Is there a higher dimensional implicit differentiation hax (sorry I'm doing maths whilst being a bit hyper) to get me a shortcut for this question
Or should I just chain rule it because z = +'ve sqrt(...)
Last edited by dan964; 8 Jun 2017 at 12:05 AM.
I was kind of hoping InteGrand would use the word 'hax' ahahah
Veni, Vidi, Vici
Last edited by InteGrand; 19 May 2016 at 11:29 PM.
Perhaps in some notations it will seem that way, but they can be unified remarkably well once one learns to think of the derivative of a function f at a point p as nothing more than a linear map between the vector space of tangents at p and the vector space of tangents at f(p) (pretty much the most flexible and abstract way to think of the classical derivative).
What it then boils down to is that if we have two smooth functions and , then is smooth and .
(Where, X,Y,Z are open subsets of arbitrary dimension Euclidean spaces, or more generally of curves/surfaces/manifolds.)
Sidebar: Spivak has a less well known book "Calculus on Manifolds", that is amazing for learning multivariable calculus. From introducing partial derivatives and Riemann integration to differential forms and the modern version of Stokes theorem. It is quite compact and terse though and considerably less accessible than his "Calculus".
Last edited by seanieg89; 20 May 2016 at 1:34 PM.
Getting stuck on how to incorporate the 0.02 correctly
Progress so far:
Last edited by leehuan; 22 May 2016 at 4:13 PM.
This thread may be packed with questions...cause my book doesn't give answers to almost anything. Sorry!
I did something like this and I'm not sure if I can break it down any further
Ah ok so you basically do leave it as is I see. So is this right...?
Cool. Ok this one is twisting with me but here's my attempt
There are currently 1 users browsing this thread. (0 members and 1 guests)
Bookmarks