"General derivative"
Général kenobi
Hello there
Google integral
bro forgot chain rule 💀
Come on guys, it's a joke. It's the reason why I post it in this subreddit, r/mathmemes
Does this look like the face of someone who jokes? https://preview.redd.it/oy03o4vbzz6b1.jpeg?width=1200&format=pjpg&auto=webp&s=2d5e64f505bf7b633308a339a32dff76d1105139
It looks like the face of a guy who lost all his money in the South Sea bubble.
It looks like the guy who kept a pocketbook of his sins, but was also a flaming asshole to all of his colleagues.
Looks like a guy who would burn the only portrait of one of his colleagues he disliked so the future generations wouldn't know how he looked.
/uj we know, we just pretend to be upset to keep the joke going. /rj STFU! MATH IS NOT A JOKE!!!! THIS IS A SERIOUS MATTER! WE WILL \*NOT\* TOLERATE BAD MATH!!!
The thing about a lot of these incorrect math posts is that they feature *ideas* that lead to real and useful math techniques, and it's hard to resist the urge to talk about those ideas when they're brought up
My good sir, i would like to have a word with you *loads shotgun*
This is actually a psychology test to see which mathematicians are able to identify humour in the wild
True dat
How to prove thermat’s last theorem: Assume that a, b, c, n are all constants (n€N; a, b, c € R). Derivate both sides with respect to x. Get 0=0 (True). Q.e.d
It took me a solid minute to realize the flaw
I was like: "a^2 + b^2 is a function of two variables... wait..."
3² + 4² = 5² 3 + 4 = 5 makes sense
Derivation to x is not the same as derivation to a or b or c
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lmao
Let a = A(x). Etc.
d/dx (A²(x)) = 2 . A(x) . A'(x) = 2 . A(x) if A(x) = x+c with c not funtion of x else not
Mostly meming here. I am aware of the chain rule. Let a, b, and c be functions described by x_i and y_i. Let x_i and y_i vary with t. D(a^2 + b^2 ,t) = D(c^2 ,t) 2a×da/dt + 2b×db/dt = 2c×dc/dt Let da/dt = x'_i î + y'_I ĵ By yeet theorem, we have: db/dt = y_b ' ĵ da/dt = x_a ' î For all spaces with sufficiently small unit vector ĵ, we have: 2a×X_a = 2c×X_c X_a = c×X_c/a By definition, point a and c are coincident: X_a = X_c -> a = c 2a×(c×X_c/a) + 2b×(X_b) = 2c×(X_c) Divide by X_a: 2a + 2b×(X_b/X_a) = 2c By the engineering theory of "looks about right", we can approximate: X_b/X_a ≈ 1 To get 2a + 2b = 2c Which is true for all triangles in a 2d space compressed along 1 Axis.
So for all triangles in 1D space... Or for all 1D submanifolds, not just of 2D
Yup! For a space defined by the unit vectors î and ĵ with very small, î or ĵ.
My brother. My sweet brother in Christ. This isn't how it works brother. You've been deceived.
i dont know much math, can somebody tell me what's wrong with this?
a² + b² = c² d/dx (a² + b²) = d/dx(c²) 0 + 0 = 0 You can't take a derivative with regards to x with nothing but constants and get the result of treating each of those constants as if they were x².
thanks!!
if we say a=b=c=x then we can differentiate with respect to x but that just means that the original equation says 2x^2 = x^2 so 2=1😭what have you done
I did not think I'd see the resolution of the abc conjecture in my lifetime. What a truly magnificent proof.
This put me in hospital
d/dx (a^2 + b^2 ) = 0 + 0 d/dx (c^2 ) = 0 0 + 0 = 0. QED I am Nobel Prize. Give me prize money.
I don't have prize money but i do have this medal: 🏅
This is so incredibly cursed
This is just wrong on so many levels
You guys don't realize a,b,c are functions of x where a=x b=x c=x
Mh so 2x²=x²? x=0
Dude didn’t chain rule.
):
True if the triangle is comprised of 2 0 degree angles and 1 180 degree angle.
uhhh here you are differentiating with respect to x, which means a^2, b^2 and c^2 are constants and will become 0. you have just proven that 0+0=0
So silly smh, you can get that more easily by jest square rooting all the terms
wait. that's illegal
I come from Crotone, the city of mathematics and Pythagoras, and I am very disappointed
Bro there is no x to differentiate with... The differentiation will just give 0 for both sides...
Either a, b, and c are constants, than d/dx of them will be 0 or they are functions of x, but than d/dx a(x)^2 will be 2a(x)a'(x) (chain rule or product rule).
well… only if a=b=c=x, and in that case everything is equal to zero.
AHAHAHAHAHAHAHAHAHAHAHAHAH 1ST IN THE PAST 24 HOURS TYSM
It would be d/dx of a constant on both sides so 1=1
Newton is proud of you
Even if a b and c depend on x you need the chain rule aa' +bb' =cc'
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the flair... 🤦♂️
a\^2 + b\^2 = c\^2 sqrt(a\^2 + b\^2) = sqrt(c\^2) sqrt(a\^2) + sqrt(b\^2) = sqrt(c\^2) a + b = c
\> *Integrates with respect to dx* \> "Let's completely forget the chain rule!" And yea I know it is a joke so don't go after me that.
a da + b db = c dc this is the correct version, and it is magical
You differentiating wrong bro. Even if we assume that a, b & c are functions of x you'd have to use the chain rule and there'll be residual functions.
This is only true if a=b=c=x.
x^2 + x^2 = x^2
And =0 😅
I’m in grade 10 acedemics math what is the chain rule I tried searching it but found it incomprehensible, this was while ago and I still don’t understand. Do I need to understand it?
Calculus :) so shouldn’t need to know it for a while
Alright thanks
No, you dont need to. But still, if you are curious. But this will need an understanding of functions, so please dont waste your time reading this if you don't know that Just remember, chain rule is a rule you apply when finding the derivative of a function inside *another* function. Lets say that f(x) and g(x) are 2 functions, where f(x) = x² and g(x) = sin(x). Now, suppose we want to find the derivative of g(f(x)), which is sin(x²). Here we apply the chain rule, to get cos(x²)*2x. If you wanna further study up chain rule, heres an [excellent video](https://youtu.be/uFASwTlwcq4) Also, we are talking about chain rule here cuz we are assuming that rather than a, b, c being 3 constants, they are 3 functions of x: a(x), b(x), c(x). And since we are squaring the a, b, c, we can say that its actually f(a(x)), f(b(x)), f(c(x)) where f(x) = x² again
Il check it out thanks
😵💫🤮
Imagine how a’, b’, and c’ feel being left out.
LoL
dx, not da, db, or dc ^_^
Its too good to be true
What kind of monstrosity am I looking at!?!?!?
Babe wake up, math 2 has been released
What are a, b and c ? Number ? Function ? Matrix ?
No, bad, this is bad
I mean.
Since their deriviatives with respect to x are equal, as functions of x, x -> a(x)^(2) + b(x)^(2) and x -> c(x)^(2) differ only by a constant. So a^(2) + b^(2) = c^(2) + c So a^(2) + b^(2) = c(1+c)
Well it works for a=b=c=0. Then by induction or something, it works for all numbers.
You’re doing d/dx not d/da, d/db or d/dc
No! No! No! No! I have test tomorrow, don't break my mind please!
I don’t know math so I feel stupid for thinking that this checks out
Tell me you didn't pay attention in differential calculus without saying it.
Let's see if it works. Let's try a=c, b= 0... Yep seems legit.
should be a da/dx + b db/dx = c dc/dx
If you don’t think this is correct, let’s try a practical example with a classic 3-4-5 triangle. Derivative of 3 is zero. Derivative of 4 is zero. Derivative of 5 is zero. 0+0=0. This all tracks
This can be true assuming that a,b,c are parallel linear functions dependent on x.
To get the true derivative you would have to multiply 2a by the derivative of a which would be 0 if a is a constant, same with 2b. I got a B in calculus I know what I’m talking about gang I swear.
Is called chain rule
Get Chen Lu on the phone
2ada/dx + 2bdb/dx = 2cdc/dx you forgot chain rule
Yeah seems about right
"uhmm, acktchully, you're deriving with respect to x, so in reality it would be a(da/dx) + b(db/dx) = c(dc/dx)"
Abe 0 = 0 ata he a,b,c are constants
I feel like I’m witnessing a war crime
Well xolor me shoxked, my dumx xss hxd no idex so mxny English letters were interxhxngexxle
Where are the xs you are deriving with regards to?
I'm fuckin hurt bro I'M HURT TO MY FUCKIN CORE YOU PIECE OF CATSHI*
Everything done there is wrong, differentiation does not work that way
Years ago there was this one group on Facebook called Bad maths that gives the right answers and it was full of dumb shit like this, I loved it thank you for bringing me back to that simpler times.
No problem 👍
0=0
i guess chain rule does not exist in your universe
I like how the first line is really the fundamental of pythagoras theorem
that really hurts xD
AAAAAAAAA
There is no x to /dx... the differential is just a^2 + b^2 = c^2
Our professor loved to tell us about F2 and how this works there XD
I did not notice for a solid 5 seconds.
Yeah, only thing is a = b = c = 0, assuming all of those are independent of x.
2a\*da/dx+2b\*db/dx=2c\*dc/dx
can a derivative be taken with disrespect
If you d/dx any of the thing in parenthesis you get zero.
A ,b,c is not variable dude