finding the derivative from an integral equation

[@txikitofandango]

f(x³+1) = 1/(3x³-1) = 1/(3(x³+1) - 4) Substitute x for x³+1 f(x) = 1/(3x - 4) No need to deal with funky inverses or "cube roots"

[@bprpcalculusbasics]

Ooooh yea!!! How did I NOT see that 😆

[@aklisilva9785]

But the integral diverge at 1.

[@fivestar5855]

Nice!

[@rithvikmuthyalapati9754]

@@aklisilva9785 There is no integral in that comment. Only a function

[@mathsx5887]

You can do that because the cube route is a bijective function from R to R, so ur doing the same

[@MattT002]

Nothing says ‘Happy Valentine’s Day!’ like some calculus.

[@bobingstern4448]

Agreed!

[@liquid_lover]

Nothing else is more relevant, obviously.

[@michaelempeigne3519]

faster way to get the function from f ( x^3 + 1 ) = 1 / ( 3x^3 - 1 ) is to make x^3 + 1 to appear everywhere : f ( x^3 + 1 ) = 1 / [ 3 ( x^3 + 1 ) - 3 - 1 ] f ( x^3 + 1 ) = 1 / [ 3 (x^3 + 1 ) - 4 ] now change the x^3 + 1 to x's : f ( x ) = 1 / ( 3x - 4 )

[@nasekiller]

i got an answer for the question at the end via substitution. however the result is weird. one the one hand it depends on f(1) (which is possible), but when i put it back into the original equation, the integrand was 1/(x-1). and obviously that will give us problems when integrating with a lower bound of 1. so maybe the problem is ill-posed and there is no solution?

[@seungukj]

Same issue for the original problem too. Subbing back f(x)=1/(3x-4) to the original integral also gives a divergent integral 😂

[@yoav613]

Yes.that explains bprp evil smile at the end🤣

[@lucas0m0james]

Yup, some functional analysis on any antiderivative of the integrand shows it'll always be divergent approaching 1

[@adrishbora]

Your videos always make my day man! Love just calculus!!!

[@bprpcalculusbasics]

Thanks!!!

[@yegorgorodzha9921]

It's nice but if we plug the function into the integral we get a problem at 1. If we try to evaluate it, we do get the lnx, but we also get +∞. Not sure how that is supposed to work, probably the lower bound of integration should have been different.

[@AtotheKres]

Ran into the same issue, tried to fix it with some additional function of the lower bound but could not get it yet. Shouldn't there be some additional degree of freedom in f due to loosing Information by differntiating? You could also write the rhs as an integral and take the Limit lowerbound=1+epsilon. 8 guess the divergencies would cancel out there.

[@jachymsetka]

what I love about these videos is that I don't understand any of it but still enjoy it

[@retueze3098]

1:02 why did we lose the bottom bound of the integral? shouldn't we subtract from the left part f(1)/(1+f(1))?

[@anshumanagrawal346]

The second question is easier than the first, you don't even need to differentiate under the Integral sign, as the integral is easily solvable by making the substitution u= 1+f(t) 😁

[@mr.atomictitan9938]

did you end up getting 1/(3(x-1)^2/3?

[@priyanshugoel3030]

1+f(x^3 +1)={1+f(1)}x Maybe f(x) can be recontructed via a power series method or iterative method At x=2.

[@AriosJentu]

Hmm. I solve this equation with substitution g(t) = f(t)/(1+f(t)), and try to solve for general solution of integral equation. I found that g(t) = 1/(3t - 3), [implies that f(t) = 1/(3t-4)] and after substituting it back, I've got an improper integral, which isn't got me ln(x) function as result, also there is an a constant ln(0)/3, which is in limit goes to -infty. Its need to be proven that this problem hasn't any correct solution, and I think this is easy to do. Maybe if integral boundary starts from 2, the result will be correct, but for this boundary there is no correct solution, I think. And if there is no solution of integral equation, there is also no solution for the derivative of function under integral sign.

[@dashingrapscallion]

I think the limits need to be changed for this to be correct. I got the same solution; however, upon plugging the solution in to check my answer, I found that you end up with a divergent integral, particularly with the lower limit of 1.

[@bprpcalculusbasics]

I think so too especially reading your comment and others’

[@dashingrapscallion]

@@bprpcalculusbasics still a really cool problem! Thank you for sharing!

[@bprpcalculusbasics]

@@dashingrapscallion thanks!

[@wojciechchudek9379]

f'(x)=1/3(x-1)^(-2/3) for x ∈(1; +∞) and f'(x)=-1/3(x-1)^(-2/3) for x ∈(0;1) , Easier is just calculate this integral with u=1+f(t) than differentiate

[@ScienceWorldz]

The answer to the question at the last is f(x) = c * exp(cuberoot(x - 1)) - 1 So, f'(x) = c * exp(cuberoot(x - 1)) / 3 * cuberoot((x - 1)^2), and since c is a constant, we can write it as f'(x) = c * exp(cuberoot(x - 1)) / cuberoot((x - 1)^2)

[@cooldude123456789098]

I think I got the answer here. Using the same methods from the first part of the video, we end up at the equation: 3x^3 * f'(x^3 + 1) = 1 + f(x^3 + 1) -> (3x-3) * f'(x) = 1 + f(x) -> (3x-3) df = (1+f)dx -> df / (1+f) = dx / (3x-3) -> ln(1+f) = (1/3) * ln(x-1) = ln[(x-1)^(1/3)] -> f(x) = (x - 1)^(1/3) -1 -> f'(x) = (1/3) * (x-1)^(-2/3)

[@elkuston7671]

Great video as always! What's the answer to the last question? I'm getting f'(x)=1/(3*(x-1)^⅔), first by integrating, then removing the f(1) by derivating, and finally solving for f'

[@adityaekbote8498]

Don't we loose solutions by differentiating on both sides?

[@Keithfert490]

The bottom limit must be 2 because, when x=1, the integral must be zero because ln(1)=0

[@naveensoh2884]

Answer for the last one I think f(x)=-1 f'(x)=0 Is it correct

[@talkgb]

Right before the inverse step you also could have just done f(x^3 + 1) = 1/[3(x^3 + 1) - 4] so f(x) = 1/(3x - 4) right?

[@juancamiloayacastano9232]

Is not possible to get ln(x) with that boundary conditions. It diverges for bound 1

[@vaithiesh.jjayasankar7910]

A really good video !!

[@benardolivier6624]

Integrating the LHS is easy, we get ln((1+f(x^3+1))/(1+f(1)) = ln(x), which is now problematic since if we suppose f is defined at x=1 we get LHS=0 and RHS tends to infinity at x=0 which contradicts that supposition so f is not defined at 1.

[@darranjurdyga1999]

idk how but I got f'(x)=0 for the last question. So I integrated the left hand side by u=1+f(t) which gives du/f'(t)=dx Then I got to integrate 1/u which gives ln(u)=ln(1+f(t)) Substitute the bounds I get ln(1+f(x**3 +1))-ln(1+f(1))=ln(x) So using the natural logarithm law I got ln( (1+f(x**3 +1) ) / (1+f(1)) =ln(x) Take both natural logs off Make x=sqrt3(x-1) and shuffle some functions to get f(x)=(1+f(1) ) (sqrt3(x-1)) -1 Make x=1 to find what f(1) is f(1)=(1+f(1) )(0)-1 f(1)=-1 Which means f(x)=(1-1)(sqrt3(x-1))-1 Being f(x)=-1 , then d/dx both sides to got f'(x)=0

[@lukaspeter718]

could be because ln(1+f(1))=infinity

[@barobabu8889]

Well I'm getting f'(x) = (C/3)(x-1)^(-2/3) , where C is any arbitrary constant. The C won't affect the parent integral though which gives ln(x). 😀😀

[@dqrksun]

曹老師新年快樂🥳

[@bprpcalculusbasics]

Thanks, you too! Btw where are you located? It is Monday 8:20 am for me.

[@dqrksun]

@@bprpcalculusbasics 剛到了元旦

[@dqrksun]

@@bprpcalculusbasics 最後那個問題有點東西

[@itdepends5304]

Your videos' thumbnails are cool How to make such ?

[@mr.inhuman7932]

Honest question: Is he saying "Chandu" instead of "Chain Rule" or is it just his accent? Or "Chandu" something I just dont know of yet? Btw I love your videos! I watch alls 3 of your channels!

[@anshumanagrawal346]

It's a joke, Dr Peyam started it and he continued it, he used to say "Chain-Rule" earlier but now he says "Chen-Lu"

[@kepler4192]

3:48 what do you mean “x to the third power” don’t you mean “three times x”

[@dianeweiss4562]

Slip of the tongue

[@bprpcalculusbasics]

Yea

[@Jha-s-kitchen]

For the last question, i am getting f ' (x) = (1 + f(x)) / (3x - 3) How to solve further (If it is needed)

[@anshumanagrawal346]

This is a seperable differential equation

[@rolfjohansen5376]

what if the integral starts at 2 instead of 1 ?

[@chrisrybak4961]

Nice video, as always, but I don’t understand why you can just ignore the lower (1) bound of the integral?

[@juancamiloayacastano9232]

Fundamental theorem of calculus. When you differenciate the integral you must evalue the integrand with the bounds and multipliy it by de derivative of the bound. (1)'=0 You can just ignore it for any constant

[@chrisrybak4961]

@@juancamiloayacastano9232 ahhh, ok. Thank you! 🙏

[@anshumanagrawal346]

The question, as it's stated is wrong and has no solution, you can check that the function he's obtained doesn't satisfy the original equation As for your question, note that the integral from 0 to x and 1 to x (just as an example) only differ by a constant, so their derivatives are the same, in general the derivative of a function of type integral from a to x (where 'a' is a constant, a given fixed real number like 1 or 0 or 1/2, NOT a function of x) is just the integrand, this is just the first fundamental theorem of Calculus, for the general case, the type of Integral from g(x) to h(x), you have something called the Newton-Leibniz Integral rule, which can be proven easily by using the second fundamental theorem of Calculus and the chain rule, and if the lower bound or upper bound is a constant, it doesn't make a difference as in the rule you multiply by the derivative of g(x) and h(x) and the derivative of a constant is 0

[@skylardeslypere9909]

Somehow I got that in the last problem, f(x) was identically equal to 1. Which isn't really possible of course, because then you'd get 0 = ln(x) which is not true. It's 1AM tho so I probably made a mistake somewhere. My solution involved a linear dirst order differential equation.

[@rshawty]

I don’t know if the answer is f’(x)=0 or if there’s no answer because I find f(x)=-1 but you can clearly see that in the original assumption f(x)≠-1, so I think I should answer “Impossible”, no ?

[@rshawty]

It’s of course about the last question

[@adityasuhane8930]

I got 3

[@rshawty]

@@adityasuhane8930 wtf, how ?

[@adityasuhane8930]

@@rshawty inside integral on the denominator we have 1+f(t) and on numerator we have f'(t)dt which is simply the differentiation of the denominator. We can substitute 1+f(t) =z Now we have, dz/z inside integral. Differentiate both sides as in previous problem, replace z by 1+f(t), put x^3+1 in place of t. Now you have 3x^2/1+f(x3+1) = 1/x Solve this like in previous problem and you have f(x) = 3x-4 So f'(x)=3

[@rshawty]

@@adityasuhane8930 And did you check your answer ?

[@dihydrogenmonoxid1337]

I study Physics, the last thing that would happen to me is someone dating me LMAO Math is better anyway

[@bringbackthedislikecount6767]

Instead of valentine, wish you a happy new year. 恭喜发财

[@bprpcalculusbasics]

Thank you. And yes “bring back the dislike count!!!

[@nidhiagrawal3354]

I don't get what valentine about blackpenredpen's question.

[@jakesboty9936]

The answer of last question is f'(x)=1/3 *(x-1)^-2/3

[@Bilbobaggiins.0]

Again I learn something new

[@Bilbobaggiins.0]

Thanks sir for this

[@lhduy_vn]

How can u find the inverse of a function like that 😇

[@juniorisrrael3184]

Can you solve d/dx (f(g(x)))= - g(x)?

[@nidhiagrawal3354]

f(x) = -x Thanks 😊 I don't get what is valintine about blackpenredpen's question.

[@juniorisrrael3184]

is a composite function must find f and g

[@nidhiagrawal3354]

@@juniorisrrael3184 f(x) = -x + c, g(x) is not restricted. ...but what is valentine about blackpenredpen's question?

[@dethmaiden1991]

Based on the title I was sure the answer would be something like sqrt(-1)

[@zgamer_5632]

when I saw that you made it for Valentine's i thought the solution was gonna be a Heart-shaped curve

[@yoav613]

Are we going to do this again or something different..well i don't know (and then evil smile hahahahaha 🤣🤣)

[@JayOnDaCob]

I can’t :(. I have no idea how to do calculus I just watch these cause I think they’re interesting

[@yashprajapati8857]

I love you!

[@tonyhaddad1394]

last challenge : Integral (from 1 to x^3 +1) [ f ' (t) /( f(t) + 1 ) ] dt = ln(x) ln ( f(t) + 1 )( evaluted fom 1 to X^3 + 1 ) = ln(x) ln( f(x^3 +1 ) + 1) - ln(f(1) + 1) = ln(x) Ln( (f(x^3 +1 )+1)/(f(1)+1) ) = ln(x) (f(x^3+1) + 1) / (f(1) + 1 ) = x f(x^3 + 1) = (f(1) +1)* x - 1 let x ----> cuberoot(x - 1) f(x) = (f(1) + 1)*cuberoot(x - 1) - 1 Let x = 1 f(1) = - 1 ( if its continous ......) f(x) = - 1 f ' (x) = 0 Im not sure if im correct !!!!

[@ThePowerOfRed21]

if f(1) = -1 then ln(f(1)+1) can't exist

[@dqrksun]

I got this too. But I realized f(1) can't exist or else you'll have a ln0

[@bernizubi5217]

Chen Lu!

[@sagarmajumder7806]

There should be log(1+x).. As,at x =0 LHS and RHS doesn't satisfactied.

[@rogerkearns8094]

Let me know if you need any help with all the St Valentine's day romance that you've just unleashed upon yourself. ;)

[@adityasuhane8930]

The last question answer.... I got 3

[@plislegalineu3005]

Idk the Q at the end is screaming "U-SUB!" to me, but idk if that would work and how that would work

[@hypnovia]

You have the right idea!!! 🌟 What’s the derivative of the bottom?

[@plislegalineu3005]

@@hypnovia I know what u should be I'm just too lazy to do things that come after this lol

[@dqrksun]

@@hypnovia f'(x)

[@aklisilva9785]

I think the integral is from 2 not from 1.

[@Сладкоежка-ж5я]

Genius

[@pedroribeiro1536]

Ironic... A math video mentioning valentine's day... Mathematicians don't have girl/boyfriends

[@alakkis]

do 10th graders take this stuff? if so, i'm interested

[@fivestar5855]

There's 666 likes on this video lol

[@chmjnationalsuperarmygener8564]

?

[@anshumanagrawal346]

Sadly, your equation is inconsistent :(