MATH117 Lecture Notes - Lecture 3: Implicit Function, Asymptote

• School: University of Waterloo
• Department: Mathematics
• Course: MATH117
• Professor: Andrew Hunt

Document Summary

Let"s try a bit more di cult versions of di erentiation. 1. 1 example 1 - implicit di erentiation with logs. Using implicit di erentiation, di erentiate and evaluate the derivative at and nd y(cid:48)(cid:18) (cid:19) This is solved using the following process: y = (3x)5x2 ln|y| = ln|(3x)5x2| 3x y y(cid:48) = y(10xln|3x| + 5x) y(cid:48) = (3x)5x2 (10xln|3x| + 5x) (cid:19) ln (cid:12)(cid:12)(cid:12)(cid:12)3 (cid:18) 1. 3 for y"(1/3), we nd that: y(cid:48)(cid:18) 1 y(cid:48)(cid:18) 1. Observe the following function: x2 + y2 = 25 (1) (2) (3) (4) (5) (6) (7) (8) (9) Note that this is the equation of a circle of radius 5 (for those who dont know this already). Let"s try using implicit di erentiation to solve this: x2 + y2 = 25. 2x + 2yy(cid:48) = 0 y(cid:48) = x y (10) (11) (12) Check for the asymptotes (as proof that this is an equation of a circle).