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It even works in a noncommutative case, though then we can’t write it like that simple formula because now we can’t commute factors: for non-commuting F G, (F G)' = F' G + F G' but not F' G + G' F.
Bri, your videos make math so much more fun! Honestly, your passion for teaching is inspiring. I’ve always thought math was intimidating, but your energy and clarity have completely changed my perspective. What got you into teaching math like this? Also… do you ever run out of clever examples, or is it just genius all the time? 😂
Implicitly, obviously, as we do whenever we encounter new inverse functions. To get d/dx arcsin(x), you say y=arcsin(x). Then sin(y)=x. Now differentiate. cos(y) dy/dx=1. dy/dx=1/cos(y)=1/√(1-sin²(y))=1/√(1-x²) To get d/dx ln(x), you say y=ln(x). Then e^y=x. Now differentiate. e^y dy/dx=1. dy/dx=1/e^y=1/x To get d/dx W(x), you say y=W(x). Then ye^y=x. Now differentiate. (e^y+ye^y) dy/dx=1. dy/dx=1/(e^y+ye^y)=y/(ye^y (1+y))=W(x)/(x(1+W(x)) I hope by now you can see, or at least figure out, the general pattern for d/dx f^-1(x). It's 1/f'(f^-1(x))
OK, I understand the proof at 2:23. Basically, the 2nd-to-last line has two addends. The 1st addend is the sum from the kth case, but just with a factor of the differentiated-new-function given to each term of that sum. The 2nd addend is simply the only other term that we need for the formula-extension -- the one where the new function is differentiated and then multiplied by all other functions.
This is eye-opening. Never bother to think about it as I thought it is like calculating the determinant of a 4 by 4 matrix, that is rudimentary arithmetic in several layers to get ot 3 by 3 matrix and then 2 by 2. Thanks!
You can't because you use the n=2 product rule for every single induction step. To go from n=1 to n=2 you need the n=2 product rule. So might as well just use the n=2 as the base case
@@yoavmor9002 I've seen tons of other induction proofs where the base case is trivial, such as the one where the sum of n integers is n(n+1)/2. The mechanics of the rule you're trying to prove is laid out in the n=k case
You would need to prove it using the definition of derivative as well as other limit laws (including epsilon-delta) which is where real analysis comes in.
Oh - that's really potentially practically useful knowledge! Thanks for making it more well known! I wonder why this isn't taught in calc courses. Is there a similar generalization for the chain rule? The only generalization of the chain rule that I currently know about is the multivariate chain rule. But here, it would have to be a rule for compositions of more than two functions. Update! Yes there is! It can be found on the wikipedia page about the chain rule! Nice!
@ from the scp foundation kind of talk. if you know scp channels. Most-recommend "Doctor Bob" considering it is a channel that tells a short animated story in addition to describing "scps".
64 people playing games and talking, Black hole threatens all of them so blue hand stops them, their name is four with x at toe, can we trust them i don't know, they Said "don't you wanna battle for a prize, you could win a bfdi" (not typing out the rest sorry :3)
talking about the script or something? Also I felt this was obvious, but one cool thing I suppose was showing a generalized equation for the product rule. But Idk, I felt this video was obvious, at least in how you'd take the product rule of multiple functions. Maybe Im missing the point tho. (Also sorry most of this meant to be its own comment but I was too lazy to do that).
Lemme do the funni Wanna get the derivative of xⁿ ? express xⁿ as x·x·x·x·...·x n times... then use the generalised product rule, considering that fₖ'(x) = 1 ⟹ the first factor of the summatory will be always 1, so we ignore it. There's only a productory inside the summation now. ∀i (fᵢ(x) = x) ⟹ the productory will be just x·x·x·...·x n times, with one term being derived (= 1). So correcting, (x·x·x·...·x n-1 times.), which is itself equal to xⁿ⁻¹. n so Σ ( nⁿ⁻¹) i=1 which is just the argument added to itself "n times", then... the final formula will be: xⁿ = n · xⁿ⁻¹ Derived the power rule.
Incredibly cursed but makes sense. It's just some arrangement of 1 * x * x ... (with n terms, where n-1 terms are x and 1 is x) n times = nx^n-1 Fun part of generalised relations is that you can always overkill proofs with them.
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I've never heard of the "produce rule"
"Eat all your greens, fool!" -- Mr T
That's the produce rule
@@neilgerace355 😆
It even works in a noncommutative case, though then we can’t write it like that simple formula because now we can’t commute factors: for non-commuting F G, (F G)' = F' G + F G' but not F' G + G' F.
good point, I'm always wondering what's gonna work or not with formulas that use products
Bri, your videos make math so much more fun! Honestly, your passion for teaching is inspiring. I’ve always thought math was intimidating, but your energy and clarity have completely changed my perspective. What got you into teaching math like this? Also… do you ever run out of clever examples, or is it just genius all the time? 😂
i love the produce rule!
totally better than the product rule.
I think the proof still works using a base case of n=1 or even n=0, although you still have to use the normal product rule in the induction step.
How do we take a derivative of the Lambert's W function ?
Implicitly, obviously, as we do whenever we encounter new inverse functions.
To get d/dx arcsin(x), you say y=arcsin(x). Then sin(y)=x. Now differentiate. cos(y) dy/dx=1. dy/dx=1/cos(y)=1/√(1-sin²(y))=1/√(1-x²)
To get d/dx ln(x), you say y=ln(x). Then e^y=x. Now differentiate. e^y dy/dx=1. dy/dx=1/e^y=1/x
To get d/dx W(x), you say y=W(x). Then ye^y=x. Now differentiate. (e^y+ye^y) dy/dx=1. dy/dx=1/(e^y+ye^y)=y/(ye^y (1+y))=W(x)/(x(1+W(x))
I hope by now you can see, or at least figure out, the general pattern for d/dx f^-1(x). It's 1/f'(f^-1(x))
@@xinpingdonohoe3978 Thank you man , I really appreciate it
Is this the reverse of the DI method for integration by parts? Basically do it until you get the final answer?
The product rule *is* the "reverse" of integration by parts.
OK, I understand the proof at 2:23. Basically, the 2nd-to-last line has two addends. The 1st addend is the sum from the kth case, but just with a factor of the differentiated-new-function given to each term of that sum. The 2nd addend is simply the only other term that we need for the formula-extension -- the one where the new function is differentiated and then multiplied by all other functions.
Very nice video, thank you for showing this :)
True.
No one in university told me this. Only a single task from the taskbook made me question it, making it not so obvious.
This is eye-opening. Never bother to think about it as I thought it is like calculating the determinant of a 4 by 4 matrix, that is rudimentary arithmetic in several layers to get ot 3 by 3 matrix and then 2 by 2. Thanks!
Amazing! You gave proven something I have proven for fun multuple months ago.
i bet you could make n=1 the base case instead of n=2, since n=k embeds the mechanics of the rule
You can't because you use the n=2 product rule for every single induction step. To go from n=1 to n=2 you need the n=2 product rule. So might as well just use the n=2 as the base case
@@yoavmor9002 I've seen tons of other induction proofs where the base case is trivial, such as the one where the sum of n integers is n(n+1)/2. The mechanics of the rule you're trying to prove is laid out in the n=k case
i never took real analysis so i didn't even realize that i never saw a proof of the 2-function case. thanks
You would need to prove it using the definition of derivative as well as other limit laws (including epsilon-delta) which is where real analysis comes in.
So interesting. Out of curiosity, what software do you use to create your videos that give it that dynamic blackboard effect?
Manim, a python library
The product rule becomes more interesting if one looks at higher derivatives.
I prefer the "logarithm rule"
Oh - that's really potentially practically useful knowledge! Thanks for making it more well known! I wonder why this isn't taught in calc courses. Is there a similar generalization for the chain rule? The only generalization of the chain rule that I currently know about is the multivariate chain rule. But here, it would have to be a rule for compositions of more than two functions.
Update! Yes there is! It can be found on the wikipedia page about the chain rule! Nice!
Why did they teach logarithmic differentiation to me instead of this in high school?
Mmmm ... Produce rule
Great video, wasn’t taught this either but always have known it if that makes sense? Like I find it common sense and, more or less, obvious
I did verify that using mat mul
Interesting!
What next video? it didn't show up
made me think "careful one doesn't produce an anomalous effect.".
Is that a saying, I don’t get it. Seems like a fun phrase though 🙂
@ from the scp foundation kind of talk. if you know scp channels. Most-recommend "Doctor Bob" considering it is a channel that tells a short animated story in addition to describing "scps".
@@God-ld6ll’ll’look into it
shoutout please sirr, im a big fann.
64 people playing games and talking, Black hole threatens all of them so blue hand stops them, their name is four with x at toe, can we trust them i don't know, they Said "don't you wanna battle for a prize, you could win a bfdi" (not typing out the rest sorry :3)
Now do the chain rule for the composition of n functions😮😮
Which AI btw? At the end it became a bit too obvious lol
talking about the script or something? Also I felt this was obvious, but one cool thing I suppose was showing a generalized equation for the product rule. But Idk, I felt this video was obvious, at least in how you'd take the product rule of multiple functions. Maybe Im missing the point tho. (Also sorry most of this meant to be its own comment but I was too lazy to do that).
what are you talking about...
Lemme do the funni
Wanna get the derivative of xⁿ ?
express xⁿ as x·x·x·x·...·x n times...
then use the generalised product rule, considering that
fₖ'(x) = 1 ⟹ the first factor of the summatory will be always 1, so we ignore it. There's only a productory inside the summation now.
∀i (fᵢ(x) = x) ⟹ the productory will be just x·x·x·...·x n times, with one term being derived (= 1). So correcting, (x·x·x·...·x n-1 times.), which is itself equal to xⁿ⁻¹.
n
so Σ ( nⁿ⁻¹)
i=1
which is just the argument added to itself "n times", then...
the final formula will be:
xⁿ = n · xⁿ⁻¹
Derived the power rule.
Incredibly cursed but makes sense.
It's just some arrangement of 1 * x * x ... (with n terms, where n-1 terms are x and 1 is x)
n times
= nx^n-1
Fun part of generalised relations is that you can always overkill proofs with them.