MATH114 Final: MATH114 Final Exam 2015 Fall Solutions

• School: University of Alberta
• Department: Mathematics
• Course: MATH114
• Professor: All

Document Summary

Y = (tan 3x)ln x(cid:20) ln(tan 3x) 3 ln x sec2 3x ln(tan(3x)) + ln x . 3 sec2 3x tan 3x y y = tan 3x (cid:21) x (c) y = 3(tan2 x2 sin x)(sec2 x2 sin x) 1. 2 (2x cos x) (d) use logarithmic di erentiation. ln y = 3 ln(sin x) + 1 y = 2 sin3 x 5x5 + 1 ex3(2 5x2)5 (cid:20)3 cot x + (3x5 + 1) 3x2 . 2x y cos(x y2) x 2y cos(x y2) (g) we use logarithmic di erentiation: taking natural log, we get ln y = (tan x) ln(sec x) 1 y y = (sec2 x) ln(sec x) + tan x sec x (h) di . implicitly. exy(y + xy ) 2yy = 1 x = . 1 yexy + xexyy 2yy = 1 x = (xexy 2y)y = 1 x yexy = y =