DIFFERENTIATING a CONTINUED FRACTION?!

"I suppose you think you're clever, don't you? Do you think I'm some kind of senseless fool!? You think I will take a deal that gives me a continued fractional amount of cheese and bread? You're out of your talc my good sir!" and with that, Baron von Grumble Guss stormed out of Chalk's Cheese and Bread Warehouse Emporium Outlet Mall Brokerage and Post Office empty handed. Chalk was stunned. All he said was that anyone who watches this video all the way through and comments "Baron von Grumble Guss watches me sleep, what a creep!" will receive good vibes from the universe.
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[@frankwilhoit]

Continued fractions never have a good place to stop.

[@GeoffryGifari]

what's next? a continued fraction *of derivatives ?*

[@MichaelPennMath]

I’ll run that by him. (Tone: Serious) could be interesting! -Stephanie MP Editor

[@ayrthhhn]

a continued fraction of continued fractions of derivatives

[@goodplacetostop2973]

4:06 Homework 11:37 Good Place To Stop

[@coaldustmarauder2071]

I don't claim to have much background in math. Please point out any logical or mathematical errors in this Working it out, I believe this same solution method works in the generalized problem, assuming the continued fraction is of the specific form given in the video Following the steps in the video Let y be the continued fraction a(x)/(b(x)+y) then cross multiply to get y(b(x)+y)=a(x) yb(x)+y^2=a(x) Complete the square to get y^2+yb(x)+1/4b(x)^2=a(x)+1/4b(x)^2 (y+1/2b(x))^2=a(x)+1/4b(x)^2 To simplify my typing let g(x)=a(x)+1/4b(x)^2 y+1/2b(x)=+-sqrt(g(x)) y=-1/2b(x)+-sqrt(g(x)) For sanity checking, lets make sure this lines up with the video so far a(x) = 1 b(x) = 2x, -1/2b(x) = -x g(x) = 1 + 1/4(2x)^2 = 1 + 1/4*4x^2 = 1 + x^2 Here, the video has y = -x+=sqrt(1+x^2), so this lines up Continuing, Take y+ and y- as in the video, y+=-1/2b(x)+sqrt(g(x)) y-=-1/2b(x)-sqrt(g(x)) To take the derivatives, we need these g'(x) = a'(x)+1/2b(x)b'(x) d/dx (sqrt(g(x)) = 1/2g'(x)/sqrt(g(x)) In the video we get g'(x) = 0 + 1/2*2x*2 = 2x d/dx(sqrt(g'(x)) = x/sqrt(1+x^2), so this lines up as well Next, we take the derivatives of y+ and y- y'+ = -1/2b'(x)+1/2g'(x)/sqrt(g(x)) y'- = -1/2b'(x)-1/2g'(x)/sqrt(g(x)) From there, we solve (y'-y'+)(y-'y'-) = 0 y'^2-(y'+ + y'-)y' + (y'+)(y'-) = 0 y'^2+b'(x)y'+(1/4b'(x)^2-1/4g'(x)^2/g(x)) = 0 That last one is a bit complicated, so lets cross check b'(x)y' = 2y', good 1/4b'(x)^2-1/4g'(x)^2/g(x) = 1/4(2)^2-1/4(2x)^2/(1+x^2) = 1 - x ^ 2/(1+x^2). Note that the order of the terms in the video are reversed, but that is because they are subtracted instead of added. This is equivalent to the video let u(x) = -(1/4b'(x)^2-1/4g'(x)^2/g(x)) Then y'^2+b'(x)y'=u(x) y'(y'+b'(x))=u(x) y'=u(x)/(b'(x)+y') The same repeated fraction as in the video In conclusion, It wasn't a concidence that 2 appeared in the denominator, because 2 = b'(x), when b(x)=2x. The numerator is a bit more complicated but it does have a closed form if my math is correct EDIT: Of course, this assumes differentiability of a(x) and b(x), and that g(x) is not 0

[@manucitomx]

I’ve never quite gotten my head around continued fractions. Thank you, professor.

[@get2113]

Me neither. I was able to do a career in statistics, econometrics, and finance without those things.

[@fabiopilnik827]

@@get2113 e has a continued fraction expansion - but I have no idea what is the continued fraction expansion of eˆ2 eˆ3 etc. Other irrational numbers have continued fraction expansion. It can be a way of saying the irrational number is approximately rational.

[@eytansuchard8640]

The correct way to understand continued fraction is to reduce it to a finite recursive term. Sometimes the recursion appears after more than one iteration, e.g. a continued fraction can appear with first numerator different than the rest. In such a case, you need to manipulate the continued fraction to become a recursive term.

[@jagatiello6900]

Check out chapter 4 of Goldman's book The queen of mathematics.

[@get2113]

@Fabio Pilnik The irrationals are by construction the limits of sequences of rationals, and thereby perforce approximately rational. I find this discussion interesting. Cheers.

[@jgilferez]

Maybe we can gain some perspective using implicit differentiation: y = 1/(2x+y) => (2x+y)y = 1 => y + (y+x)y' = 0 => y' = -1/(1+ x/y), from where we obtain a continued fraction expression for y' in terms of the continued fraction expression for y. Following a similar argument with y = 1/(z + y), where z = z(x), we would obtain y' = -z'/(2+z/y).

[@adandap]

Ah, good. I came here to say this but now I don't have to type it! One thing that puzzles me is that when you do that there doesn't seem to be any sign of the sqrt(x^2+1). There seems to be at least two different continued fraction expressions for y'

[@pyropulseIXXI]

That is what I did; I used implicit differentiation

[@pyropulseIXXI]

If y = 1 / (2x + y), then y' = - y / (x + y) = -1/ ( (x/y) +1 ), where y = 1/(2x+1/(2x+1/(2x+1/(2x+...

[@paulkohl9267]

Had same idea but I think there is a missing 2 factor of x in your formula. (2x+y)y = 1, implic. diff. y' (2x + y) + y (2 + y') = 0. Sub in (2x + y), y' / y + 2y + y y' = 0 implies, y' (1/y + y) = -2y, y' = -2y^2 / (1 + y^2). Plug in continued fraction (or quadratic) for y in terms of x. voila

[@jgilferez]

@@paulkohl9267 Hi, if you are referring to my third equation, I believe it is correct. After implicit differentiation, distribute the products over the sums. Putting together like terms, one sees that 2 is a common factor of all them, so we can divide by it, and take y' as a common factor of y and x.

[@ruferd]

Absolute yes for a playlist of Galois theory. Also, these video descriptions are on another level. Absolutely masterful.

[@michaelguenther7105]

Just use implicit differentiation on y^2 + 2xy = 1.

[@akshatjain2775]

The editing is superb!

[@CglravgHRjsksgS]

The thumbnail is all the money 🔥🔥🔥

[@ImaPilotC]

To find a generic way to write the derivative as a continued fraction, you can use the same technique as the video, substituting 2x = g(x) (i.e., f(x) = 1/(g(x)+1/(g(x)+...)) = 1/(g(x)+f(x)). This gives f'(x) = u(x) / (g'(x) + u(x)/(g'(x)+...)), where u(x) = - g'(x)^2 / (g(x)^2 + 4). You can verify this leads to the same solution for g(x) = 2x.

[@opfromthestart3645]

I got something different for u, u=g'^2*g^2/(4g^2+16)-g'^2/4, which also gets the correct answer with g=2x

[@opfromthestart3645]

Nevermind it is equivalent

[@GeoffryGifari]

what kinds of functions can we expand in this way? is it always possible to find a continued fraction expansion?

[@SurajSingh-nx7yj]

Thank you sir 🙏

[@pointlesssentience3987]

In the general case, I’d try to represent the partial continued fraction as a partial sum (via Euler’s identity on continued fractions) and prove uniform convergence of the partial sums on some closed interval. Then after differentiating Termwise, perhaps there will be some closed form after simplifying the sum…

[@josehurtado4929]

Could it be solve by just derivate both sides of the expretion written in 0:55 with the chain rule an solve for y’ ?

[@wiseSYW]

plotting Y=1/(2X+Y) in Desmos is pretty cool. looks like a rotated hyperbola

[@КириллЙошкин]

It is a rotated hyperbola exactly. We can transform the equation to (y+x)² - x² = 1 which is standart hyperbola equation in coordinates x + y and x

[@JonAndKatyAnderson]

Id be interested in seeing the derivation of a general form for the derivative of a continued fraction, where 2x is replaced with a polynomial of degree n.

[@Jason-ot6jv]

great video, love your videos!!!

[@ScouseRobert]

Upcoming MathMajor courses on Lie Algebras and Galois Theory. I'm psyched! 😀👍🏼

[@mtaur4113]

Since f(x)*(bx+f(x)) = a, we can differentiate both sides and maybe stuff happens. Can't really try right now though.

[@spicymickfool]

Thats the continued fraction for sqrt(x^2+1)-x, so the derivative would be x/sqrt(x^2+1)-1. Generally sqrt(x+a)=sqrt(x)+a/(2sqrt(x)+a/(2sqrt(x)+a/...) as a continued fraction. Interestingly, this method is essnentially the Babylonian root finding algorithm in disguise.

[@thomasdalton1508]

He said that, but you missed the ± that causes problems.

[@maxvangulik1988]

essnentially

[@thejuliomario1]

Baron von Grumble Guss watches me sleep, what a creep!

[@pacolibre5411]

Excellent video, but is anyone else a little sad that we didn't see an infinite iterated chain rule application?

[@sujitsivadanam]

Interestingly, I was kinda half-expecting that.

[@maxvangulik1988]

That would diverge due to infinite 2xs multiplying the whole thing by 2

[@GeodesicBruh]

@@maxvangulik1988 no

[@GeodesicBruh]

@@maxvangulik1988 it you take take the derivative of something well defined it should be ok 1) write out the fraction to a depth of N 2) take the derivative and use induction to produce a formula for the N case 3) let N go to infinity to recover the case in the video

[@maxvangulik1988]

@@GeodesicBruh are you saying something would cancel out the infinite 2s?

[@MrDynamite110]

Love the videos but recently the descriptions have been *fire*. Also, Baron von Grumble Guss watches me sleep, what a creep!

[@ralvarezb78]

Hi, since y^2 +2xy = 1, we can take implicit derivative: 2yy' + 2y + 2xy' = 0 so y' = -y/(x+y) right?

[@КириллЙошкин]

How to represent y' as a continued fraction from there?

[@ralvarezb78]

​​​@@КириллЙошкин y is already continued fraction, so by direct substitution. The trick is to express - y/(x+y) as - 1/(1+x/y) then substitute ANYWAY he asks for a derivative of continued fraction which can have multiple représentations, I don't see anywhere asking for a derivative in continued fraction form.

[@GeoffryGifari]

I noticed something weird: we can write y = 1/(2x + y) , the derivative y' = u/(2 + y') with the function u = -1/(x^2 + 1) first, the pattern that keeps coming up is something/(something + something) more interestingly, the function u has x^2 in the denominator. differentiate once we get 2x (which is in y) differentiate again we get 2 (which is in y') coincidence?

[@pyrotas]

I am not sure if I get it correct, however (at least over the field of real numbers) I would say that y_{+}(x) is equivalent to the fraction for x\geq0, y_{-}(x) is equivalent for x

[@Kram1032]

A fun exercise that might help a bit here is the derivative of an iterated function: f' = df/dx (f(g))' = df/dg dg/dx (f(g(h(x)))' = df/dg dg/h dh/dx ... f_(i+1)(x) = f(f_i(x)) f_0(x) = x df_n/dx = product(df_i/df_(i-1), {i, 0, n}) (I hope I made no error in the notation / this makes sense) I think it might be possible to use this to derive a more general result for any continued fraction of as simple a form as the one in the video. It might be more complicated if it didn't repeat so easily.

[@hra1465]

Consider f as the limite of the sequence (fn) of odd functions defined on R* as: f0(x) = 0. fn+1 (x) = 1/(2x + fn). NB: This denominator can't be zero since fn and x always have the same sign. Note that f is an odd function with being positive when x>0, and a discontinuity on 0: f(0-) = - f(0+) = -1. Thus: x>0: f=y+ x

[@gp-ht7ug]

Michael, please now show us how to integrate such functions! Please

[@GandalfTheWise0002]

The generalization is fairly simple (assuming I didn't overlook something). Replacing x with g(x) so that y = 1/(2g(x)+y) ends up working out much the same except that u = -g'(x)/(1+g(x)^2). The factor of 2 in front of the g(x) eliminates a bunch of factors of 1/2 and 4 floating around. The steps are pretty much identical to the derivation in the video. I'd been half expecting some ugly complication to arise.

[@ssaamil]

Love the thumbnail hahaₕₐₕₐ...

[@tieger638]

How can I get a chalkboard like you have?

[@somename-xz2zm]

At the intermediate step y^2 + 2xy = 1 we can do implicit differentiation to get 2yy’ + 2xy’ = 0 which leads to the bizarre result x = -y. I would like to understand what is wrong with this implicit differentiation that is leading to superfluous results?

[@aricwong1906]

u did not apply the chain rule when differentiating 2xy. It should instead be 2yy’ + (2y + 2xy’) = 0

[@somename-xz2zm]

@@aricwong1906 Many thanks. So then we can continue and get a different recursion as follows, which feels much simpler than solving a quadratic equation: 2yy' + 2y + 2xy' = 0 yy' + y + xy' = 0 yy' + y + xy' + x = x y(y' + 1) + x(y' +1) = x (y' + 1)(x + y) = x y' = x/(x + y) - 1 y' = (x - x - y)/(x + y) y' = -y/(x + y) = -y/y(1 + x/y) y' = -1/(1 + x/y)

[@fierydino9402]

Really interesting!

[@adrienanderson7439]

I tried to differentiate f(x)=( 2x+f(x) )^-1 with the chain rule and after some simplification got f '(x) = -f(x)^2 * ( 2 - f '(x) ) , so a continued multiplication of squared continued fractions, also could be considered some kind of differential equation.

[@byronwatkins2565]

If the general case is valid, then several particular cases must also work out... I would start there. Then I would try f(x) = a(x) / (b(x) + a(x) /...) I cannot imagine a more general form than this, but this will also get complicated quickly.

[@БорисНазаров-х7к]

The very first question that pops in my mind is this function at least differentiable? And what happens on the border of two different domains where its closed form is different?

[@robshaw2639]

I forgot to ask - was this topic inspired by something you've seen, or was it an original idea to look at differentiating a continued fraction function? Also, many thumbs up on the Galois Theory course! 👍👍 That's what I am most looking forward to... Does Randolph have summer courses? Or, is the Galois Theory course KZbin-only? Ah. maybe during summer break, you are making your own courses (along with Lie Algebras) on the channel?

[@derendohoda3891]

If the continued fraction is nice, could try to undo Euler's continued fraction formula to convert the continued fraction into a power series, then differentiate term-by-term, then re-apply it. Unfortunately this won't work in general.

[@alexanderpavlov2558]

why do the derivatives at 4:02 not have 2x in the numerator

[@txikitofandango]

Ooh setting up (y' - y+')(y' - y-') = 0, I don't think I would've ever thought of that. I did quotient rule on y = 1/(2x + y), good enough 8)

[@jacemandt]

Yes, I noticed this too! Is that a standard technique? If I know that either x=a or x=b, to then write that (x-a)(x-b)=0? The converse is obviously used all the time, but I don't think I've ever seen it used in that direction either.

[@anthonypazo1872]

Let F_f(x) be defined as the continued fraction that has f(x) in place of "2x". By this exact method, we can show that F'_f(x) equals the continued fraction with u in the numerators and f'(x) + u/... in the denominators. Furthermore, u = [f'(x)]² / 4+f(x)² by this derivation. I'd love to show my work, but i don't feel like typing it out in a comment. Is there a better way to send it to you? Also, I'm not sure how to test when continued fractions like this converge or not.

[@marc-andredesrosiers523]

Would be nice to attempt this with the definitition. The fact that infinite fractions are not as thoroughly defined as limits asinfinite series blocks partly those efforts, as far as I can tell. 🙂

[@sugarland1729]

It would have made the question more interesting to ask to write the derivative of f in continued fraction form whose elements are only made of simple functions (constants, x , x^2) .

[@VibratorDefibrilator]

I guess that last question is rhetorical, because the answer (and the approach) is self-evident. In general case our continued fraction y will look like: y = g / (f + y), where g = g(x), f = f(x), y = y(x) are functions of x. Consequently, the question will be how would it look like the derivative y' where it is represented as continued fraction: y' = u / (v + y'). Again, u = u(x) and v = v(x) are functions of x. If we use the approach from the video, we wil obtain u = u(f, f', g, g' ) and v = v(f, f', g, g' ), namely u = (f*f'*g' + (g')^2 - g*(f')^2) / (f^2 + 4*g) v = f' In the video example f(x) = 2*x, g(x) = 1, so we have u = -1/(x^2 + 1), v = 2, which is exactly the case. Very nice problem, I liked it.

[@MathOrient]

Amazing stuff :)

[@gavasiarobinssson5108]

By the way, this kind of repeated fractions were known to children in the 1930s. They show up in old school books. But they didnt repeat ad infinitum.

[@ScienceTalkwithJimMassa]

How would you graph this thing? To the question of domain...at first glance, one might think that the domain would be all reals except for zero. But, if one were to substitute zero and evaluate ad nauseum, we see that the function appears to be continuous at zero with the result f(0) = 1. What am I missing? When you have y = 1/(2x + y), couldn't one simply utilize implicit differentiation? Doing this, I obtained: y' = -2/[(2x + y)^2 +1] and then we can keep substituting the y that was defined. Does this help at all in the evaluation or is this an ill advised approach? I am not a mathematician. I am an oceanographer. Partial derivatives, multiple integrals, DEs, vector calculus were my thing.

[@dugong369]

the fraction is defined as the limit of 1/2x , 1/(2x+1/2x) , 1/(2x+1/(2x+1/2x))), etc. For zero all the terms are undefined. You may have assumed it used 1/(2x+1), 1/(2x+1/(2x+1)), etc. Mathologer has a good video kzbin.info/www/bejne/opaplqNvrtlmgLs where he states, around 8:00, that continued fractions could have been defined either way, but this way is more useful.

[@shmuelzehavi4940]

The correct solution is: f(x) = -x + sign(x) √(x^2+1) . Therefore, f(0-) = - 1 , f(0+) = 1 and your intuition is right. I believe my previous comment may help (i don't believe implicit differentiation is a proper way to solve this problem).

[@NewtonMD]

Wait is he expecting us to answer the last question? 😅

[@sjswitzer1]

Was that _really_ a good place to stop? :)

[@us9434]

Did anyone figure out what the domain would be?

[@shmuelzehavi4940]

I believe my previous comment gives the answer (every real value except zero).

[@daveintucson8541]

So you're just assuming the original continued fraction converges to something reasonable? Looking at partial versions in Desmos, every other one has a point where they diverge to ±∞ It sorta looks like they're converging, but I feel like this is a point that should be addressed.

[@chrislankford7939]

I'm getting for a continued fraction f(x) = 1/(g(x)+f(x)), f' = -g'/(2+g/f). Pretty straightforward from the implicit differentiation of the definition. This would be easy to construct in practice, but isn't the uniform continued fraction we solved for here.

[@LifeIsBeautiful-ki9ky]

this is an easy problem in Continued Fraction Theory.

[@gagadaddy8713]

This function has a blind point --- if x equal to zero, from the equation, we will obtain y = ±1 . However, if we go back to the original function. It is ridiculous that a continue fraction of all zero turn out to be ±1. 😅 Anyway, though y doesn't exist when x tend to zero. However, the derivative of the function still exists when x tend to zero. Interesting!

[@gagadaddy8713]

So, professor you answer of derivate still hold for x belongs to real domain.😘

[@paul21353]

I disagree. The value of a continued fraction is the limit of its partial fractions. At x=0 the first partial fraction, (1/2x) already does not exist. So f makes no sense at x=0. An easy calculation shows all non-zero values of x make sense and f(x) exists. So imho the domain of f is all non-zero real values of x.

[@gagadaddy8713]

@@paul21353 Yes! y, the function of the continued fraction doesn't has limit when x tend to zero. This one I have no disagreement. But my point is the 1st derivative of the function(y') has limit when x tend to zero. Check it out yourselves, and you can find that y' is equal to -1 when x tend to zero, no matter you approach it from the positive side or the negative side.

[@shmuelzehavi4940]

This problem may be looked as a convergence problem of a sequence of functions φ_n (x) to a limit function f(x) (which is to be found out) when n⟶∞ , where φ_1 (x) may be any arbitrary differentiable function, and for n>1 , φ_n (x) is given by: φ_n (x) = 1 / ( 2x+φ_(n-1) (x) ) for n > 1 Therefore, f(x) must satisfy the equation: f(x) = 1 / ( 2x+f(x) ) or: (f(x))^2 + 2x(f(x)) - 1 = 0 which gives: F(x,k) = -x + k √(x^2+1) , where k ∈ {1,-1} and: F ' (x,k) = (kx - √(x^2+1)) / √(x^2+1) The necessary and sufficient condition for convergence is: | f ' (x) | < 1 , which implies: k = sign(x) . Therefore we have: f(x) = -x + sign(x) √(x^2+1) and: f ' (x) = (x sign(x) - √(x^2+1)) / √(x^2+1) = (|x| - √(x^2+1)) / √(x^2+1) = |x| / √(x^2+1) - 1 f'(0) = -1 and consequently φ_n (0) does not converge to f(0) . Therefore, the solution domain is: x ∈ (-∞,0) ∪ (0,∞) (homework exercise). In other words, x may take every real value except zero.

[@MacHooolahan]

Incidentally - does anyone else (phps just in UK, late 80s) recall the math books using sin^-1(x) for inverse sine? Don't know why this still bothers me.... But it does!

[@gavasiarobinssson5108]

It is not obvious to me that the continued fraction is well defined nor that it is equal to one of the earlier fractions. Can anyone elaborate?

[@vihaankharia2829]

the domain of y is [-1,1]

[@maxvangulik1988]

Just do it implicitly. 1/y=2x+y (-1/y^2)dy/dx=2+dy/dx (-(1+y^2)/y^2)dy/dx=2 dy/dx=-2y^2/(1+y^2)

[@MacHooolahan]

Notation for this (in general math) is awful - an upside-down cap-PI phps?

[@xaxuser5033]

again, the rigor is absent

[@ayrthhhn]

-1/(1+x^2) is the derivative of arccot(x)... Hmm...

[@digxx]

Another valid endless fraction would be f'(x)=-1/(1+x/f(x))=--1/(1+2x^2)+x^2/2x^2+x^2/2x^2+x^2/2x^2+....

[@pyropulseIXXI]

I used implicit differentiation

[@txikitofandango]

All this sqrt(1+x^2) stuff makes me wonder if there isn't a triggy solution to this.

[@КириллЙошкин]

There is cosh-ish function involved when you check convergence of sequence of fractions

[@ayrthhhn]

I think there is, I saw that and that was the first thing that popped in my head Also, "u" is the derivative of arccot(x), this definitely smells triggy

[@oddlyspecificmath]

I actually stopped watching another math video because that thumbnail is _awesome_

[@pavelievlev9730]

i don't think there actually is this plus-minus sign problem. if x > 0, then the truncated continuous fractions converge to sqrt(x^2+1)-x and if x 0, truncated fractions are clearly positive, whereas the second root -x-sqrt(x^2+1) is negative, so f can only be the sqrt(x^2+1)-x root, which is positive. same for the second branch. P.S. the fact that both functions are defined for x in R doesn't matter: the behavior of sqrt(x^2+1)-x for negative x has nothing to do with the continued fraction in question. same for the other branch.

[@sabriath]

I got -y/(x+y) 🤷‍♂

[@xl000]

At which level is this usually studied? I mean the average content of this channel, in terms of either age range or in the EU. I’m not familiar at all with how math are taught in the US

[@JirivandenAssem]

University. Just look up math minors on universities and compare the subjects,

[@jacobwakem7607]

It would be nicer if you found the derivative as a continued fraction itself.

[@farfa2937]

Maybe I'm dumb, but can't you just implicitly differentiate 1:30 and call it a day?

[@mathunt1130]

You could have done an implicit derivative on y=1/(2x+y), this would have led to an equation for y', which is what you wanted.

[@КириллЙошкин]

It's harder to represent the derivative as continuous fraction from implicit derivative

[@chrisglosser7318]

y^2+2xy=1 => 2yy’+2y+2xy’ =0. So y’=-y/(x+y)=-1/(1+x/y)

[@chrisglosser7318]

You absolutely can get a get a general function (swap 2x->f(x)) by following the same steps

[@chrisglosser7318]

So basically, if y=1/(f+1/(f+…)), Then 1/y=1/(-f+1/(-f+…)). Then you can show y’ = -f’/(2+f/y) =-f’/(2+f/(-f+1/-f+…))

[@axbs4863]

y = 1/(2x+1/(2x+....)) y=1/(2x+y) y^2 + 2xy = 1 y^2 + 2xy + x^2 = 1 + x^2 (y+x)^2 = 1 + x^2 y + x = sqrt(1 + x^2) y = sqrt(1 + x^2) - x

[@rob876]

f(x) = 1/(2x + f(x)) f'(x) = -(2 + f'(x))/(2x + f(x))^2 (2x + f(x))^2f'(x) = -2 - f'(x) [1 + (2x + f(x))^2]f'(x) = -2 f'(x) = -2/[1 + (2x + f(x))^2]

[@sharpnova2]

why not implicit differentiation at the start?

[@doctorb9264]

Same idea here.

[@camilocagliolo]

Baron von Grumble Guss watches me sleep, what a creep!

[@neilwoller]

Baron von Grumble Guss watches me sleep, what a creep!

[@ronbannon]

Baron von Grumble Guss watches me sleep, what a creep!

[@OmnipotentEntity]

Baron von Grumble Guss watches me sleep, what a creep!