I have officially passed 2 of my classes with good grades thanks to you. I appreciate your clear and quick explanation. All people can learn but not a lot of people can teach. You have a gift. Thanks again!
yes! if dy/dt was negative, you'd end up with a difference instead of a sum. We just put the plus sign there, because it still allows you to consider the sign. Because if dy/dt is negative, then you'd get +(-dy/dt), and you'd end up with the negative sign in the end. so it still works. i hope that makes sense, and i'm glad you liked the video!! :D
I only took the very basic differential and integral calc in college, and although I did well in the classes, I didn't quite have a fundamental grasp of anything higher level. I happened to stumble upon this video, and somehow, so many things clicked in place. I really understand the math now. Thank you.
because du/dt means that you're taking the derivative of u with respect to t, but you can only do that when you have a single variable. in this case, because we have two variables (x and y), which means you can only take partial derivatives of u, one with respect to x and another with respect to y. when you add the partials together to show the derivative of u with respect to t, you still have to indicate that it's made up of partial derivatives, so you use partial u / partial t.
This is remarkable; an extremely impressive method not only for remembering the chain rule for taking partial derivatives of multi-variables but also understanding it. Right at the end when you have written down all 3 equations of the partial derivatives; How about showing how they are all connected together? In other words telling us how those 3 equations are related to the total differential du? Do you simply sum the 3 equations together to get du? Many thanks, what an amazing teacher!
I'm 99% sure you multiply your derivatives at the end. So you would have (du/dr)(du/ds)(du/dt)= ((df/du)?). I just don't know the notation for the total derivative sorry.
It's the sum of those 3 equations in the video with each multiplied by their respective independent differential to give the total differential du (sorry didn't mean to say total derivative). I actually knew this was the answer. But I wanted to know if Krista had a neat proof or intuitive understanding? This website proves that it is the sum with each multiplied by it's independent respective differential: www.solitaryroad.com/c353.html I've never heard of any reasoning behind multiplying derivatives like that before. I'm not suggesting that you are wrong; I've just never seen it. Thanks for your reply!
So I know for sure that if you have a function z=f(u,v) where u=f(x,y) and v=f(x,y) then the derivative of z = (dz/dx)(dz/dy) and (dz/dx) = (dz/du)(du/dx)+(dz/dv)(dv/dx) and (du/dy) = (dz/du)(du/dy)+(dz/dv)(dv/dy) I don't know if that helps at all? I'm still learning it myself. I'll have a read through that link tomorrow as it's quite late here, so thankyou :)
It's difficult to answer this question as I cannot tell if dz/dx is an ordinary derivatve or a partial derivative but I assume from the context that it's partial. But what do you mean by "then the derivative of of z = (dz/dx)(dz/dy)"? Why are you multiplying partials together like that? I'm curious now as someone who has been using partial derivatives and the chain rule for years, and yet I have never seen a reason to multiply them together like that. As said before; I am not hinting that you are incorrect, I may be misinterpreting what it is you are writing. Kindest regards, get some sleep and look at that link lol ;)
Name suggestion: at a couple points in this video Krista's explanation was "straddling" Partial Derivatives and..... dx/dy. I suggest calling dx/dy "full derivatives".
great video! I don't know why my teacher haven't showed the tree diagram method, it's very systematic. One question though! Could you still use the tree diagrams if for example the variable r = r(p,q) or something? So generally speaking: If one of the independent variables in your case 2 example where actually dependent upon p and q, could one still use the tree diagram systematically?
Need Help badly: At 7:06,case 2 type: What if only the X-variable had one independent variable (let's say S); would the differential of X with respect to S be a partial or full differential? Thanks in advance
This is a helpful video, except for that there is a mistake. in case 1, du/dt is supposed to be a total derivative, not a partial. There is only one independent.
4 people failed their exams and blamed this video rather than themselves. Or maybe 4 college professors lost their jobs as this video did it for them. Either way
+trainedtiger That sounds like three separate equations for r(theta), s(theta), and t(theta). In that case, you're dealing with parametric equations, which is a totally separate thing from what we're doing here.
No not parametrizing the equation, I mean a third "level". z=f(x,y), x= f(s,t) y= f(s,t), where s is a function of theta, and t is a function of theta, finding dz/dtheta.
I have officially passed 2 of my classes with good grades thanks to you. I appreciate your clear and quick explanation. All people can learn but not a lot of people can teach. You have a gift. Thanks again!
Aw thanks! I'm glad I can help. :)
Isn't she great...
@@hg2. she is.
I spent 50 minutes with my professor without real understanding and I couldn't be better without you.
I wish I had a teacher like you at my college! Love how you explain and your handwriting is beautiful! ♡
Thank you so much, Aashima! :D
Watching this video exactly eight years after it was published. Still very helpful, thank you
You're welcome, Zachary, I'm glad it helped! :)
yes! if dy/dt was negative, you'd end up with a difference instead of a sum. We just put the plus sign there, because it still allows you to consider the sign. Because if dy/dt is negative, then you'd get +(-dy/dt), and you'd end up with the negative sign in the end. so it still works. i hope that makes sense, and i'm glad you liked the video!! :D
This is really clear, way better than my prof, thanks!
I am watching this in 2021 from Indian and thanks for Google get an amazing tutor like you❤️ thanks a lot mam❤️❤️
Made in 2013, still helpful to this date! THANKS A BUNCH!
Your website Math services are outstanding! I wish you much success Krista! you deserve the best! Thanks for sharing too
I only took the very basic differential and integral calc in college, and although I did well in the classes, I didn't quite have a fundamental grasp of anything higher level.
I happened to stumble upon this video, and somehow, so many things clicked in place. I really understand the math now. Thank you.
This helped me a lot, thank you.
You're welcome, Tazmeen! I'm so glad it helped! :)
Very nice videos. Of your videos I have watched, your videos are MUCH more concise and helpful then the majority of calculus videos out there! Thanks
+Trein Veracity Aw thanks! I'm so glad you like them.
You such an amazing explainer. Thank you for this video.
You're welcome, Chriv, I'm so glad it made sense! :D
because du/dt means that you're taking the derivative of u with respect to t, but you can only do that when you have a single variable. in this case, because we have two variables (x and y), which means you can only take partial derivatives of u, one with respect to x and another with respect to y. when you add the partials together to show the derivative of u with respect to t, you still have to indicate that it's made up of partial derivatives, so you use partial u / partial t.
Just discovered your videos and they are SUPER helpful!! I especially like how detailed you are. THANK YOU!
Great video and break down of how to use the chain rule in each circumstance!
+Megan L Thanks! Glad you liked it!
+Megan L Thanks!
you're welcome, i'm so glad they're helpful!! :D
Thank you Krista
You’re so welcome! ❤️
great explaination!
Right back at ya! Glad you liked it! :D
Good luck on Tuesday! :D I plan to have more MV videos, but not likely before your exam.
I understand partial derivatives now, thank you!
Very interesting! That's just what we learned today in AB Calculus!
Very clear explanation! Thank you!
+Nahom Girmay You're welcome!
Thanks For the review, very refreshing!
You Rock Krista
Just in time for my Multi Variable exam next Tuesday. This weekend would be a great time to put up some new MV videos, too! Haha Thanks!
This lesson was very useful
Glad to hear that, John! :D
thank you for such nice explanation :)
You're welcome, Kamran! :)
This is remarkable; an extremely impressive method not only for remembering the chain rule for taking partial derivatives of multi-variables but also understanding it. Right at the end when you have written down all 3 equations of the partial derivatives; How about showing how they are all connected together? In other words telling us how those 3 equations are related to the total differential du? Do you simply sum the 3 equations together to get du? Many thanks, what an amazing teacher!
I'm 99% sure you multiply your derivatives at the end. So you would have (du/dr)(du/ds)(du/dt)= ((df/du)?). I just don't know the notation for the total derivative sorry.
It's the sum of those 3 equations in the video with each multiplied by their respective independent differential to give the total differential du (sorry didn't mean to say total derivative). I actually knew this was the answer. But I wanted to know if Krista had a neat proof or intuitive understanding? This website proves that it is the sum with each multiplied by it's independent respective differential:
www.solitaryroad.com/c353.html
I've never heard of any reasoning behind multiplying derivatives like that before. I'm not suggesting that you are wrong; I've just never seen it. Thanks for your reply!
So I know for sure that if you have a function z=f(u,v) where u=f(x,y) and v=f(x,y)
then the derivative of z = (dz/dx)(dz/dy)
and (dz/dx) = (dz/du)(du/dx)+(dz/dv)(dv/dx)
and (du/dy) = (dz/du)(du/dy)+(dz/dv)(dv/dy)
I don't know if that helps at all? I'm still learning it myself.
I'll have a read through that link tomorrow as it's quite late here, so thankyou :)
It's difficult to answer this question as I cannot tell if dz/dx is an ordinary derivatve or a partial derivative but I assume from the context that it's partial. But what do you mean by "then the derivative of of z = (dz/dx)(dz/dy)"? Why are you multiplying partials together like that? I'm curious now as someone who has been using partial derivatives and the chain rule for years, and yet I have never seen a reason to multiply them together like that. As said before; I am not hinting that you are incorrect, I may be misinterpreting what it is you are writing. Kindest regards, get some sleep and look at that link lol ;)
amazing tutor
Thank you so much. You were very clear and helpful!
You're welcome, I'm glad it helped!!
Days of confusion fixed in 8 minutes, Thanks!
wow! perfect timing, that's awesome! :D
Name suggestion: at a couple points in this video Krista's explanation was "straddling" Partial Derivatives and..... dx/dy. I suggest calling dx/dy "full derivatives".
you're the best..ur handwriting is sooo beautiful..u explain clearly...❤❤❤
Thanks, Sizwe!
Thanks!
Yay! Glad I could help. :)
Love all ur ved... Thanks
thanks... very helpful
Glad you liked it! :)
Very easy to understand! Thank you!!
Love from Pakistan! :)
Glad it could help!
Thank you very much! This was very helpful!
You're welcome, I'm so glad it helped!
great video! I don't know why my teacher haven't showed the tree diagram method, it's very systematic.
One question though!
Could you still use the tree diagrams if for example the variable r = r(p,q) or something? So generally speaking: If one of the independent variables in your case 2 example where actually dependent upon p and q, could one still use the tree diagram systematically?
Thankq
You're welcome, Sandeep! :)
Thank you mam!
You're welcome, Nikhil! :)
Need Help badly: At 7:06,case 2 type: What if only the X-variable had one independent variable (let's say S); would the differential of X with respect to S be a partial or full differential? Thanks in advance
In case 1 it should be du/dt (ordinary derivative) instead of partial derivative, shouldn’t it?
oh good! thanks! :D
This is excellent!
Thanks!
In case 1, why is it partial u by partial t instead of just du/dt ?
OK, thanks.
I love you sooooooo much
Yay!! :D
grat!
how do I calculate for second order derivatives??
you voice your accent the way you say things is making me fall for you.. i cant concentrate
This is a helpful video, except for that there is a mistake. in case 1, du/dt is supposed to be a total derivative, not a partial. There is only one independent.
this is awesome yho
4 people failed their exams and blamed this video rather than themselves.
Or maybe 4 college professors lost their jobs as this video did it for them.
Either way
I can't find anything anywhere that would describe if r,s, and t all depended on a variable like theta
+trainedtiger That sounds like three separate equations for r(theta), s(theta), and t(theta). In that case, you're dealing with parametric equations, which is a totally separate thing from what we're doing here.
No not parametrizing the equation, I mean a third "level". z=f(x,y), x= f(s,t) y= f(s,t), where s is a function of theta, and t is a function of theta, finding dz/dtheta.
I have a facebook page here :) facebook . com/integralCALC however I can't help with specific problems.
What if `x` depends on `t`, but then `t` depends on `x` again? :J
Video was awesome no doubt, but I didn't know you were so beautiful until at the end :) Cheers
You should seriously consider becoming a university teacher, if you're not one already.
Your subtitles hides the solution
how can I communicate whit you ? Do you have a facebook account ?
my lord-- please have mercy on me-- please teach me contour integration --
thanking u madam