Sir I like you. I really like you. I watched videos of all you-tube maths channels I didn't get what you have explained here. This explanation satisfied me and thank you so much really God bless you🙏🙏❤❤
u and v are simply substitutions for g(x) and h(x) respectively, just like y is a substitution for f(x). So, if: y = f(x) u = g(x) v = h(x) it stands to reason that for every Δx there is a corresponding Δy, Δu and Δv (because each of their equivalent functions are changing as x is changing), and therefore: y + Δy = f(x + Δx) u + Δu = g(x + Δx) v + Δv = h(x + Δx) Don't think of g(x) so much as a function. Think of it as an output value on the vertical axis, u.
Thanks for the explanation. Now that I look at it again after taking a calculus course, I understand. I don't know why I didn't get this before considering it was nicely explained in the video. Thank you very much for the explanation though. :) Sorry for the extremely late reply.
how can you expand (u+deltau)(v+deltav) if they are really f(u+deltau)+f(v+deltav) you are just adding the side lengths of each in the first case but finding the y coordinate of the lengths added together in the second case
Yes, look at videos 3.07a and following. Hopefully that will give you what you are looking for. In my mind, the intuitive approach to the Chain Rule involves treating the individual differential quantities as distinct algebraic entities. The Chain Rule would then be stated: If u is function of v, and v is a function of x, then du/dx = du/dv * dv/dx simply because the dv's cancel out. If you think of u and v as outer and inner functions, then du/dx is "the derivative of the outer with respect to the inner, times the derivative of the inner with respect to x." Some mathematicians and math teachers would strongly disagree with treating differential quantities such as du and dv as distinct algebraic entities. But that's how Leibniz thought of it, and to the extent that it is controversial, I'm siding with Leibniz.
When you get to the point where the derivative of y with respect to x is equal to the product of the derivative of the composite functions (dy/dx = dy/du * du/dx), are they multiplied because at any point on the function f(x) the rate of change at any point will comprise of both the rate of change in dy/du and du/dx at any single point? I like to picture things visually and it helps me to think of it like that. It's nice to hear more about Leibniz. Newton is usually the poster boy. I live in England, btw.
This is amusing. The zero of the numerator is deemed to be smaller than the zero of the denominator. I call this "shoe-horning" logic. Gotta be a better way of cancelling the last term than this (and there is).
you need to state what the delta v and u represent, also there a couple times you cant tell whats going on when its written out. also you need to explain in the writing why delta u multiplied by delta v over delta x goes to zero. this is the feedback i received from my prof
It is quite strange to say this doesn't hold up in class. It is a mathematical proof so it is either correct or it isn't. It cannot be correct on YT but incorrect in class. In my career I have too often witnessed teachers who when presented with a method they have not been previously aware of, reject it. There's more than one way to skin a cat. In this video, a valid sequence of operations take us from y=uv to the product rule. It is proof. Therefore if it doesn't hold up in class consider keeping the proof and rejecting the professor.
