Implicit Differentiation

[@AangContreras]

"The Nike method: You just do it", hahaha. I loved it. Nice video. Gracias!

[@blackpenredpen]

Anything with Chen lu is good.

[@drpeyam]

Agreed hahaha

[@Happy_Abe]

And produlu

[@aldricbenalan4755]

@@Happy_Abe Haha

[@victorpaesplinio2865]

I saw the proof of the Theorem in my calculus 2 classes. Sadly, I wasn't able to apply it in the exam, but I passed the course! Thank you Dr Peyam for explain thing so well! Your calculus 3 videos are saving me in this quarantine because vector calculus is confusing

[@MrCigarro50]

Gracias por este tema. Como siempre, claro, sucinto, explícito, suficiente,...etc.

[@aldricbenalan4755]

Nice Job Dr. Peyam, I learned a lot!!

[@aldricbenalan4755]

Thanks!!

[@nozack5612]

Good Dr, I have a question for you from another Dr (Aero Engr; math minor, retired). Although I am retired I still like playing with math and therefore greatly appreciate your presentations. So conceptually, I like to think of the process you described in method 1 as operating in an x-z plane while holding y constant, so therefore any delta-y (dy) is zero, you can then take directional derivatives (conceptually) constrained to such a plane.. Since you can choose any y-plane, you can span 3-space (except the origin in this case) and all is fine. But I wanted to try another method, akin to a directional derivative in the direction of a specified line, where I hold, say y and z constant, and am then constrained to operate on a line of varying x; dy and dz are now zero on that line and I proceed: F(x,y,z) = x^3 + y^3 + z^3 - 6xyz pF/px = 3x^2 - 6yz where p is the partial operator. Then holding x,y constant along a line of varying z (dx, dy = 0) pF/pz = 3z^2 - 6xy So I suppose I must choose an intersection of these two lines and say at that point: (pF/px) / (pF/pz) = (pF/px) (pz/pF) = pz/px = (3x^2 - 6yz)/(3z^2 -6xy) = (x^2 - 2yz)/(z^2 -2xy) and this is the answer provided except for a minus sign !?? So where is the flaw and where is the missing sign change, given that the magnitude is correct?

[@SpoonPhysics]

it appears to be a first integral of a system of euler-lagrange equations

[@nozack5612]

Since the original function is symmetric wrt x,y,z pXi/pXk = (2XjXk-Xi^2)/(Xk^2-2XiXj) where p is partial differentiation operator and i,j,k are subscripted indices that may be permutated. Or in other words you have also solved for px/py, px/pz, py/px, py/pz, pz/py.

[@stabgan]

I love your energy

[@ajedrezandres11]

I'm really really thankful! This was an awesome video, and tomorrow I have my Analysis 2 exam on implicit diff! so thanks a lot, a in-the nick on time video :)

[@ukaszpawlak6953]

I had this theorem proven on my calculus lecture just 4 weeks ago, AGAIN your video was there just a ONE day earlier. Are you collaborating with my calculus professor to achieve that ?? O.o I really need to watch your videos as soon as they come out :D

[@foreachepsilon]

Having a bit of a brain fart, but why is y allowed to be treated like a constant but z isn’t? If we wanted to find del y/del x would z be treated as a constant?

[@SefJen]

I asked the same question

[@drpeyam]

Precisely! For dz/dx treat y as a constant, think z = f(x,y)

[@Danicker]

Yes it's because were differentiating z with respect to x so the value of y is not changing. You can think of a partial derivative as analysing vertical 2D slice of a surface and finding the gradient.

[@elvisp5456]

holy shit, that was good. thank you very much, sir

[@阿陽-j2w]

I saw the proof of that theorm before, it's super difficult.

[@SefJen]

Why doesn't y depend on x and z does ?

[@krispibean6775]

Think about z as f(x,y) instead of an independent variable, the same way that y is actually f(x) on a regular xy plane. Z is a function of both x and y.

[@nozack5612]

y is held fixed. The partials wrt y can also be calculated in the same way by holding one of the other variables constant. So for py/pz you hold x constant, and for py/px you hold z constant.

[@tgx3529]

F(x,y,z)is the function of three variables, you can calculate partial derivatives. If for some point F(Xo,Yo,Zo )=0, for very small surroudings of this point variables are dependent, this function f always exists for examp variables z, x , z=f(x), (you can write this dependence ) then dz/dx=-Fx/Fz...... or for z=g(y) is dz/dy=-Fy/Fz.......or y=h(x) ...... x=p(y)........(ifFz0.....). I think, there is some similar idea.

[@jayakhandelwal3900]

USE THE CHEN LU... WHENEVER I SEE YOUR CHANNEL I AM REMINDED OF "USE THE CHEN LU"

[@leswhynin913]

A good tune-up lesson!

[@LHC-ik7zr]

Can you do some videos on the fundamentals of Riemann integration? such as what it means for a function to be integrable, box integrals in R^n, and deducing whether some interesting functions are integrable over a box B, where the function is not continuous on B?

[@LHC-ik7zr]

Also maybe some stokes' theorem and differential form? (I found these topics very interesting but I wished I could understand them better)

[@shandyverdyo7688]

I love the second method. What's the name of the second method btw?

[@drpeyam]

Believe it or not it’s the Chen lu 😂

[@aldricbenalan4755]

@@drpeyam Lol, nice job incorporating the Chen Lu into almost everything you do!! Stay safe, and keep up the amazing work Dr. Peyam!!

[@martinsanchez-hw4fi]

This always confuses me :/ Isn't y (or shouldn't we consider it as) also an implicit function of x?

[@drpeyam]

Here no, x and y are independent variables, and z depends on x and y. It’s not quite the same thing as in single variable

[@Happy_Abe]

Is there a way to solve for dz/dx just in terms of x and y?

[@drpeyam]

Not really since you cannot solve for z in terms of x and y

[@gtatomek]

Thanks Dr Peyam! :3

[@tajpa100]

your class was A +

[@salemhernandezrios1147]

Grax bro

[@周品宏-o7w]

7:07 2yz-x^2

[@luca7253]

Dini's theorem

[@ebog4841]

assuming y is non dependent on x is silly... just go full generality

[@jesusalej1]

Hi Dr. Sorry, but I think you are calculating dz/dx, it means total derivative of z respect to x... partial derivative of z respect to x is 0 by deffinition, it doesn't mean that it does not exist z(x,y)... but its derivatives are total. I know you have notice that. You are a genius. Anybody makes mistakes, even if the results are correct. Go on with maths... the best distraction for a quarintine.

[@drpeyam]

???

[@drpeyam]

I’m calculating partial z / partial x not the total derivative

[@jesusalej1]

I saw that, but only F(x,y,z) has partial derivatives respect to x, y and z. Given that definition. At fact, also exist dy/dx=-Fx/Fy, if Fy is not 0... don't be upset about, I learned that way 23 years ago. If I am not correct, I would like to konw. Even so, I think you are a total genius and I am very happy to connect you. You still owe me a challenge about spherical coordinates.

[@drpeyam]

Thanks!

[@nozack5612]

@@jesusalej1 Implicit differentiation. You are specifically and intentionally holding y constant. In other words, you are calculating the other derivatives in that y-plane only, but ... you will notice that y is in the answer so you may select whatever y value or y plane you want to evaluate those derivatives anywhere in 3-space (except the origin). The same process is applied in the other two (orthogonal) planes x and z for those respective derivatives. Think of it as a directional derivative, where the value of the derivative depends on the direction in n-space (here 3) the derivative is to be taken; choose a direction orthogonal to a coordinate plane if you choose.