Let's change the quotient rule!

[@fattimiv]

I guess the "dream" chain rule would be [f(g(x))]' = f'(g'(x)) which seems like a pretty gnarly functional equation to solve.

[@mu11668B]

lol I was gonna to comment for this :D

[@allozovsky]

Say, f(x) = x³, g(x) = 1/(1-x) works just fine.

[@fattimiv]

@@allozovsky I'm sure you could find some nice solutions, but my hunch is that it will be much harder to come up with a nice relationship between f and g as in this video and the product rule video.

[@NotYourAverageNothing]

f'(g)*g' = f'(g') f'(g) = f’(g')/g' int[f'(g) dx] = int[f’(g')/g' dx] ???

[@allozovsky]

There are some "good" functions f(x), for which g(x) is also '"good". For example: f(x) = (x^(n+2))/(n+2), f'(x) = x^(n+1), so f'(g(x)) g'(x) = f'(g'(x)) turns into (g(x))^(n+1) g'(x) = (g'(x))^(n+1), thus (g'(x))^n = (g(x))^(n+1) and g'(x) = (g(x))^((n+1)/n). The trivial part: dg/dx = g^((n+1)/n), ∫dg/g^((n+1)/n) = ∫dx, where ∫dg/g^((n+1)/n) = ∫g^(-(n+1)/n) dg = = [1 - (n+1)/n = -1/n] = -n g^(-1/n) + C, so -n g^(-1/n) = x - C, g^(-1/n) = -(x - C)/n, g(x) = ((C - x )/n)^(-n), g(x) = (n^n)/(C - x)^n. And for different values of n we get: n = 1: f(x) = x^3, g(x) = 1/(C - x), n = 2: f(x) = x^4, g(x) = 4/(C - x)^2, n = 3: f(x) = x^5, g(x) = 27/(C - x)^3, and so on. Polynomials (as function f(x)) may also produce "good" results.

[@randomjin9392]

I never remembered the quotient rule to be honest. I always derived it from the product rule + chain rule when needed.

[@ranani20]

I just remember D-high, and I know the rest

[@donaldbiden7927]

Same here

[@rajbunsha8834]

Same here

[@birdbeakbeardneck3617]

thats frustrating

[@herbcruz4697]

"Low d-high minus high d-low, all over low-squared."

[@matthartley2471]

I liked how you didn't need to use the quotient rule when calculating the actual derivative, but you did when calculating the quotient rule free derivative.

[@yassinezaoui4555]

This is awesome 😁 Note for those who struggle with the quotien rule for (f/g): We can refer to a special case to generalize for the numerator ( is it (1)f'g-fg' or (2)fg'-f'g ? ) Let's consider f=1 and g=x (1) => (f/g)' = -1/x² (2) => (f/g)' = 1/x² But we know that (1/x)' = -1/x² or we know that it is a decreasing function so its derivative is < 0 We can conclude that the 1st case is the right one :)

[@General12th]

16:00 I noticed how when we get to *a^2 / (a - 1)* this implies that *a* can't be equal to 1. I like this because if we try to find the domain of *f* we need to make sure *g* isn't 0 (which is obvious from our original goal), but it turns out we also need to make sure *g' - g* not equal to zero either. In other words, *g* can't be *e^x* specifically. It has to be *e^ax* where *a* is any other number but 1, and that's exactly what you confirmed.

[@beatoriche7301]

Unpopular opinion: I have no idea why calculus teachers introduce the quotient rule at all; it's really just a trivial corollary of the product rule and the chain rule, and there's no further mathematical structure behind it. In a class where most students will inadvertently end up memorizing at least the most common formulae because deriving them each time is a lot of work, they should not be given more formulae to memorize, but shown how to apply the existing tools to deal with different problems. Worse still, as Michael mentioned in the video, the quotient rule is by far not as simple as the product rule or the power rule, with the sign in particular being a source of error; by contrast, just applying the chain rule and the product rule is simpler, less error-prone, and usually much cleaner, especially when dealing with polynomial terms. Lastly, another problem with treating the quotient rule as a separate rule is that this framing arbitrarily makes multiplication and division seem like different processes when, from an algebraic perspective, division is just the inverse operation of multiplication. There are two basic arithmetic operations, not four, and the labels we put to things should reflect this truth. While we're at it, why not introduce a difference rule for differentiating differences of functions or a root rule for differentiating roots? That'd really be just as arbitrary. Introducing the quotient rule, in my view, isn't just unnecessary, but also makes calculus seem needlessly complicated and obfuscates students' intuitions for subjects like abstract algebra.

[@Le_Tchouck]

We can easily think this way once we have your hindsight, but the way some rules are "created" and taught mostly comes from the order in which they are taught, the knowledge the pupils already have, and that's something you learn from teaching. More than teachers, it's a job for "instructionnal program designers" (not sure how it is called in English), and student from different country may learn different rules at different ages, solve the same problems with different tools because of the learning order. You evoke the "two" basic operations. When we start learning difference and divisions, we only know positive integers, barely decimal. In this set, we admit at that time that some operations are impossible. Then we introduce new sets in which we can find equivalence between operations... I don't disagree with what you said though, just wanted to put things in perspective.

[@VoidFame]

I totally agree. It's utterly pointless to even teach the quotient rule a separate concept. I taught myself to just use the product rule after always forgetting how to use the quotient rule. Ironically, I had this seemingly inconsequential moment a while back. I was taking an optics class and it ended. I had nowhere I really had to be so I just started my homework. Another class came in shortly after and it became pretty obvious it was calc 1. The professor was going over the chain, product, and quotient rule. I let him finish his explanation, and then he asked for questions. I asked him: "isn't the quotient rule is really just the product rule?" Probably not the sliest way to act like a calc 1 student. "no, no, what do you mean?" probably caught off guard. "Can't you get the quotient rule from the product rule?" He couldn't do it. He was utterly confused. I was trying to play a bit coy and not directly say "please show these kids that it's just a special case of the product rule" It went nowhere, and I didn't want to interrupt any more so I just let him do his thing. Most people kinda knew what was up at that point. I left a few minutes afterwards. But this is the problem with our education system. The mundane parroting of facts without any clear instruction or insight.

[@coreymonsta7505]

Maybe the teacher should just show a proof of the quotient rule because, the rule is useful for differentiating quotients quickly? I suppose after performing the differentiation the product rule and chain rule way, you’d basically arrive at the same expression except the two fractions will probably need to be combined.

[@wompastompa3692]

Because "lo d hi minus hi d lo all over lo lo" is super easy to remember and is a tiny bit easier than rewriting f/g as f×g-¹, doing product rule, and then not forgetting to do chain rule when taking the derivative of g-¹.

[@integralboi2900]

Calculus teachers should give an exercise like this: Prove the quotient rule using only the product and chain rule. This may build some intuition.

[@VerSalieri]

Just a note, instead of identification when doing the partial fraction, just substitute some value for x. For example, x=0 gives r^2=Ar, so A = r. Similarly, taking x=r gives B=r as well.

[@udic01]

3:26 you could do on rhs the partial fraction decomposition and get that it is (g'/g)+g'/(g'-g) and when you anti derivate it you get that Ffx)=C*g(x)*exp(antiderivative(g'/(g'-g))) Most of the times it's easier to calculate.

[@goodplacetostop2973]

19:17

[@user_2793]

It took me a minute to realise

[@teeweezeven]

Maybe it is yeah

[@yonil256]

Great video, Thanks! Another method to think about it is that f(x)/g(x) is the same as f(x)*g-1(x) and the derivative is then the product rule.

[@spacelightning6587]

Awesome video! Also, if you ever want an easy way to remember the quotient rule, there is a cute rhyme for it: "low dee high minus high dee low, square the bottom and away we go!" Here "high" means the numerator function, "low" means the denominator function, and "dee" means the derivative. It has helped a lot of my students.

[@JohnWick-xd5zu]

Best quarantine content 😍 love from Egypt

[@lovingfoe1444]

I usually set y=f/g. And I take the natural log of both sides and differentiate that. And you get y’/y = f’/f - g’/g. And it is extremely easy to remember.

[@lovingfoe1444]

I feel that is the best way to remember the formula. And it is much more elegant than the traditional formula.

[@marcrenard515]

A little trick to remembre the order : when we derivate f over g, we said first f and then g, so the first which is derivated in the numerator is f. To remember the sign : in (f x g)' there is a cross for the multiplication, and in the formula, we still have a cross for the addition. For f over g we have the horizontal line for the fraction, and in the formula, we have the horizontal line for the minus. I dont know if it's clear, but it has already help some of my students

[@thephysicistcuber175]

8:18 divide numerator and denominator by x^2 for the objectively optimal way of integrating that rational function.

[@hoodedR]

Nice

[@sirlight-ljij]

I fail to see how exactly it is easier

[@hoodedR]

@@sirlight-ljij it immediately takes the form of 1/linear basically

[@sirlight-ljij]

​@@hoodedR No, it does not. It requires a substitution, which is not much different from what was done, resulting in unnecessary complications. We can also add and subtract rx do divide into pairs, but there is only one way that works everywhere, so why bother creating additional items that only complicate matters?

[@thephysicistcuber175]

It doesn't complicate, one can immediately see that after doing the manipulation the numerator is a multiple of the derivative of the denominator.

[@sirlight-ljij]

Next would be a dream sum rule... oh, wait, it's already as simple as it can be

[@seanshameless0]

But what if it was all a dream?

[@sydr570]

And that’s on linearity :)

[@cobino2133]

Not if your functions have severals parameters/variables 😉

[@kingkartabyo6206]

Might work on dream product rule, easier

[@sirlight-ljij]

@@kingkartabyo6206 There is already one about it, kzbin.info/www/bejne/m4qpe4lpe8anp80

[@ChefSalad]

You could get a cleaner result by rewriting f(x) as f(x)=(r*x^(-1)-1)^(-r) by dividing the top and bottom of the fraction by x and rewriting all the reciprocals as negative powers. If you do, then finding f'(x) doesn't involve the quotient/product rule.

[@danuttall]

I think fixing f(x) =0 would make this "freshman's quotient rule" be true. You got to love the trivial case! For me, I memorized the product rule in the same order as the quotient rule, then just slapped in a couple of changes (plus -> minus, all divided by the square of the denominator) to get the quotient rule.

[@vladislavlukmanov4998]

To check the quotient rule it's easier to take the derivative of 1/x which is -1/x^2 (here f = 1, g = x)

[@drpkmath12345]

By far, this is my favorite title among your videos haha

[@zakirreshi6737]

Same...always forgets the order of numerator then i was remembering it by differentiating 1/x .

[@valemontgomery9401]

0 is not an allowable value for a because that would make e^ax=1. This means that g(x)=1. g'(x) would therefore = 0, but because we divide by g'(x) in the new rule, the answer would be undefined. a cannot equal 0

[@pcbenutzer6651]

And what if g is the Zeta function ?

[@davidrubio.24]

What I do is remember that the minus comes from applying the product rule using g to the -1. So the minus has to go with the derivative of g.

[@ivanklimov7078]

maybe it could also make for an interesting video if you played around with ideas of these "dream rules" for integration, since that, in my opinion, causes a lot more frustration for students than differentiation.

[@Debg91]

a=0 would not be valid either: g(x) would collapse to 1, and so g' = 0, causing division by zero.

[@mu11668B]

Also for a = 1.

[@coreymonsta7505]

was just thinking about a good way to remember the order of f'g and g'f, just consider f(x) = x and g(x) = 1 to tell the correct ordering

[@TJStellmach]

Assuming you know the quotient rule but just forget the sign (and you _don't_ want to re-derive it from the product rule and chain rule every time), just considering the g(x)=1 case will error-check for you, with arbitrary f(x).

[@kevingregoryjohnston8134]

Just being a bit pedantic. The antiderivative of the LHS should be ln(|f(x)|), as the antiderivative is defined on any open interval not containing 0. As stated, the coefficient C is constrained to be positive, however since the f is in absolute values, we can remove the absolute values by allowing C to be either positive or negative. We also note that dividing by f(x) assumes that f is not identically zero. Thus the solution f=0 must be checked manually, which we incorporate into the solution by allowing C=0 (as f=0 is a solution for almost all g). Mathematicians note all of these ideas automatically, as they are common to virtually all separable differential equation problems, but it might be worth noting.

[@billf7585]

[@arimermelstein9167]

I learned it as “the bottom times the derivative of the top - the top time’s the derivative of the bottom all over the bottom squared” and we were made to memorize this until we got it right. So we never used f and g

[@H0tinNYC]

Lo-deHi minus Hi-deLo all over LoSquared.

[@General12th]

@@H0tinNYC That doesn't really rhyme.

[@irockyou1337]

Dream version of integration by parts: ∫udv = uv 😂

[@cerwe8861]

Na, more like ∫u'dv=uv So that you can just like with adding split the Integrals

[@factorization4845]

Then integration of vdu =0, so that's not too hard

[@MrRyanroberson1]

int u dv = uv for all u not f(v); basically: if u is constant. integral 3 dx = 3 x

[@twistedsector]

This is like the opposite of the physicist's IBP rule

[@LundBrandon]

my professor taught me "HOdHI - HIdHO / HOHO" It's got a good rhythm to it which helps make it easy to remember

[@exedlore]

This! We also use "low dee-high minus high dee-low", then we anticlimactically state the denominator as "low squared". 😂

[@twistedsector]

@@exedlore try low-Dee-hi, hi-Dee-low, with a minus sign and a square below. Its catchier

[@exedlore]

@@twistedsector Catchy indeed. I love it!

[@danielw9542]

I never had trouble with the quotient rule. Not because i could remember it but insteas because it was obsolete. You may as well remember the product rule and the chain rule. From those, you can derive the quotient rule as well as other parts of calculus. If you have been practicing differentiation this is an easy process. Treat f(x)/g(x) = f(x)(g(x)^-1) The rest is just calc. No need yo remember which way to subtract

[@Mathelite-ii4hd]

you are brilliant man...it was a splendid video:)

[@elrichardo1337]

Heh, in my AP Calc AB class, we used the mnemonic "low dee high minus high dee low over low low" to help remember the quotient rule.

[@7177YT]

I enjoyed this very much! Thank you!

[@IoT_]

8:40 there is much way easier method that you can do in that case (non-repetitive non-complex zeros) even in your head. r²/(x*(r-x)) first step: get rid of the first zero: x=0 and put this root in the expression: r²/(r-0)=r. Which means A=r ; second step: r-x=0 => x=r ; r²/r=r . Which means B=r. Done

[@MrRyanroberson1]

while this method is much easier, it does have some limitations when generalized, and is a special case. though i do love to use it

[@IoT_]

@@MrRyanroberson1 yes. I've already mentioned these limitations : zeros should be non-complex and non-repetitive

[@sjoerdo6988]

Is it really safe to divide by (g'g-g^2)? What if that is 0 for some value of x?

[@sjoerdo6988]

it doesn't have to be for all x, but for some x we might be dividing by 0. Say for example g(x) = x^2. Then for x=2, we're dividing by 0.

[@sjoerdo6988]

if g=g' would have to be true FOR ALL x, then g would indeed have to be the exponential function. But what if g=g' FOR SOME x? In that case we're dividing by 0 for that x. There are many such functions that have that property, like x^2, which I described above.

[@Ninjamaster222333]

is there some reason to why michael says anti-derivative instead of integral?

[@AGENTX506]

It's a technicality, mostly: An antiderivative (also known as an indefinite integral) is just doing the 'opposite' of differentiating: For some function f, the antiderivative finds a function F such that F' = f. More concretely, if you are given a function f and interpret f(x) to be the slope of some other function F at the point x, then the antiderivative finds this function F. An integral is an area under a curve, so it must have limits associated with it, i.e. to integrate f(x) from a to b is to find the area under f(x) from a to b. It obviously doesn't make sense to integrate without a start and end point. Keep in mind that so far, these two ideas are completely independent. The antiderivative just does the 'opposite' of a derivative, so it finds a function that has a particular slope at each point. An integral is just about finding the area under a curve between two points. It just so happens that (because of the fundamental theorem of calculus) the antiderivative of f (i.e. F) satisfies the criteria that for any a,b, the integral of f(x) from a to b is exactly equal to F(b)-F(a) Basically, an integral is an antiderivative combined with a start and end point. You evaluate an antiderivative at two points (i.e. F(b) and F(a)), take their difference and boom, that's an integral.

[@claudeabraham2347]

Very good!

[@djvalentedochp]

good job

[@jacobshort2468]

I'm supposed to be cramming for a calc ii test. I haven't even begun to think about differential equations yet. And here I am, engrossed in this. What are you doing to me, Michael?

[@General12th]

How did the test go?

[@AyushMishra-lt5yv]

Good Job !

[@cleverswarm2379]

8:26 - r squared!

[@roberttelarket4934]

I love the "ear muffs"!

[@factorization4845]

17:58 if a is 0, then f(x) and g(x) are both one. L.H.S.=0 but R.H.S.=0/0

[@MrRyanroberson1]

0/0 = 0 confirmed O.O

[@teeweezeven]

Of course there's also the trivial answer of f(x)=0 and g(x) being any non-constant function (I think that's what you get if you try to solve for g(x)=e^x)

[@Jukka70]

I found the quotient rule easy to remember. Vdu-Udv/V squared. My calculus professor was nuts he made us remember this by saying there was a song called the voodoo that you used to do, so that made us always remember the beginning was Vdu - Udv Vdu sounds like Voodoo, Udv was that you used to do, and used to also meant the past or subtraction. LOL

[@QuasiChameleon]

Great job! You can generalize further in your specific examples by keeping the constant capital C as well as another additive constant from the indefinite integral exponent to e in your function f.

[@Andreyy98]

Is c*e^x the only "nice" function family that is not a solution to this differential equation? As this family is the complete set of the solutions of g'g-g^2=0.

[@arielfuxman8868]

What does it mean to fix g(x)?

[@arielfuxman8868]

And why when integrating f'(x)/f(x) you did not put an absolute value?

[@ivanklimov7078]

a lot of people make this mistake: [f(g(x))]'=f'(x)g'(x)] when using the chain rule, so maybe that could be an interesting functional equation

[@jimallysonnevado3973]

the constant 0 function paired with any g works

[@chinmaytrivedi9249]

Thanks a lot Finally !!!!!!!!!!

[@xd-os7jl]

Good

[@mcwulf25]

A good piece of mathematics. But the example will almost never occur, as the starting point of the quotient is f/g which is already simplified as he demonstrates. The rule is only useful if f doesn't divide by g.

[@alexp6013]

I don't think we can have a = 0.

[@marc_zjn]

It is quite funny, that we need to use the quotient rule to evaluate the "special, easier" quotient rule...

[@mouadenoua6277]

It was funny at 16:27😂🤷‍♂️

[@hoodedR]

Penn.exe has crashed. 😂 Love that he doesn't edit out such mistakes

[@fedem8229]

By the way, shouldn't the result be exp(a⁴/(a³-1))? Because g'(x)=a*a*exp(ax)=a²exp(ax), so the numerator should have been a⁴exp(2ax), and in the denominator g*g'(x)=a(exp(ax))*a²(exp(ax)=a³exp(2ax)

[@jkstudyroom]

When I was a senior in HS, I had such a hard time with the product and quotient rule, I found a pattern. I memorized the product rule as f'g+g'f. Then the quotient rule becomes minus and all is honky dory! I actually found one book that actually uses my method. Really, separable equation?

[@MrRyanroberson1]

the product rule gets even prettier if you look at it as: f * g = e^(ln(f) + ln(g)), and then take the derivative of that with the chain rule: e^(ln(f) + ln(g)) * (f'/f + g'/g) = f * g * (f'/f + g'/g) = f' g + g' f, which works for any number of functions, as in f * g * h * ...

[@hassanalihusseini1717]

Haha, I think I prefer the normal chain rule.

[@bsharpmajorscale]

The dream chain rule is that it doesn't exist.

[@twistedsector]

1v1 Flammy

[@hoodedR]

Hell yeah

[@jacemandt]

Wait..."d-u earmuffs"?? 😆

[@raystinger6261]

I think Michael Penn missed the point by not solving that integral with g(x) (4:48). If we try, we get: ln(f) = ln(g) + (some integral) => f(x) = C*exp(ln(g) + (some function)) =>f(x) = C*exp(ln(g))*exp(some function) =>f(x) = C*g(x)*exp(some function) However, we're looking for a quotient rule here, so the g(x) gets cancelled out in the division: (f/g) = (C*g(x)*exp(some function))/g(x) => (f/g) = C*exp(some function) The problem with that is that the quotient is merely incidental, we're taking some fuction, multiplying it by g(x)/g(x), therefore multiplying it by 1, deriving it and calling that "quotient rule". So my conclusion for that would be that (f/g)' = f'/g' is impossible.

[@sirlight-ljij]

But due to what was shown in the second half of the video, such pairs of functions do exist

[@raystinger6261]

@@sirlight-ljij We can always find pairs of g(x) and f(x)=g(x)h(x), but that's no quotient rule, it's just the derivative of h(x) multiplied by 1.

[@sirlight-ljij]

@@raystinger6261 Any function can be written that way, you are just defining h(x)=f(x)/g(x). There still is a differential equation to solve for h(x) in this case, it is an alternative approach that is just less direct.

[@raystinger6261]

@@sirlight-ljij I agree with you on that one, my problem is that the solution found just states that back to us. That's a pointless solution in my books, but this is why I claimed that was "my conclusion", I figured others might have a different interpretation of the results.

[@raystinger6261]

@@sirlight-ljij Here's the deal: in my original comment, that part which I called "some integral" is actually Integral((g'/(g'-g))dx) If you're willing to explore that integral further and find an interesting solution, let me know. Other than that, it's been a pleasure talking to you, bruh.

[@roberttelarket4934]

Another Calculus students' dream but a professor's nightmare!

[@seanfraser3125]

The dream chain rule would be [f(g(x))]’ = f’(g’(x)) I have no idea what pairs of functions, if any, would satisfy this equation.

[@allozovsky]

E.g., f(x) = x^(n+2) and g(x) = (n^n)/((C - x)^n).

[@stephenbeck7222]

U sub ear muffs. Wow!

[@paulkarch3318]

(f g^-1)' = f' g^-1 + f(g^-1)'

[@Superman37891]

Let f(x) = hi and g(x) = ho. Then d/dx(f(x)/g(x)) = (ho de hi - hi de ho)/(ho ho)

[@NyscanRohid]

The video title is a lie.

[@isambo400]

Low D High minus High D Low all overLow Low

[@Bayerwaldler]

Dream-ChainLu schould be: (f(g(x))' = f'(g'(x)). Lets try f(x) = e^x, g(x) = ln(x). f(g(x) = x, so (f(g(x)) = 1. OTOH f'(x) = e^x, g'(x) = 1/x. So f'(g'(x)) = e^(1/x). Hmmm... 🤔... Doesn't work! But: It's a good place to start! 😄😄

[@allozovsky]

Just let f(x) = x^(n+1) and then solve (g(n))^n = (g'(x))^(n-1) for g(x).

[@funkysam1345]

Great video. A very deep insight into the quotient rule. Would really love if u could start solving IIT JEE Advanced math problems. (Like this comment so that he sees)

[@JoseMendez-ud6hj]

lo di hi hi di lo gang

[@basedvortigaunt2406]

Chain Rule: [f(g(x))]' = f'(g(x))*g'(x) Dream Version of Chain Rule: dy/dx = (dy/du)*(du/dx)