Have you seen that way of differentiating x^x by letting f(x,y)=y^x and add the partial derivative with respect to y and the partial derivative with respect to x leaving us with x*y^x-1 +y^x*ln(x) then set x=y and u get the usual result! Magic!
Last time on dr peyam: gradients and line integrals This time on dr peyam: wHaTS A pARtiAL DeRiVAtiVe They are definitely a whole lot of fun. Especially considering other coordinate systems. Use the chen lu!
Thank you :) Will be good also to tell about full (or complete, idk) differentials of this functions (idk, how does they calling on english, but I know, that ODEs with it calling as exact ODE's), and what they means :) I know something about it, and I can easily calculate them, but what they are really means - I haven't thought about it :D
So can you take a partial derivative in a direction other than x or y? E.g. along the line y=2x? Is there a way to get an equation in x,y,θ where θ is an arbitrary direction? Can you take the derivative along a curve? Do these have value other than as answers to "can you"?
you can, but to understand how you need to do a bit of vector calculus. its not as bad as it sounds though, i just dont remember how off the top of my head
in some direction from the origin x = at, y = bt, z = ct ... where t is some real parameter plug that into f(x,y,z...) to get f(at,bt,ct,...) then find the derivative of that with respect to t. Only caveat is you have to rescale your answer because you could make those constants arbitrarily large so I think you have to divide it all by ||grad(at,bt,ct,...)_t|| = ||(a,b,c,...)|| = sqrt(a^2 + b^2 + c^2 + ...) so to find the partial derivative of f(x,y,z,...) in the direction (a,b,c...) from the origin it's (d/dt (f(at,bt,ct,...)))/(sqrt(a^2+b^2+c^2+...)) would be more complicated if it's not the origin we want to find the directional derivative at, but we can just shift all the variables, let's say we are trying to find the derivative in the direction of (a,b,c...) at the point (A,B,C...) then x = at+A, y = bt+B, z = ct+C ... ||grad(at+A,bt+B,ct+C,...)_t|| = ||(a,b,c,...)|| = sqrt(a^2 + b^2 + c^2 + ...) so it's (d/dt (f(at+A,bt+B,ct+C,...)))/(sqrt(a^2+b^2+c^2+...)) so if we want to find this derivative as a function of f(x,y,z...) then (A,B,C...) = input (x,y,z) so we get (d/dt (f(x+at,y+bt,z+ct,...)))/(sqrt(a^2+b^2+c^2+...))
I find the fx and fy too confusing. I prefer the long hand method. The way I learnt at university was to write the partial differential with a long line by the side with the other variables on the right hand side. This way you know what you are differentiate with and what you are keeping constant. Important when you are differentiate with 4 variables. Short hand can be too confusing.
Next video: What is a number? . . . . Euclid: A number is a multitude composed of units, where a unit is that by virtue of which each of the things that exist is called one.
@@drpeyam That'd be cool, like constructing the natural numbers from set-theoretic axioms? You should mention Euclid's definition cause it doesn't even apply to the integers haha.
Dr Peyam, I don't understand the end of the lecture. If fxy=fyx, I'm not able say anything about continuity mixed derivatives using Clairaut theory.What does this example prove? Thank you.
What I meant is that if fxy and fyx are continuous then by clairaut’s theorem fxy = fyx. Most functions in a multivariable course satisfy that assumption
Dr Peyam, could you please do a video explaining why is the gradient perpendicular to level sets? I would really love to see your perspective on that :)
the board:
fₓᵧ = 6y²
fᵧₓ = 6y²
Peyam the Explorer: WHICH ONE OF THOSE 4 ARE EQUAL? (dum dum dum dum)
Have you seen that way of differentiating x^x by letting f(x,y)=y^x and add the partial derivative with respect to y and the partial derivative with respect to x leaving us with x*y^x-1 +y^x*ln(x) then set x=y and u get the usual result!
Magic!
That just seems like a coincidence lol
Hmmmm! Or maybe it isn’t???
@@drpeyam the plot thickens...
Really wonderfully taught. Thank you Dr. Peyam!
Last time on dr peyam: gradients and line integrals
This time on dr peyam: wHaTS A pARtiAL DeRiVAtiVe
They are definitely a whole lot of fun. Especially considering other coordinate systems. Use the chen lu!
I like to mix it up haha
You’re awesome! Thank you some much for bring us that knowledge and share with us this amazing videos! :)
Why didn't I find this video three years ago?
Because it was uploaded 3 weeks ago?
Next video will be partial integration. I am 95% sure about that.🙂
Hahaha
Next video: What is a Maths?
Machen Sie ein Video auf Deutsch, bitte!
Hab ich schon gemacht!
Es heißt Ladder Problem
Danke schön
I was eating sand when I was 12
That's a very intuitive, easy to follow explanation!
BRUH HOW DID U EVEN PUT THE FUNCTION TO PICTURE SO WELLL?!?!? MY GOD...
You are such a great storyteller!! 😂👍
Great video. Exciting to watch
Thank you :)
Will be good also to tell about full (or complete, idk) differentials of this functions (idk, how does they calling on english, but I know, that ODEs with it calling as exact ODE's), and what they means :) I know something about it, and I can easily calculate them, but what they are really means - I haven't thought about it :D
not *thinked* but *thought* !
you probably mean total derivative
Yes, a video on total derivatives please! :)
@@Apollorion ye, thank you, I'm not good enough for making nice comments on english :)
So can you take a partial derivative in a direction other than x or y? E.g. along the line y=2x? Is there a way to get an equation in x,y,θ where θ is an arbitrary direction? Can you take the derivative along a curve? Do these have value other than as answers to "can you"?
Of course! It’s called a directional derivative
Nice username btw!
you can, but to understand how you need to do a bit of vector calculus. its not as bad as it sounds though, i just dont remember how off the top of my head
in some direction from the origin
x = at, y = bt, z = ct ... where t is some real parameter
plug that into f(x,y,z...) to get f(at,bt,ct,...)
then find the derivative of that with respect to t.
Only caveat is you have to rescale your answer because you could make those constants arbitrarily large so I think you have to divide it all by ||grad(at,bt,ct,...)_t|| = ||(a,b,c,...)|| = sqrt(a^2 + b^2 + c^2 + ...)
so to find the partial derivative of f(x,y,z,...) in the direction (a,b,c...) from the origin
it's (d/dt (f(at,bt,ct,...)))/(sqrt(a^2+b^2+c^2+...))
would be more complicated if it's not the origin we want to find the directional derivative at, but we can just shift all the variables, let's say we are trying to find the derivative in the direction of (a,b,c...) at the point (A,B,C...)
then x = at+A, y = bt+B, z = ct+C ...
||grad(at+A,bt+B,ct+C,...)_t|| = ||(a,b,c,...)|| = sqrt(a^2 + b^2 + c^2 + ...)
so it's (d/dt (f(at+A,bt+B,ct+C,...)))/(sqrt(a^2+b^2+c^2+...))
so if we want to find this derivative as a function of f(x,y,z...) then (A,B,C...) = input (x,y,z)
so we get (d/dt (f(x+at,y+bt,z+ct,...)))/(sqrt(a^2+b^2+c^2+...))
Bring on the Jacobian!
Thanks sirr
Thank you good Dr :)
I find the fx and fy too confusing. I prefer the long hand method. The way I learnt at university was to write the partial differential with a long line by the side with the other variables on the right hand side. This way you know what you are differentiate with and what you are keeping constant. Important when you are differentiate with 4 variables. Short hand can be too confusing.
Cool video!!! Wish you could do a total differential video
Thank you for explaining all this calc 2 topics!
Nice! I just registered for PDE class in autumn.
Next video: What is a number?
.
.
.
.
Euclid: A number is a multitude composed of units, where a unit is that by virtue of which each of the things that exist is called one.
I’ll actually make a video on the natural numbers soon 😂
@@drpeyam That'd be cool, like constructing the natural numbers from set-theoretic axioms? You should mention Euclid's definition cause it doesn't even apply to the integers haha.
Dr Peyam, I don't understand the end of the lecture. If fxy=fyx, I'm not able say anything about continuity mixed derivatives using Clairaut theory.What does this example prove? Thank you.
What I meant is that if fxy and fyx are continuous then by clairaut’s theorem fxy = fyx. Most functions in a multivariable course satisfy that assumption
Stay safe Dr.!
🤓
Dr Peyam, could you please do a video explaining why is the gradient perpendicular to level sets? I would really love to see your perspective on that :)
Oh that’s easy, consider any curve r(t) on the level set f = c, then f(r(t)) = c for all t and then just use the Chen lu
@@drpeyam Yes, indeed, but I was referring to the geometric view. Tbh I cannot visualize how and/or why that works.