Differential Calculus. The Derivative and Rules of Differentiation

Differential Calculus: The Derivative and Rules of Differentiation 2020 Limits Question 1: Find limx→3 f (x): f (x) = x2 − 9 x−3 (A) +∞ (B) -6 (C) 6 (D) Does not exist! (E) None of the above 2 −9 Answer: (C) Note the the function f (x) = xx−3 = (x−3)(x+3) = x + 3 is actually a line. However it x−3 is important to note the this function is undefined at x = 3. Why? x = 3 requires dividing by zero (which is inadmissible). As x approaches 3 from below and from above, the value of the function f (x) approaches f (3) = 6. Thus the limit limx→3 f (x) = 6. Question 2: Find limx→2 f (x): f (x) = 1776 (A) +∞ (B) 1770 (C) −∞ (D) Does not exist! (E) None of the above Answer: (E) The limit of any constant function at any point, say f (x) = C, where C is an arbitrary constant, is simply C. Thus the correct answer is limx→2 f (x) = 1776. Question 3: Find limx→4 f (x): f (x) = ax2 + bx + c (A) +∞ (B) 16a + 4b + c (C) −∞ (D) Does not exist! (E) None of the above 1 Answer: (B) Applying the rules of limits: limx→4 ax2 + bx + c = limx→4 ax2 + limx→4 bx + limx→4 c = a [limx→4 x]2 + blimx→4 x + c = 16a + 4b + c Question 4: Find the limits in each case: x2 |x| x→0 (i) lim 2x+3 x→3 4x−9 (ii) lim x2 −3x x→6 x+3 (iii) lim Answer: (i) (ii) (iii) lim 2x+3 x→3 4x−9 (|x|)2 x→0 |x| x2 |x| x→0 lim = 2 lim x −3x x→6 x+3 = lim 2·3+3 4·3−9 = = lim | x |= 0 x→0 =3 62 −3·6 6+3 =2 Question 5: Show that lim sin x = 0 (Hint: −x ≤ sin x ≤ x for all x ≥ 0.) x→0 Answer: Given hint and squeeze theorem we have lim −x = 0 ≤ lim sin x ≤ 0 = lim x hence, x→0 x→0 x→0 lim sin x = 0 x to0 Question 6: Show that lim x sin( x1 ) = 0 x→0 Answer: Note first that for any real number t we have −1 ≤ sin t ≤ 1 so −1 ≤ sin( x1 ) ≤ 1. Therefore, −x ≤ x sin( x1 ) ≤ x and by squeeze theorem lim x sin x1 = 0. x→0 Continuity and Differentiability Question 7: Which of the following functions are NOT everywhere continuous: (A) f (x) = x2 −4 x+2 (B) f (x) = (x + 3)4 (C) f (x) = 1066 (D) f (x) = mx + b (E) None of the above Answer: (A) Remember that, informally at least, a continuous function is one in which there are no breaks its curve. A continuous function can be drawn without lifting your pencil from the paper. More formally, a function f (x) is continuous at the point x = a if and only if: 1. f (x) is defined at the point x = a, 2. the limit limx→a f (x) exists, 3. limx→a f (x) = f (a) 2 −4 is not everywhere continuous because the function is not defined at the point The function f (x) = xx+2 x = −2. It is worth noting that limx→−2 f (x) does in fact exist! The existence of a limit at a point does not guarantee that the function is continuous at that point! 2 Question 8: Which of the following functions are continuous: (A) f (x) = |x|  3 (B) f (x) = 1 2 (C) f (x) = x<4 x≥4 x+3 1 x  (D) f (x) = ln x 0 x<0 x=0 (E) None of the above Answer: (A) The absolute value function f (x) = |x| is defined as:  x x≥0 f (x) = −x x<0 Does this function satisfy the requirements for continuity? Yes! The critical point to check is x = 0. Note that the function is defined at x = 0; the limx→0 f (x) exists; and that limx→0 f (x) = 0 = f (0). Question 9: Which of the following functions are NOT differentiable: (A) f (x) = |x| (B) f (x) = (x + 3)4 (C) f (x) = 1066 (D) f (x) = mx + b (E) None of the above Answer: (A) Remember that continuity is a necessary condition for differentiability (i.e., every differentiable function is continuous), but continuity is not a sufficient condition to ensure differentiability (i.e., not every continuous function is differentiable). Case in point is f (x) = |x|. This function is in fact continuous (see previous question). It is not however differentiable at the point x = 0. Why? The point x = 0 is a cusp (or kink). There are an infinite number of lines that could be tangent to the function f (x) = |x| at the point x = 0, and thus the derivative of f (x) would have an infinite number of possible values. Question 10: Is function  f (x) = 0 :x=0 x sin(1/x) : x 6= 0 continuous at point 0? Answer: Note that f is continuous at a point a if lim f (x) = f (lim x). x→a x→a In this case, we take a = 0 and lim f (x) = lim x sin(1/x) = 0 x→0 x→0 by question 6. Moreover, f (lim x) = f (0) = 0 x→0 thus, f is continuous at 0. 3 Derivatives Question 11: Find the derivative of the following function: f (x) = 1963 (A) +∞ (B) 1963 (C) −∞ (D) 0 (E) None of the above Answer: (D) The derivative of a constant function is always zero. Question 12: Find the derivative of the following function: f (x) = x2 + 6x + 9 (A) f 0 (x) = 2x + 6 + 9 (B) f 0 (x) = x2 + 6 (C) f 0 (x) = 2x + 6 (D) f 0 (x) = 2x (E) None of the above Answer: (C) Remember that 1) the derivative of a sum of functions is simply the sum of the derivatives of each of the functions, and 2) the power rule for derivatives says that if f (x) = kxn , then f 0 (x) = nkxn−1 . Thus f 0 (x) = 2x2−1 + 6x1−1 + 0 = 2x + 6. Question 13: Find the derivative of the following function: 1 f (x) = x 2 (A) f 0 (x) = − 2√1 x (B) f 0 (x) = √1 x (C) f 0 (x) = 1 √ 2 x (D) f 0 (x) = √ x (E) None of the above Answer: (C) Remember that the power rule for derivatives works with fractional exponents as well! 1 1 Thus f 0 (x) = 21 x 2 −1 = 12 x− 2 = 2√1 x . Question 14: Find the derivative of the following function: f (x) = 5x2 (x + 47) (A) f 0 (x) = 15x2 + 470x (B) f 0 (x) = 5x2 + 470x (C) f 0 (x) = 10x (D) f 0 (x) = 15x2 − 470x 4 (E) None of the above Answer: (A) Ideally, you would solve this problem by applying the product rule. Set g(x) = 5x2 and h(x) = (x + 47), then f (x) = g(x)h(x). Apply the product rule: f 0 (x) = g 0 (x)h(x) + g(x)h0 (x) = 10x(x + 47) + 5x2 (1) = 10x2 + 470x + 5x2 = 15x2 + 470x Question 15: Find the derivative of the following function: f (x) = (A) f 0 (x) = 5x2 −470x (x+47)2 (B) f 0 (x) = 10x2 +470x (x+47) 5x2 x + 47 (C) f 0 (x) = 10x (D) f 0 (x) = 5x2 +470 (x+47)2 (E) None of the above Answer: (A) Ideally, you would solve this problem by applying the quotient rule. Set g(x) = 5x2 and g(x) h(x) = (x + 47), then f (x) = h(x) . Apply the quotient rule: g 0 (x)h(x) − g(x)h0 (x) h(x)2 10x(x + 47) − 5x2 (1) = (x + 47)2 10x2 + 470x − 5x2 = (x + 47)2 5x2 + 470x = (x + 47)2 f 0 (x) = Question 16: Find the derivative of the following function: f (x) = 5(x + 47)2 (A) f 0 (x) = 15x2 + 470x (B) f 0 (x) = 10x − 470 (C) f 0 (x) = 10x + 470 (D) f 0 (x) = 15x2 − 470x (E) None of the above Answer: (C) Ideally, you would solve this problem by applying the chain rule. Set g(h) = 5h2 and h(x) = (x + 47), then f (x) = g(h(x)). Apply the chain rule: f 0 (x) = g 0 (h)h0 (x) = 10h = 10(x + 47) = 10x + 470 5 Higher Order Derivatives Question 17: Find the second derivative of the following function: f (x) = 5x2 (x + 47) (A) f 00 (x) = 30x − 470 (B) f 00 (x) = 30x + 470 (C) f 00 (x) = 15x2 + 235 (D) f 00 (x) = 15x2 + 470x (E) None of the above Answer: (B) The second derivative is just the derivative of the first derivative. Simplest solution would be to multiply to re-write the function as f (x) = 5x2 (x + 47) = 5x3 + 235x2 . Now take the derivative: f 0 (x) = 15x2 + 470x. Taking the derivative again yields the second derivative: f 00 (x) = 30x + 470. Question 18: Find the third derivative of the following function: f (x) = 5x2 (x + 47) (A) 15 (B) 15 + x (C) 30x (D) 30x + 470 (E) None of the above Answer: (E) Just take the derivative of your answer to Question 12 to get the third derivative of f (x) = 5x2 (x + 47). Answer: f 000 (x) = 30. Question 19: Suppose that you have the following utility function: √ u(x) = x 00 (x) Find − uu0 (x) . (A) 1 2x 1 (B) − 2x (C) 2x (D) −2x (E) None of the above 00 (x) Answer: (A) The ratio − uu0 (x) is called the Arrow-Pratt measure of relative risk aversion and you will encounter it in core microeconomics. The first derivative of the utility function (otherwise known as 3 marginal utility) is u0 (x) = 2√1 x (see Question 9 above). The second derivative is u00 (x) = − 14 x− 2 = − 4√1x3 . Thus the Arrow-Pratt measure of relative risk aversion is: √ − 4√1x3 u00 (x) 2 x 1 − 0 =− 1 = √ = √ u (x) 2x 4 x3 2 x 6