you’re a LIFESAVER. I’m in Calc 2 and had all the wrong intuitions and ideas about Leibniz’s notation, this is the exact context and understanding I needed!
I have been struggling with understanding the weird notations for a while now, thank you for this! You are one of the rare excellent teachers who actually have passion for what they teach. Keep it up!
I sincerely can say this video just increased my understanding of calculus which unfortunately is a rare thing. A must watch for those that want to get better at calculus.
that was just mind-blowing. whole my life I was just on automation writing dy/dx without understanding what it stands for and how is i exactly used. thank you! keep up the good work!
This notation is awesome to show the chen lu (chain rule) for example but it's not always true (the fact that we can break it apart like a fraction especially) so in my country, we always use it in Physics.
Thank you. My college teaches Calculus in 8 weeks. We brushed over all of this notation information. I picked up enough to get by until we got past the chain rule and into implicit differentiation. I did not understand why or how you could complicate an inside/outside derivative with substitution because we did not explain what was going on with the different derivative notation. THANK YOU.
I think in Leibnitz's own papers, he made the _d_ of _dy_ and _dx_ a bit funny, with a tail to the right of the stem, over the pronumeral. This helps you remember that the _d_ is not itself a variable (pronumeral) to be multiplied by _x_ or _y._
Might as well call them the same thing it's a reductive process with derivative variants? That's so old. And he taught you to do that? And you were just like it it's ok. The school was like a prison.
I don't think the triangle on top of chemical equations is a delta indicating change. I'm pretty sure it's the alchemical symbol for "fire", and it means adding heat, increasing temperature or adding other forms of energy to the reaction.
I was just going to say that, I don't remember exactly but i'm also pretty sure that putting a ∆ on top of the arrow signifies a temperature change (we never really got into thermochemistry before the break so i dont know for sure). My bet is on that it means that we add heat to cause the reaction such as the reaction is endothermic (needs energy to start (such as lighting a candle)). Pretty sure its just another way of saying + heat (so that you dont write reactant + reactant + heat -> product )
@@emiloberg2110 Are you a 9th grade student? Yeah that sign is used for endothermic reactions, specifically to denote change in temperature or "Heating". E.g., 2KClO3 ---∆---> 2KCl + 3O2 It doesn't give much info without temperature, pressure, etc being written.
@@Astro-ms6zo I'm in the second year of gymnasiet (Swedish) so 17y/o so highschool? Idk what year is equivalent to second grade in the Australian school system
Isnt the first advantage a but debatable? It basically works like a fraction but its not because fraction means part of a whole. Which is not what dy and dx actually is. Though i guess for new learners, it is a simple way of explaining it.
I wish professors would make the connection of the simple slope equation (y2-y1)/(x2-x1) which ends up being (f(x+h)-f(x))/(x+h)-x) this is for me really easy to remember
Holy shit, it all just clicked. I've been in Calculus for about a month and a half now, and kinda just going with the flow without really understanding what d/dx or dy/dx really meant (also I didn't want to ask because I didn't want to make myself seem stupid or something).
i went to see the next video on the series and when i coudn't find it I looked at when the video was uploaded and got disappointed. anyway, waiting for the next video!
actually... d and (d/dx) seem fine by themselves, as long as you respect the type. (d/dt)(d/db) does not commute, for example. dd = d^2. y=x^3. dx=1. d^2x = 0. ddy = d^2y = d^2[x^3] = d[d[x^3]] = d[ 3 x^2 d[x] ]. = 3 d[ x^2 dx ]. = 3 * (2 * x ddx). = 3 * (2 * x d[1]). = 3 * 2 * 0 = 0. note that we DID NOT DIVIDE BY dx after applying d. (d/dx)^2[x^3] = (d/dx)[d[x^3]/dx] = (d/dx)[ 3 x^2 dx/dx] = (d/dx)[ 3 x^2 ] = d[3 x^2]/dx = 3 * 2 * x dx/dx = 6 x. When we manipulate d, we know that its type is different. "(d/dt)" isn't really a fraction. It's a function with unapplied arguments. But "df/dt" really is a fraction. dt is not zero. df and dt aren't exactly numbers. (d /dx) is a an operator that applies operator d, implicit diff, and divides by implicit diff of x. (d/dx)^2 seems especially useful. They don't commute though! (d/dx)[y] = (dy)/(dx) ... literally the implicit diff of y divided by the implicit diff of x. and it does work as fractions, as long as you don't assume that in f(b) that d^2b==0. Assume: f = 3b + 5. b = (1/2)t^2. dt = 1 ... t is a line of slope 1 ddt = 0 df = d[3b + 5] = d[3b] + d[5] = 3 d[b] + 0 = 3 d[(1/2)t^2] = 3 (1/2) d[t^2] = 3 (1/2) * 2 t dt = 3 t dt. df = 3 t d[t]. df/dt = (d/dt)f = 3 t note that d^2b is NOT 0. note the multiplication rule on differentials. it's very subtle. d[a * b] = a * d[a] + d[a] * b ... notice that the multiplication rule was a sum of partials d^2b = ddb = d[d[ (1/2)t^2]] = d[t * dt] = (t * d[dt]) + (d[t] * dt) = t d^2t + dt^2 = t ddt + dt^2 = 0 + dt^2 = dt^2. d^2b = dt^2. really? check again... is d^2b/(dt^2) = 1, and d^2t = 0? ddt = 0 = d^2t. db = t dt. db/dt = t. (d/dt)^2 b = (d/dt)[(d/dt)b]] = (d/dt)[ db/dt ] = (d/dt)t = dt/dt = 1. !!! Check f with respect to b.... (d/db)f = d[3b + 5]/db = (3 db)/db = 3. (df/db)(db/dt) = df/dt = d[3b + 5]/dt = 3 db/dt = (3 d[(1/2)t^2])/dt = 3(t dt)/dt = 3t. (d/dt)f = 3t. Second derivative (d/dt)^2: (d/dt)[d/dt]f = (d/dt)[d[f]/dt] = (d/dt)[(d[3b + 5])/dt] = (d/dt)[3 db/dt] = (d/dt)[3 d[(1/2)t^2]/dt] = (d/dt)[3 t dt/dt] = (d/dt)[3 t] = 3 (d/dt)t = 3 dt/dt = 3. ... careful! d^2t isn't always zero for second derivative! (d/dt)[d/dt]f = (d/dt)[df/dt] = d[ df/dt ]/dt = d[ df * dt^{-1}]/dt = ... = d^2f/(dt)^2 - (df/dt)(d^2t/(dt)^2) but note that ddt = 0 as our assumption that "t is a line of slope 1" so... (d/dt)^2 seems a meaningful second derivative notation. it fully expands to.... d^2f/(dt)^2 - (df/dt)(d^2t/(dt)^2) which seems to be a fix to the literally wrong Leibniz notation of second derivative: " d^2a/(db)^2" ... presumes that d^2b = 0. don't forget the subtracted term, and it continues to work like fractions.
[d/dt]^2 f ... be careful... this is a "parenthesis" with different precedence. ... [d/dx]a means to apply a to d, and divide by dx. = [d/dt][d/dt]f = [d/dt][d[f]/dt] = d[ df/dt ]/dt ... you can't jump to [d^2/dt^2], f must be applied first. THIS is the main standard notation bug. = d[ df/dt ]/dt ... follow this literally, and you find a subtracted term. = d[ d[f] * d[t]^{-1} ] / d[t] = d[ df * (dt)^{-1} ] / dt ... definition of inverse d[a b]/dt = (d[df] * (dt)^{-1} + df * d[(dt)^{-1}] )/dt ... (da b + a db)/dt = .... note that people often get d[ 1/ dt] wrong! = (d^2f/dt + df * (-1 * dt^{-2}) * d[dt])/dt // 1/da = -(1/da^2)d[da] = d^2f/dt^2 + (-df/dt)*(d^2t/dt^2) Look at that very carefully. The actual "acceleration" is not "d^2f/dt^2". That is a bug in the notation that ASSUMES that d^2t = 0. This is only true if t is a straight line. When this parameter is time, then that would be correct. When t is time, d[t[ = 1. ... ie: time is a straight line. "acceleration" is really the term we usually associate with acceleration, minus "velocity" times that odd term that people assume is zero. d^2f/dt^2 - (df/dt)(d^2t/dt^2) ... this is TRUE acceleration. d^2t = d[dt] = d[d[t]] is USUALLY 0, such as in the case of time; but not in general.
@@pierrejoseph4797 d/dx is a verb, dy/dx is a noun. d/dx expects to have a term that follows, to indicate the expression it differentiates relative to x.
Those are two notations that two different people independently coined for representing the same concept, that we use in modern times. Sometimes there are advantages to Leibnitz's notation, other times there are advantages to Lagrange's notation. Newton's dot notation (ẏ) fell out of favor historically, but has made a renaissance in recent years, for derivatives with time as an implied variable of differentiation. dy/dx is Leibnitz's notation for indicating the derivative of y with respect to x. Each d means "infinitesimal change in", with the d standing for difference. Or rather, differenz, as Leibnitz would've spelled it. f'(x) is Lagrange's notation for indicating the derivative of function f(x), implied to be relative to x. The number of apostrophes indicates the degree of differentiation. If there are more than three, we typically use Roman numerals in superscript, so that you don't lose track of counting them. If there is an unspecified degree of differentiation that we indicate with a variable, we often put that variable in parenthesis in superscript, such as f^(n) (x), as you see in the definition of the Taylor series.
Leibniz was not the co-inventor of the calculus, he was THE ONLY discoverer of the calculus, which came about from his work with Huygens, the Dutch mathematician, that started on solving math with Series and Sequence. Newton was a fraud of the British Empire, namely as a honorary member of its intelligence and political intrigue office for the emerging scientists in Europe, the British Royal Society. Newton was a deranged ALCHEMIST we now know from the discovery of his infamous trunk that was auctioned off to British Economist (British Empire Fabian Society member} John Maynard Keynes. There's a great lecture on this from an American prof. William R. Newman (no, not Newton!) that EVERYONE should watch, especially if you are a proud american because due to the propaganda system (including public schooling) everyone in today's world is made to feel inferior intellectually to British/English people (because their accent is associated with highly educated people, more educated than Americans could be!!) . Not true at all, of course. And the fraud of Newton reveals this tactic. By the way, ...yes...the British Empire exists still to this day...in the Square Mile of the Corp. of the City of London (that is not under the jurisdiction of the UK Parliament) it's not a relic of a bygone era....it's banks are those that sit on Wall Street, and its banks that like HSBC that run the Narco Terrorism racket....and the offshore banking system. It is the foreign power that controls Washington, DC thru Wall Street. ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-NUhL1cli4ug.html
