Student Number:_{} No._{} Marks _{}
1. You have 50 minutes for this exam. You are not allowed to use any calculators.
2. This exam contains 3 questions and has a total of 100 marks.
3. Show all of your work. Your work must justify the answer that you
give.
1 (30 Marks) Find the derivative of each of the following functions.
(1) f(x) = 4Öx+(1[1/x])^{2}.
Solution: f¢(x) = [4/(2Öx)]+2(1[1/x])[1/(x^{2})] = [2/(Öx)]+[(2(x1))/(x^{3})].
(2) f(x) = cos(5x+3).
Solution: f¢(x) = sin(5x+3)(5x+3)¢ = 5sin(5x+3).
(3) f(x) = sec^{2}(x+1).
Solution: f¢(x) = 2sec(x+1)(sec(x+1))¢ = 2sec(x+1)tan(x+1)sec(x+1) = 2sec^{2}(x+1)tan(x+1).
(4) f(x) = cot^{4}(5x^{2}2).
Solution:

2 (20 Marks) Find the second derivative of the following implicitly defined function.

Solution: Taking the derivative of this equation relative to x, we have 6x^{2}+3y^{2}y¢ = 0, that is,
 (1) 
Hence, y¢ = [(2x^{2})/(y^{2})]. Taking the derivative of (1), we obtain

This implies

3 (50 Marks) Let f(x) = [(x^{2})/(1+x)].
(1) Find the domain of f.
(2) Find all the critical points of f and the second derivative of f.
(3) Find the intervals on which f is increasing, decreasing, concave down or concave up.
(4) Evaluate the local maximum and local minimum values of f.
(5) Find all the asymptotes of f.
(6) Find the xintercepts and yintercept of f.
(7) Sketch the graph of f.
Solution: (1) The domain of f is (¥, 1)È(1,¥).
(2) f¢(x) = [(x(x+2))/((1+x)^{2})]. Let f¢(x) = 0. Then x = 0 and x = 2. Moreover, f¢(1) does not exist. Hence, all the critical points are x = 1, x = 2 and x = 0.
f¢¢(x) = [2/((1+x)^{3})]. f¢¢(x) ¹ 0 for x ¹ 1 and f¢¢(1) does not exist.
(3) We list the following table.

f is increasing on (¥, 2) and (0, ¥) and decreasing on (2, 1) and (1, 0). f is concave up on (¥, 1) and concave down on (1, ¥).
(4) The local maximum value f(2) = 4 and the local minimum value f(0) = 0
(5) a = lim_{x®¥}[f(x)/x] = [(x^{2})/(x^{2}+x)] = 1 and b = lim_{x® ¥}(f(x)ax) = lim_{x® ¥}([(x^{2})/(1+x)]x) = 1. Hence, y = ax+b = x1 is an asymptote.
lim_{x® 1}f(x) = ¥ and lim_{x® 1}f(x) = ¥. Hence, x = 1 is a vertical asymptote.
(6) x = 0 is the xintercept of f and f(0) = 0 is the yintercept of f.
(7) Sketch the graph of f as follows.