MATH116 Lecture Notes - Lecture 16: Hypotenuse, Implicit Function

• School: University of Waterloo
• Department: Mathematics
• Course: MATH116
• Professor: Fiona Dunbar

Document Summary

Practice solutions 5: find dy/dx by implicit di erentiation. (a) x2 + xy + y2 = 7 (b) sin(x) + cos(y) = sin(x) cos(y) 2x + (1y + xy(cid:48)) + 2yy(cid:48) = 0 2x + y + xy(cid:48) + 2yy(cid:48) = 0. Xy(cid:48) + 2yy(cid:48) = 2x y y(cid:48)(x + 2y) = 2x y y(cid:48) = 2x y x + 2y (b) cos(x) sin(y)y(cid:48) = cos(x) cos(y) sin(x) sin(y)y(cid:48) sin(y)y(cid:48)+sin(x) sin(y)y(cid:48) = cos(x) cos(y) cos(x) Y(cid:48)(sin(y))(sin(x) 1) = cos(x)(cos(y) 1) y(cid:48) = cos(x)(cos(y) 1) sin(y)(sin(x) 1: find y(cid:48)(cid:48) by implicit di erentiation. y2 xy = 3. 2yy(cid:48) (y + xy(cid:48)) = 0 2yy(cid:48) xy(cid:48) = y y(cid:48) = y. Di erentiating again, y(cid:48)(cid:48) = y(cid:48)(2y x) y(2y(cid:48) 1) (2y x)2 y. 2y x 1) (2y x)2 y(2y x) y(2y (2y x)) (2y x)3. 2y2 xy xy (2y x)3 = 2y(y x) (2y x)3: prove that d dx arccos(x) = 1 . Then cos(y) = x (with x [0, ]).