The chain rule with constraints | MIT 18.02SC Multivariable Calculus, Fall 2010

The chain rule with constraints
Instructor: David Jordan
View the complete course: ocw.mit.edu/18-...
License: Creative Commons BY-NC-SA
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[@mihaelgolob8739]

The simplest solution for this specific problem is to just subtitute t with xy/z and then do the normal partial differential.

[@bobkameron]

Great content!

[@xsli2876]

Super! This is what MIT for! By the way, very handsome.

[@LAnonHubbard]

Saw the lecture. Made my brain hurt. I'm not understanding it all completely and think I'll need to watch the lecture and this recitation video a few more times perhaps. Thanks.

[@narutouzumakix9201]

do you understand it now?

[@32_gurjotsingh82]

@@narutouzumakix9201 does he now?

[@clockwork4068]

@@32_gurjotsingh82 Does he now?

[@mathadventuress]

Do you understand now

[@advaithkumar5966]

Correct me if i'm wrong here, but can't we directly write zt = c for some constant c so z*dt + t*dz = 0
and also since x and y are constants dw is directly -2zt*dz - z^2*dt and we get the same answer ( -zt )?
w is essentially f(z,t) here so chain rule directly gives the latter expression

[@xnonqme3716]

that's pretty much what he did by setting dy = dx = 0. when doing the problem you should be aware of the fact that x and y don't change and not waste time doing their partial derivatives. he did it this way for clarity i suppose.

[@aashsyed1277]

ggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggreat content!