Best Explanation of Partial Derivatives and Gradients
290K views · Oct 7, 2025 · Education
Comments · 152
@satishgoda · 8 months ago
This video needs a Nobel Prize! Thank you for breaking it down visually!
28
@PratyushKhandelwal-n3n · 11 months ago
This is more than awesome :face-blue-smiling::face-blue-smiling::face-purple-wide-eyes:
21
@popogast · 11 months ago (edited)
<a href="https://www.youtube.com/watch?v=TYLyAfFn_ME&t=96">1:36</a> z = T(0,0) = 10 is shown correctly by the graph. But z = T(-10,-10) should be -190.<br>Also the gradient should become steeper and steeper with increasing distance from (0,0) if the function were correct.
7
@moonlightis-h2j · 11 months ago
U always make hard topics easy to understand....
59
@kazukawasaki97 · 11 months ago
more I NEED MORE
23
@DmitryBogachev-f2l · 8 months ago (edited)
The visualization is a little unlucky. The gradient of the two-args function is a vector in the (x, y) plain, it's not a black arrow pointing upwards as shown at <a href="https://www.youtube.com/watch?v=TYLyAfFn_ME&t=268">04:28</a>.
2
@calebchencs · 5 months ago
This tutorial is awesome! Thank you so much!
1
@rugeneus · 9 months ago
It's briliant explanation. Now everything is crystal clear.
3
@thinkmath4270 · 9 months ago
This is one I needed. i couldn't wrap head around during lectures, how all these corelates each other especially on 3 dimensional graph. amazing video. thanks so much !!
1
@etherealvox2010 · 8 months ago
This video illustrates the difference between teaching out of passion vs ... whatever reasons teachers and professors teach for , cause it sure ain't passion ...
4
@ruszkait · 10 months ago
The black gradient vector also has a length. At the flat field, it shall shrink to a null vector (instead of pointing upwards, or anywhere at all), and only when it climbs the hill or descends into the hole, then it shall have a length. Growing till the middle of the hill and decrease again to the summit.
1
@kyh2617 · 6 months ago
Thank you, professor. Your lessons are very useful for my AI studies.
2
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