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That was a great video on the topic!
Nope… not having this…
No .
I think the answer is at the start of the video. Infinity times, anything isn't infinity. Infinity times, anything positive is infity and infinity times anything negative is minus infinity. Infinity times zero is neither. Like any other number times 0, infinity times 0 should be one.
Thing is, the chain rule itself is most easily justifiable by thinking of the derivative like a fraction: like, if I have dy/dx = y'(x) => dy = y'(x) dx Suppose now that dx/dt = x'(t) Meaning that: dx = x'(t) dt Therefore dy = y'(x) * x'(t) dt => dy/dt = y'(x) x'(t) Notice that at more or less every step I'm manipulating differential elements of x and t like they're fractions? I don't think it's sufficient to say that fractional behaviour can be satisfactorily explained with the chain-rule, because it's not clear that whatever's behind the chain-rule isn't itself just treating derivatives like fractions.
Why did i get 1/3
😂😂😂😂😂 simply its 1 dont complicate it ,weird mathematicians 😂😂
there is a much simpler way to solve it. When you have the ln(tan(theta)+1)) just use king property to get ln(tan(pi/4-theta)+1) and it just simplify like butter. You can also try x = (1-t)/(1+t) it work good with log.
I have got a doubt here... 1-1+1-1+1-1..... is just 0 right? grouping 1-1 together. Guys i have no clue about it so please help. Nice video anyways😊😊
*How* is dy/dx not a fraction? Is a derivative not defined as lim (x2 -> x1+) (f(x2) - f(x1)) / (x2 - x1)) ? Is dy not the numerator and dx the denominator?
It's n-1 on n and n=that number
EZ y=0 x=98
Aren't the first equations approximations?
I would say that the answer to the derivative at a known value IS a fractional value of the instantaneous change in y compared to x. As an example it could be the slope of a function where y depends upon x. So writing it that way actually makes sense. That’s the best I’ve got today. (On the integral side I think most people use the dx notation so it’s a bit symmetric.)
It's been decades since I took calculus in high school and college, but how is dy/dx NOT a fraction? It's the change in y over the change in x.
Because it's a limit. When a limit is undetermined you can't rearrange the terms in an equation since Lim(f/g) is not necessarily Lim(f)/Lim(g), and this notation implies that.
Calculus without limits. Once you realize that dx^2 = 0 is an EASY construction (dx = [[0 1] [0 0]]) you can do all derivatives without the use of limits at all. It's really a SIMPLE THING. f(x+dx) is just a 2x2 matrix. NO LIMITS
When we are first taught about derivatives.. we are told that dy/dx comes into shape when ∆x in ∆y/∆x approaches zero.. Here ∆y/∆x is indeed a fraction!! Its just y2-y1/x2-x1... So why is xy/dx not considered a fraction??? Dy and dx are two separate quantities aren't they?
Because Lim(f/g) doesn't equal necessarily Lim(f)/Lim(g) when the limit is undetermined (and this is an indetermination 0/0).
x% of y is the same as y% of x
thank you. However, there is a problem in this representation. When Leibniz introduced the infinitesimal notation he was not following your path of justification. This is not just a historical anecdote - it reveals the weakness of this representation: things are magically getting right so that we can act as if we can multiply or divide infinitesimals even though there are no infinitesimals. But this was not a magical coincidence but a system that has an inner logic behind that makes it work. This is missed by the presentation. It is also worth mentioning that Abraham Robinson has developed a modern Leibniz approach - nonstandard analysis.
Derivatives are not quotients in Robinson's approach, either.
Wouldn't you just find the Common denominator of each? Which would be six, so that would just be 6/6 + 3/6 + 2/6 = 11/6
This is just the Fibonacci seguence. 60 seconds will never explain mathematics.
So 1^∞︎︎
This sounds like an AI voice.....
x=8,Y=10 .
I dub this ' reductio ad absurdum'....QED! 🤔
So sad I wasn't shown this in Calculus!
Why? He just making it harder than it has to be.
@@reh3884 In the early Calculus days it felt like they were just making up the algebraic rules for what could and could not be done with Calculus. For anyone who thinks deeply about what is happening during each step of a problem, blindly copying an example that doesn't make sense is the harder way.
In engineering textbooks it is a common practice
It is impossible to find.
That is basically for ratios too, yeah ratios fractions are confussing if u think about it deeply,
No, it isn't confusing. A ratio is a comparison. The value of a ratio is not the ratio. It's like the area of a square is not the square itself. There are objects with magnitude, but the magnitude is not the object.
@@CliffSedge-nu5fv the way of thinking that magnitude is not an object is kinda a wrong way if thinking since cm is obviously a measurement and cm² is obviously measurement * measurement. That's the problem that u can't really be good at physics without thinking of it as an object instead of a magnitude
We use magnitudes only in math to avoid writting measurements, objects etc... That's why ratios aren't as simple
@@CliffSedge-nu5fv more over 1l/1000ml just cancels out to be 1 and if you only thought of it as a ratio and magnitude that would of been IMPOSSIBLE
or let x=1/2+1/4+1/8+..., so i factor out 1/2 so i have 1/2(1+1/2+1/4+1/8+..)=x, i replace x in the new expression in order to get 1/2(1+x)=x, so 1/2+1/2x=x, multiply by 2 the two sides and obtain 1+x=2x, bring the x on the other side so x=1. x=1=1/2+1/4+1/8+...
Are there any situations where treating it as a fraction is incorrect. I was taught that you can usually treat dx/dy as a fraction but NOT ∂x/∂y due to rules about partial derivatives. But in engineering classes we CONSTANTLY treated dx (or any variable) as separable from a respective derivative to integrate across an equation and never ran into any problems. I always thought of the “d” as an infinitesimal version of “Δ” or difference, and dx/dy being the infinitesimal rate of change. Since a rate is by definition a quotient, it’s ok to separate them across an equation. Maybe this seems like a primitive understanding compared to the pure math way of interpreting it but it clearly works in engineering and physics.
And to follow up about that exception with partials… taking a “partial derivative” seems just as janky as treating dx/dy as a fraction. It’s literally just a technique we invented to deal with multivariable functions, just as representing a derivative with an infinitesimal quotient is also a technique. I guess I fundamentally disagree with this notion that “it’s just notation, not an actual fraction” because I can’t imagine a situation (besides the aforementioned multi-variate exception) where it can’t be treated as such. Isn’t that how Leibniz designed the notation to be used?
@@samgrattan5465 The thing with partials is that it's still also kind of a fraction, we just can't separate it because the notation is bad. If we have a function of two inputs, z(x, y) then the derivative with respect to x is ∂z/∂x and the derivative with respect to y is ∂z/∂y but the key point is that the ∂z's in those two fractions are not equal. I think if you specified _where_ the ∂z came from, like ∂_x(z) for the differential caused by ∂x, then you wouldn't need the fractional notation to specify where the differential came from and you could separate them. IDK, something I need to look into more.
5:05 isn't this nothing but a simple u-sub? on the left integral we just let u=y(x), and hence du=(dy/dx)dx. after substitution we get ∫f(u)du which is the integral on the right if we switch dummy variable u back to y. Anyway, I think the simple explanation for this topic is that it is indeed wrong to treat dy/dx as fraction and manipulate differentials. It's just notation, not a literal fraction. However, the formal argument of what you are trying to do when incorrectly manipulating these things as fraction follows the same fundamental logic as what you are doing when you are treating it incorrectly, so it's basically just a shortcut. This is why it's okay to mistreat it as a fraction because you'll end up most likely with correct solution; and it's NOT accidentally. Again, if you did it formally you'll see you're doing same sort of logic , e.g. swap dy/dx with limit of delta y / delta x where you can now treat these as fraction etc..
The highest Math I ever took was precalculus 6 years ago when I got my associates. I didn't really want to keep going because I didn't want to take on student loan debt, but I think I would like to now. I was going to retake precalculus in the spring to refresh my memory before calculus but I took one look at the first assignment and I was like "I don't remember any of this." The class was clearly designed for people who just took college algebra a month before. I was going to take college algebra over the summer so I could go back even further, but the class was full. Starting next week I'm taking intermediate algebra. The plan is to take college algebra in the fall, precalculus next spring, then calculus either next summer or in the fall. I figured if i have to build up my skills all over again so be it. I'm already in a good career. This isn't like a barrier to entry to get a good job. I just want to improve my chances of securing something higher later.
Wait, but isnt chain rule literally derivatives also treated as a fraction?
"It is NOT a Fraction >:( " Chain Rule: Are you sure about that?
Proof of chain rule: (f(g(x)))' = Lim ((f(g(x+dx))-f(g(x))/dx). Multiply and divide by g(x+dx) - g(x) and you obtain the rule. Where is the limit treated like a fraction? You are just doing valid manipulations, since g(x+dx) - g(x) ≠ 0 for some dx (g is not a constant function).
@@SergioLopez-yu4cu I meant eg: dy/dx = dy/dt × dt/dx and that's all it was for me in College, I bet in Uni it goes more rigorous and all that, but I don't know about that. It was also mainly a joke.
To repeat my reply to another comment, the problem of not being a fraction is not at all just in partial derivatives. Any derivative higher than the first in dy/dx = 1/(dx/dy) is untrue. Try differentiating this yourself to see why You will find d^2x/dy^2 = -f’’(x)/(f’(x))^3 Where f’(x) = dy/dx As you can see there is a differential relationship between a second derivative (or higher) and its reciprocal, so it is not a fraction Still shocks a lot of people
Bruh why not show how to compute C. Useless video for refreshing this topic
So you've just proof that it works like fraction, so it's a fraction 🙃
dy/dx works as a fraction. No, just because the chain rule also works, it doesn't mean the original is false. That's not how maths works.
This is an HP and as far as i know there exists no formula or any method to find sum of hp
dy/dx isn't an fraction, it's a ratio. Ratio of how a function changes in y, for a small change in x. A fraction is a/b where a & b both are integers, but a ratio doesn't have to have integers, for example π:1 or e:π is a ratio.
Close. You might be confusing ratio with rate. A ratio is not a number, it is a comparison. The value of a ratio could be expressed as a fraction.
Can you help me please, how to solve integral f(x) dlnx. If you saw this please reply to this comment as soon as possible. Its very urgent
Is there a typo? It seems like you need to divide -2.7231 by -Ln(3) to get the nontrivial solution. Looks like two steps were missed.
the chain rule itself is derived by treating it as a fraction. In fact, it is a fraction because it refers to slope which is small change in y divided by small change in x
That's just a shortcut. There are limit proofs of the chain rule that are more bulletproof than proofs involving treating the notation as a fraction.
No, it isn't
The problem with the dy/dx notation is also that y'' = d²y/dx². This is almost never clearly explained when said or used and y'' looks so simple.
Well this is also short hand, and it's using a lot of shenanigans to get to that state y'' = the second derivative of y = the derivative of the derivative of y = the derivative of dy/dx = d[dy/dx]/dx the first weird step is grabbing the dx in the numerator out y'' = d[dy](1/dx)/dx y'' = d[dy]/dx • 1/dx the next weird step is changing d[dy] to d²y, as if we're multiplying the 'd's y'' = d²y/dx • 1/dx then we multiply the 'dx's y'' = d²y/(dx)² Now the part that makes no sense: remove the parentheses. It makes no sense but we do it because nobody's got time for parenthses y'' = d²y/dx² and that's how you get this weird notation
@@xXJ4FARGAMERXx Exactly, but it's never explicitly told that way.
Doesnt the chain rule itself treat derivatives as fractions? Like if dy/dx = dy/du x du/dx, the du's cancel out and thats why it works. Thats how i interpreted it, what am i missing here?
That's not how the chain rule is formally proven. There are limit proofs of the chain rule that are much more rigorous than treating the notation as a fraction.
@@carultch That's doesn't change the fact it works as above.
@@reh3884, of course, it works, but that doesn't mean you can treat like a fraction every derivative just because the composition of functions verifies that (which is pure luck, since the theorem is proven with a limit).
@@reh3884Well, your argument is: -If my bicycle were a car then it would have wheels. -My bicycle has wheels. -Therefore my bicycle is a car. That is: -If dy/dx were a quotient then the chain rule dy/dx = dy/dt • dt/dx would be correct. -The chain rule is correct. -Therefore dy/dx is a quotient.
If we raise both sides of the equation x=°√A (A is some consonant) by the power of 0 we get: x^0=A For A=1 we get x=any number. For any A other than 1, no x satisfies the equation.
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Leibniz had everything worked out when he wrote derivatives as a ratio. That said, it isn't a fraction in the usual sense... But in every form of (total, not partial) derivatives, the very definition of the derivative is a fraction. Hence the chain rule is just abuse of notation.
The chain rule is a theorem, the chain rule using df, dy and this bs is an abuse of notation.
the answer can easily obtained by trial and error method ANSWER: 4