MATH 241 Lecture 13: MATH241_Leture13

• School: University of Maryland
• Department: Mathematics
• Course: MATH 241
• Professor: Christian Rosendal

Document Summary

Math241 lecture 13 higher point partial derivatives. Just as regular derivative, we can take second, third etc, partial derivatives, however, there are options and a notation note. There are four second partial derivatives in this case. Case 4: (cid:1877) then (cid:1876) (cid:4666)(cid:1876),(cid:1877)(cid:4667)=(cid:1876)(cid:2870)(cid:1877)+(cid:884)(cid:1876)(cid:1877)(cid:2871)+(cid:1877) (cid:3051)=(cid:1876)=(cid:884)(cid:1876)(cid:1877)+(cid:884)(cid:1877)(cid:2871) (cid:3052)=(cid:1877)=(cid:1876)(cid:2870)+(cid:888)(cid:1876)(cid:1877)(cid:2870)+(cid:883) (cid:3051)(cid:3051)=(cid:2870)(cid:1876)(cid:2870)=(cid:884)(cid:1877) (cid:3051)(cid:3052)= (cid:2870)(cid:1877)(cid:1876)=(cid:884)(cid:1876)+(cid:888)(cid:1877)(cid:2870) (cid:3052)(cid:3052)=(cid:2870)(cid:1877)(cid:2870)=(cid:883)(cid:884)(cid:1876)(cid:1877) (cid:3052)(cid:3051)= (cid:2870)(cid:1876)(cid:1877)=(cid:884)(cid:1876)+(cid:888)(cid:1877)(cid:2870) (cid:3051)(cid:3052)= (cid:2870)(cid:1877)(cid:1876) Note for us we will always have (cid:3051)(cid:3052)=(cid:3052)(cid:3051) and similarly for other variables. Comment 1: goal is to deal with functions which are functions of variables which depend on other variables. Comment 2: chain rule in 13. 4 has mostly abstract uses but we will use it with calculations to practice. Comment 3: you can ignore most of what is in the text, we will focus on the core idea. Example 1: suppose (cid:1878)=(cid:1876)(cid:2870)(cid:1877)+(cid:1876) (cid:1876)=(cid:1871)(cid:2870)(cid:1872) (cid:1877)=(cid:1871)+(cid:1872) (cid:1871),(cid:1872) we might ask: We observes that (cid:1878) depends (cid:1876) and (cid:1877) which then depends on (cid:1871) and (cid:1872). The chain rule says we can find this as follows.