Chapter 2: Differentiation

Size: px
Start display at page:

Download "Chapter 2: Differentiation"

Transcription

1 Chapter 2: Differentiation Spring 2018 Department of Mathematics Hong Kong Baptist University 1 / 82

2 2.1 Tangent Lines and Their Slopes This section deals with the problem of finding a straight line L that is tangent to a curve C at a point P. Let C be the graph of y = f (x), and P be the point (x 0, y 0 ) on C so that y 0 = f (x 0 ). Also assume that P is not an endpoint of C. A reasonable definition of tangency can be stated in terms of limits (of the slopes of secant lines). 2 / 82

3 3 / 82

4 4 / 82

5 Given the curve y = f (x) and the point P = (x 0, f (x 0 )), we choose a different point Q on the curve C so that Q = (x 0 + h, f (x 0 + h)). Note that h can be positive or negative, depending on whether Q is to the right or left of P. The line through P and Q is called a secant line to the curve. The slope of the line PQ is f (x 0 + h) f (x 0 ). h This expression is called the Newton quotient or difference quotient for f at x 0. 5 / 82

6 Nonvertical tangent line Definition Suppose that the function f is continuous at x = x 0 and that f (x 0 + h) f (x 0 ) lim = m (m is finite) h 0 h exists. Then the straight line having slope m and passing through the point P = (x 0, f (x 0 )) is called the tangent line (or simply the tangent) to the graph of y = f (x) at P. An equation of this tangent line is y = m(x x 0 ) + y 0. 6 / 82

7 Example 1 Find an equation of the tangent line to the curve y = x 2 at the point (1, 1). Solution: Note that f (x) = x 2, x 0 = 1 and y 0 = f (x 0 ) = 1. By definition, the slope of the tangent line is m = f (1 + h) f (1) lim h 0 h = (1 + h) 2 1 lim h 0 h = lim (2 + h) h 0 = 2. Accordingly, the equation of the tangent line is y = m(x x 0 ) + y 0 = 2(x 1) + 1 = 2x 1. 7 / 82

8 8 / 82

9 (a) Vertical tangents or (b) no tangent line Definition (a) If f is continuous at P = (x 0, y 0 ), where y 0 = f (x 0 ), and if either f (x 0 + h) f (x 0 ) f (x 0 + h) f (x 0 ) lim = or lim =, h 0 h h 0 h then the vertical line x = x 0 is tangent to the graph y = f (x) at P. (b) If the limit of the Newton quotient fails to exist in any other way than by being or, the graph y = f (x) has no tangent line at P. 9 / 82

10 Example 2 Find the tangent line to the curve y = x 1/3 at the point (0, 0). Solution: For this curve, the limit of the Newton quotient for f at x = 0 is f (0 + h) f (0) h 1/3 lim = lim h 0 h h 0 h = lim 1 =. h 0 h2/3 Thus by definition, the vertical line x = 0 (i.e., the y-axis) is the tangent line to the curve y = x 1/3 at the point (0, 0). 10 / 82

11 11 / 82

12 Example 3 Does the graph of y = x have a tangent line at x = 0? Solution: For this curve, the Newton quotient is f (x 0 + h) f (x 0 ) h = 0 + h 0 h = sgn(h). Now since sgn(h) has different right and left limits at 0, the Newton quotient has no limit at h 0. This implies that y = x has no tangent line at (0, 0). Remark: The curve suddenly changes the direction and is not smooth at x = / 82

13 13 / 82

14 The slope of a curve Definition The slope of a curve C at a point P is the slope of the tangent line to C at P if such a tangent line exists. In particular, the slope of the graph of y = f (x) at the point x 0 is lim h 0 f (x 0 + h) f (x 0 ). h 14 / 82

15 Example 4 Find the slope of the curve y = x/(2x + 3) at the point x = 3. Solution: The slope of the curve at x = 3 is f (3 + h) f (3) m = lim h 0 h = lim h 0 3+h 9+2h 1 3 h = lim h h = / 82

16 2.2 The Derivative Definition The derivative of a function f is another function f defined by f (x) = lim h 0 f (x + h) f (x) h at all points x for which the limit exists (i.e., is a finite real number). If f (x) exists, we say that f is differentiable at x. Remark: The derivative f is read as f prime. The derivative of f is essentially the slope function of f. The process of calculating the derivative f of a given function f is called differentiation. 16 / 82

17 Differentiability on Intervals A function f is differentiable on an open interval (a, b) if f (x) exists for all x in (a, b). A function f is differentiable on a closed interval [a, b] if f (x) exists for all x in (a, b) and f +(a) and f (b) both exist, where f +(a) = lim h 0+ f (a + h) f (a), h and f (b) = lim h 0 f (b + h) f (b). h 17 / 82

18 Example 5: (The derivative of a linear function) Show that if f (x) = ax + b, then f (x) = a. Solution: By definition, we have f (x) = f (x + h) f (x) lim h 0 h = a(x + h) + b (ax + b) lim h 0 h = ah lim h 0 h = a. A special case: If g(x) = c (constant), then g (x) = / 82

19 Theorem (Power Rule) If f (x) = x n, then f (x) = nx n 1. Proof (when n is a positive integer): To prove the theorem, we need the following factorization formula: Then a n b n = (a b)(a n 1 + a n 2 b + + ab n 2 + b n 1 ). f (x) = lim h 0 (x + h) n x n h = lim h 0 [ (x + h) n 1 + (x + h) n 2 x + + (x + h)x n 2 + x n 1] = nx n / 82

20 Table 1: Some elementary functions and their derivatives f(x) f (x) c 0 x 1 x x x 2 (x 0) 1 x 2 x (x > 0) x r rx r 1 (r 0 and x r 1 is real) x 2x sgn(x) = x x 20 / 82

21 Leibniz Notation 21 / 82

22 By the definition of derivative, we define the Leibniz notation as dy dx = lim y x 0 x. d The symbol dx is a differential operator and is read as the derivative with respect to x of... Remarks: (i) d is not a variable, so do not cancel it out from the numerator and denominator! (ii) There are situations in which it may be convenient to consider dx and dy as separate quantities. However, they are NOT real numbers and do not always follow the usual rules of arithmetic. 22 / 82

23 The following notations are equivalent: dy dx = y = D x y = d dx f (x) = f (x) = D x f (x) = Df (x). The following notations are equivalent: dy = y x=x0 = D x y dx x=x0 x=x0 = d dx f (x) x=x0 = f (x 0 ) = D x f (x 0 ). The symbol x=x0 is called an evaluation symbol, i.e., the expression is evaluated at x = x 0. For instance, d dx x 3 x= 4 = 3x 2 x= 4 = 3( 4) 2 = / 82

24 2.3 Differentiation Rules Theorem If f is differentiable at x, then f is continuous at x. Proof: Since f is differentiable at x, by definition we have f (x + h) f (x) lim = f (x), h 0 h where f (x) is finite. Using the limit rules, ( ) f (x + h) f (x) lim (f (x + h) f (x)) = lim (h) h 0 h 0 h = f (x) lim h 0 h = 0. This leads to lim h 0 f (x + h) = f (x). That is, f is continuous at x. 24 / 82

25 Linear Rules of Differentiation Theorem If f and g are differentiable at x, and if c is a constant, then the functions f + g, f g and cf are all differentiable at x and (f (x) + g(x)) = f (x) + g (x), (f (x) g(x)) = f (x) g (x), (cf (x)) = cf (x). 25 / 82

26 Proof: (i) For the sum rule, by definition we have (f (x) + g(x)) (f (x + h) + g(x + h)) (f (x) + g(x)) = lim h 0 ( h ) f (x + h) f (x) g(x + h) g(x) = lim + h 0 h h f (x + h) f (x) g(x + h) g(x) = lim + lim h 0 h h 0 h = f (x) + g (x). (ii) The proofs of the difference rule and constant rule are similar and are thus omitted. 26 / 82

27 Product Rule of Differentiation Theorem [Product Rule] If f and g are differentiable at x, then their product fg is also differentiable at x, and (f (x)g(x)) = f (x)g(x) + f (x)g (x). 27 / 82

28 Proof: For the Product rule, by definition we have (f (x)g(x)) = lim h 0 f (x + h)g(x + h) f (x)g(x) h f (x + h)g(x + h) f (x)g(x + h) + f (x)g(x + h) f (x)g(x) = lim h 0 h ( ) f (x + h) f (x) g(x + h) g(x) = lim g(x + h) + f (x) h 0 h h f (x + h) f (x) g(x + h) g(x) = lim lim g(x + h) + f (x) lim h 0 h h 0 h 0 h = f (x)g(x) + f (x)g (x). This proves the theorem. 28 / 82

29 Example 6 Find the derivative of (x 2 + 1)(x 3 + 4) using and without using the Product rule. Solution: (i) Using the Product rule, we have d dx (x 2 + 1)(x 3 + 4) = 2x(x 3 + 4) + (x 2 + 1)(3x 2 ) = 5x 4 + 3x 2 + 8x. (ii) Without using the Product rule, we have d dx (x 2 + 1)(x 3 + 4) = d dx (x 5 + x 3 + 4x 2 + 4) = 5x 4 + 3x 2 + 8x. We note that the two solutions are the same. 29 / 82

30 Quotient Rule of Differentiation Theorem [Quotient Rule] If f and g are differentiable at x, and if g(x) 0, then the quotient f /g is differentiable at x and ( ) f (x) = f (x)g(x) f (x)g (x) g(x) g 2. (x) 30 / 82

31 Proof: For the Quotient rule, by definition we have ( ) f (x) f (x + h)/g(x + h) f (x)/g(x) = lim g(x) h 0 h f (x + h)g(x) f (x)g(x + h) = lim h 0 hg(x)g(x + h) 1 = g 2 (x) lim f (x + h)g(x) f (x)g(x + h) h 0 h = 1 g 2 (x) lim h 0 = f (x)g(x) f (x)g (x). g 2 (x) This proves the theorem. [ f (x + h) f (x) g(x) f (x) h ] g(x + h) g(x) h 31 / 82

32 Reciprocal Rule of Differentiation Theorem [Reciprocal Rule] If g is differentiable at x and if g(x) 0, then the reciprocal 1/g is also differentiable at x and ( ) 1 = g (x) g(x) g 2 (x). Remarks: (i) The Reciprocal rule is a special case of the Quotient rule with f (x) = 1. (ii) By the Reciprocal rule, we can generalize the Power rule to allow negative powers: d dx x n = nx n / 82

33 Example 7 Find the derivatives of (a) y = x 2 + x + 1 x 3 ; and (b) y = x 2 x + 1. Solution: 33 / 82

34 Example 7 Find the derivatives of (a) y = x 2 + x + 1 x 3 ; and (b) y = x 2 x + 1. Solution: (a) By the generalized Power rule, we have dy dx = d ( x 1 + x 2 + x 3) = (x 2 + 2x 3 + 3x 4 ). dx 34 / 82

35 Example 7 Find the derivatives of (a) y = x 2 + x + 1 x 3 ; and (b) y = x 2 x + 1. Solution: (a) By the generalized Power rule, we have dy dx = d ( x 1 + x 2 + x 3) = (x 2 + 2x 3 + 3x 4 ). dx (b) By the Quotient rule, we have dy dx = (x 2 ) (x + 1) x 2 (x + 1) (x + 1) 2 = 2x(x + 1) x 2 (x + 1) 2 = x 2 + 2x (x + 1) / 82

36 2.4 The Chain Rule Theorem [Chain Rule] If f (u) is differentiable at u = g(x), and g(x) is differentiable at x, then the composite function f g(x) = f (g(x)) is differentiable at x, and (f (g(x))) = f (g(x))g (x). In Leibniz s notation, by letting y = f (u) where u = g(x), we have y = f (g(x)) and dy dx = dy du du dx. Remark: The proof of the Chain rule is not required for the course. 36 / 82

37 The Chain rule is designed for composite functions and is the most commonly used rule of all the differentiation rules. To understand the Chain rule, we can treat the Leibniz notations dy/dx, dy/du, and du/dx as if they were quotients of two quantities. The Chain rule can then be written as dy dx = dy du du dx. Assume that y = f (u), u = g(v) and v = h(x). Then y = f (g(h(x))) and the Chain rule is dy dx = dy du dv du dv dx. The Chain rule can be extended to composite functions with any finite number of intermediate functions. 37 / 82

38 Example 8 Verify the Chain rule using the function y = 1/(x 2 4). Proof: 38 / 82

39 Example 8 Verify the Chain rule using the function y = 1/(x 2 4). Proof: (i) By the Reciprocal rule, we have dy dx = (x 2 4) (x 2 4) 2 = 2x (x 2 4) / 82

40 Example 8 Verify the Chain rule using the function y = 1/(x 2 4). Proof: (i) By the Reciprocal rule, we have dy dx = (x 2 4) (x 2 4) 2 = 2x (x 2 4) 2. (ii) To verify the Chain rule, we let y = f (g(x)) with f (u) = 1/u and u = g(x) = x 2 4. Then by the Chain rule, dy dx = dy du du dx = 1 u 2 (2x) = 2x (x 2 4) 2. Thus, we obtain the same answer with the Chain rule. 40 / 82

41 Example 9 Find the derivative of the function f (x) = (x + 1 x )4. Solution: Writing y = u 4 where u = x + 1/x, we get f (x) = dy dx = dy du du dx = (4u 3 )(1 1 x 2 ) = 4(x + 1 x )3 (1 1 x 2 ). Remark: those who are used to the chain rule will often simply write f (x) = 4(x + 1 [ x )3 x + 1 ] = 4(x + 1 x x )3 (1 1 x 2 ). 41 / 82

42 Example 10 Find the derivative of the function f (x) = (1 + 2x 3 + 1) 3. Solution: Let y = f (x). Then we have y = u 3, u = 1 + v, v = 2x 3 + 1, dy du = 3u2, By the Chain rule, we have du dv = 1 2 v, dv dx = 6x 2. f (x) = dy du dv du dv dx = 1 3u2 2 v 6x 2 = 9x 2 (1 + 2x 3 + 1) 3. 2x / 82

43 Some Built-in Chain Rules If u is a differentiable function of x, then we have the built-in Chain rules for the following functions: ( ) d 1 = 1 du dx u u 2 dx (Reciprocal Rule) d 1 u = dx 2 du u dx (Square Root Rule) d dx ur r 1 du = ru dx (General Power Rule) d u = sgn(u)du dx dx (Absolute Value Rule) 43 / 82

44 Derivatives of Basic Elementary Functions Recall that the elementary functions are built from the basic elementary functions and constants through composition (f g) and combinations using operations (+,, and /). By the above rules of differentiation, we conclude that the derivatives of elementary functions are all available, given that the derivatives of basic elementary functions are provided. We have learned the differentiation rules for power functions. In 2.5, we study the differentiation rules for trigonometric functions. The differentiation rules for (i) exponential functions, (ii) logarithmic functions, and (iii) inverse trigonometric functions will be introduced in Chapter / 82

45 2.5 Derivatives of Trigonometric Functions The trigonometric functions, especially sine and cosine, play a very important role in the mathematical modeling of real-world phenomena. In particular, they arise whenever quantities fluctuate in a periodic way. This section provides details in calculating the derivatives of sine and cosine functions. The other four will then follow from known identities and the differentiation rules in Section / 82

46 Some Special Limits Theorem The functions sin(θ) and cos(θ) are continuous at every value of θ. In particular, at θ = 0 we have lim sin(θ) = sin(0) = 0 θ 0 and lim cos(θ) = cos(0) = 1. θ 0 Theorem For the sine function, we have sin(θ) lim = 1. θ 0 θ 46 / 82

47 Example 11 Show that cos(h) 1 lim = 0. h 0 h Proof: Let θ = h/2, then h = 2θ. By P.7, we have This leads to cos(2θ) = cos 2 (θ) sin 2 (θ) = 1 2 sin 2 (θ). cos(h) 1 lim h 0 h cos(2θ) 1 = lim θ 0 2θ sin 2 (θ) = lim θ 0 θ sin(θ) = lim θ 0 θ = 0. lim θ 0 sin(θ) 47 / 82

48 Theorem The derivative of the sine function is d sin(x) = cos(x). dx Proof: We have d sin(x + h) sin(x) sin(x) = lim dx h 0 h h 2 cos(x + = lim 2 ) sin(h 2 ) h 0 = lim h 0 cos(x + h 2 ) lim h 0 = cos(x). h sin(h/2) h/2 48 / 82

49 Theorem The derivative of the cosine function is d cos(x) = sin(x). dx Proof: We have d cos(x + h) cos(x) cos(x) = lim dx h 0 h h 2 sin(x + = lim 2 ) sin(h 2 ) h 0 = lim h 0 sin(x + h 2 ) lim h 0 = sin(x). h sin(h/2) h/2 Another simple proof is to apply for the Chain rule. Specifically, d dx cos(x) = d dx sin(π 2 x) = cos(π 2 x) = sin(x). 49 / 82

50 Derivatives of Other Trigonometric Functions tan(x) = sin(x) cos(x) cot(x) = cos(x) sin(x) sec(x) = 1 cos(x) csc(x) = 1 sin(x) d dx tan(x) = sec2 (x) d dx cot(x) = csc2 (x) d sec(x) = sec(x) tan(x) dx d csc(x) = csc(x) cot(x) dx 50 / 82

51 2.6 High-order Derivatives Definition The second derivative of a function is the derivative of its derivative. If y = f (x), the second derivative is defined by y = f (x). The following notations are referred to as the same: y = f (x) = d 2 y dx 2 = d d dx dx f (x) = d 2 dx 2 f (x) = D2 x y = Dx 2 f (x). 51 / 82

52 Example 12 Find the second derivative of f (x) = 5x 4 3x 2 + 2x 7. Solution: The first derivative is f (x) = 20x 3 6x + 2. The second derivative is then f (x) = 60x / 82

53 Definition For any positive integer n, the nth derivative of a function is obtained from the function by differentiating successively n times. If the original function is y = f (x), the nth derivative is denoted by Specifically, we have f (k) (x) = d dx d n y dx n or f (n) (x). [ ] f (k 1) (x), k = 1, 2,, n, if f (0) (x), f (1) (x),, f (n) (x) are all differentiable. 53 / 82

54 Example 13 Calculate the first 3 derivatives of f (x) = x Solution: Note that f (x) = (x 2 + 1) 1/2. We have f (x) = 1 2 (x 2 + 1) 1/2 (2x) = x(x 2 + 1) 1/2, f (x) = (x 2 + 1) 1/2 + x( 1 2 )(x 2 + 1) 3/2 (2x) = (x 2 + 1) 3/2, f (3) (x) = ( 3 2 )(x 2 + 1) 5/2 (2x) = 3x(x 2 + 1) 5/2. 54 / 82

55 By induction, we have the following higher order derivatives for some important functions: n! 1) (x n ) (k) = (n k)! x n k k = 1, 2,, n (2.5.1) 0 k > n ( 1) k 1 cos(x) n = 2k 1 2) sin (n) (x) = ( 1) k sin(x) n = 2k ( 1) k sin(x) n = 2k 1 3) cos (n) (x) = ( 1) k cos(x) n = 2k (2.5.2) (2.5.3) 55 / 82

56 2.8 The Mean-Value Theorem Theorem (Mean-Value Theorem) Suppose that the function f is continuous on the closed, finite interval [a, b] and that it is differentiable on the open interval (a, b). Then there exists a point c in the open interval (a, b) such that f (b) f (a) = f (c). b a This says that the slope of the chord line joining the points (a, f (a)) and (b, f (b)) is equal to the slope of the tangent line to the curve y = f (x) at the point (c, f (c)), so the two lines are parallel. 56 / 82

57 57 / 82

58 The assumptions of the Mean-Value theorem are all necessary for the conclusion; see counterexamples on the next slide. The Mean-Value theorem gives no indication of how many points C there may be on the curve between A and B where the tangent is parallel to AB. The Mean-Value theorem gives us no information on how to find the point c, even though such a point must exist. This type of theoretical tool is known as an existence theorem. 58 / 82

59 59 / 82

60 60 / 82

61 Example 14 Show that sin(x) < x for all x > 0. Proof: (i) If x > π/2, then sin(x) 1 < π/2 < x. (ii) If 0 < x π/2, then by the Mean-Value theorem, there exists a point c (0, π/2) such that sin(x) x = sin(x) sin(0) x 0 = d dx sin(x) x=c = cos(c) < 1. By (i) and (ii), sin(x) < x for all x > / 82

62 Increasing and Decreasing Functions Definition Let f be a function defined on an interval I and x 1, x 2 be two points in I. (a) If f (x 1 ) < f (x 2 ) whenever x 1 < x 2, then f is increasing on I. (b) If f (x 1 ) f (x 2 ) whenever x 1 < x 2, then f is nondecreasing on I. (c) If f (x 1 ) > f (x 2 ) whenever x 1 < x 2, then f is decreasing on I. (d) If f (x 1 ) f (x 2 ) whenever x 1 < x 2, then f is nonincreasing on I. 62 / 82

63 Increasing and Decreasing Functions 63 / 82

64 Increasing and Decreasing Functions Theorem Let J be an open interval, and let I be an interval consisting of all points in J and possibly one or both of the endpoints of J. Suppose that f is continuous on I and differentiable on J. (a) If f (x) > 0 for all x in J, then f is increasing on I. (b) If f (x) < 0 for all x in J, then f is decreasing on I. (c) If f (x) 0 for all x in J, then f is nondecreasing on I. (d) If f (x) 0 for all x in J, then f is nonincreasing on I. 64 / 82

65 Example 15 On what intervals is the function f (x) = x 3 12x + 1 increasing? On what intervals is it decreasing? Solution: The derivative of f is f (x) = 3x 2 12 = 3(x 2)(x + 2). Obviously, f (x) > 0 if x < 2 or x > 2; and f (x) < 0 if 2 < x < 2. Therefore, f is increasing on the intervals (, 2) and (2, ) and is decreasing on the interval ( 2, 2). 65 / 82

66 66 / 82

67 Example 16 On what intervals is the function f (x) = what intervals is it decreasing? Solution: The derivative of f is x x increasing? On f (x) = (x) (x 2 + 1) x(x 2 + 1) (x 2 + 1) 2 = 1 x 2 (x 2 + 1) 2. Obviously, f (x) > 0 if 1 < x < 1, and f (x) < 0 if x < 1 or x > 1. Therefore, f is increasing on the interval ( 1, 1), and is decreasing on the intervals (, 1) and (1, ). 67 / 82

68 2.9 Implicit Differentiation If the dependent variable y can be written as an explicit formula of x, say y = f (x), then the derivative of y can be obtained using the differentiation rules. Sometimes y cannot be written as a function of x explicitly. In this case, we say y is an implicit function of x. Write it as the equation F (x, y) = 0. The derivative dy/dx of implicit functions can be obtained by a technique called implicit differentiation. The idea is to differentiate the equation with respect to x, regarding y as a function of x having derivative dy/dx or y. 68 / 82

69 Example 17: Calculate dy/dx at x = 3/5, where y is a function of x defined implicitly by x 2 + y 2 = 1. Solution: The equation represents the unit circle, which is not the graph of a function. However, it defines two differentiable functions y = f (x) and y = g(x), corresponding to the upper and lower branches of the graph: f (x) = 1 x 2, g(x) = 1 x 2. When calculating dy/dx, one must specify which branch of the graph we are interested in. This is usually done by also specifying the y value. 69 / 82

70 1 (3/5, 4/5) (3/5, 4/5) / 82

71 Example 17: Calculate dy/dx at the point (3/5, 4/5), where y is defined implicitly by x 2 + y 2 = 1. Solution: 1 Check that the given point indeed lies on the curve: ( 3 ) 2 ( ) 2 5 = 1. 2 Define the function F (x) = x 2 + (y(x)) 2 1, which is identically zero because y(x) is assumed to satisfy the equation for all x near 3/5. We now differentiate F (x) using the chain rule: df dx = 2x + 2y dy dx = / 82

72 3 Isolate dy/dx to get dy dx = x y dy = dx = x=3/ = 3 4. Remarks: We could have obtained the same answer by evaluating f (3/5), where f (x) = 1 x 2. If we had used the point (3/5, 4/5) instead, the implicit function would have represented g(x) in the lower branch, where dy dx = x=3/ = Thus, the dependence on y in the implicit derivative reflects the fact that the equation defines more than one function. 72 / 82

73 Example 18: Find dy dx if y sin(x) = x 3 + cos(y). Solution: For this equation, y cannot be expressed as an explicit function of x. In what follows we use implicit differentiation. Specifically, we take the derivatives on both sides w.r.t x, This leads to d dx (y sin(x)) = d dx (x 3 ) + d dx cos(y). sin(x) dy dx + y cos(x) = 3x 2 sin(y) dy dx. Solving this equation, we have dy dx = 3x 2 y cos(x) sin(x) + sin(y). 73 / 82

74 Example 19: Find dy dx if xy 2 = y x. Solution: Treat this as an implicit function. Taking the derivatives on both sides w.r.t x, we have This leads to Further, y 2 + 2xy dy dx = dy dx 1 (1 2xy) dy dx = y dy dx = y xy. 74 / 82

75 2.10 Antiderivatives In the previous sections we have concerned with the problem of finding the derivative f of a given function f. The reverse problem, given the derivative f, find f, is also interesting and important. The reverse problem is related to integral calculus. We take a preliminary look at this problem in this section and will return to it with more detail in Chapter / 82

76 Definition A function F (x) is said to be an antiderivative of function f (x) if F (x) = f (x) for every x in the domain of f (x). The process of finding antiderivatives is called anti-differentiation or indefinite integration. 76 / 82

77 Example 20 Verify that F (x) = x 3 + 5x is an antiderivative of f (x) = 3x Solution: F (x) is an antiderivative of f (x) if F (x) = f (x). Differentiating F (x) we get F (x) = 3x = f (x). Remark: Let G(x) = x 3 + 5x + 2. It is easy to verify that G(x) is also an antiderivative of f (x). This shows that the antiderivative of f (x) is not unique. 77 / 82

78 Theorem (Fundamental Property of Antiderivatives) If F (x) is an antiderivative of f (x), then the general expression for antiderivatives of f (x) is where C is an arbitrary constant. F (x) + C, Proof: Suppose that F (x) and G(x) are two antiderivatives of f (x). Let g(x) = G(x) F (x). We have g (x) = G (x) F (x) = f (x) f (x) = 0. This leads to g(x) = C. Further, G(x) = F (x) + g(x) = F (x) + C. 78 / 82

79 Some examples Function Antiderivatives k kx + C x a 1 a + 1 x a+1 + C (a 1) sin(x) cos(x) + C cos(x) 1 cos 2 (x) 1 sin 2 (x) sin(x) + C tan(x) + C cot(x) + C 79 / 82

80 Definition The set of all antiderivatives of f (x) is called the indefinite integral, or simply the integral of f (x) with respect to x, denoted by f (x)dx = F (x) + C, where the symbol is the integral sign, f (x) is the integrand of integral, x is the variable of integration, and C is the constant of integration. 80 / 82

81 Example 21 Find (3 x 2 )dx. Solution: We have (3 x 2 )dx = 3 dx x 2 dx = 3x 1 3 x 3 + C. 81 / 82

82 Example 22 Find sin(2x)dx. Solution: We seek a function whose derivative is sin(2x). From the antiderivative tables, we guess a function of the type g(x) = cos(2x). But then the chain rule implies Dividing both sides by 2 gives g (x) = ( sin(2x)) 2 = 2 sin(2x). d dx ( 1 2 g(x) ) = sin(2x). So the required antiderivative is sin(2x) dx = 1 cos(2x) + C / 82

Chapter 2: Differentiation

Chapter 2: Differentiation Chapter 2: Differentiation Winter 2016 Department of Mathematics Hong Kong Baptist University 1 / 75 2.1 Tangent Lines and Their Slopes This section deals with the problem of finding a straight line L

More information

Chapter P: Preliminaries

Chapter P: Preliminaries Chapter P: Preliminaries Winter 2016 Department of Mathematics Hong Kong Baptist University 1 / 59 Preliminaries The preliminary chapter reviews the most important things that you should know before beginning

More information

2.1 The derivative. Rates of change. m sec = y f (a + h) f (a)

2.1 The derivative. Rates of change. m sec = y f (a + h) f (a) 2.1 The derivative Rates of change 1 The slope of a secant line is m sec = y f (b) f (a) = x b a and represents the average rate of change over [a, b]. Letting b = a + h, we can express the slope of the

More information

Chapter P: Preliminaries

Chapter P: Preliminaries Chapter P: Preliminaries Spring 2018 Department of Mathematics Hong Kong Baptist University 1 / 67 Preliminaries The preliminary chapter reviews the most important things that you should know before beginning

More information

Limit. Chapter Introduction

Limit. Chapter Introduction Chapter 9 Limit Limit is the foundation of calculus that it is so useful to understand more complicating chapters of calculus. Besides, Mathematics has black hole scenarios (dividing by zero, going to

More information

2.2 The derivative as a Function

2.2 The derivative as a Function 2.2 The derivative as a Function Recall: The derivative of a function f at a fixed number a: f a f a+h f(a) = lim h 0 h Definition (Derivative of f) For any number x, the derivative of f is f x f x+h f(x)

More information

DRAFT - Math 101 Lecture Note - Dr. Said Algarni

DRAFT - Math 101 Lecture Note - Dr. Said Algarni 3 Differentiation Rules 3.1 The Derivative of Polynomial and Exponential Functions In this section we learn how to differentiate constant functions, power functions, polynomials, and exponential functions.

More information

CHAPTER 3 DIFFERENTIATION

CHAPTER 3 DIFFERENTIATION CHAPTER 3 DIFFERENTIATION 3.1 THE DERIVATIVE AND THE TANGENT LINE PROBLEM You will be able to: - Find the slope of the tangent line to a curve at a point - Use the limit definition to find the derivative

More information

a x a y = a x+y a x a = y ax y (a x ) r = a rx and log a (xy) = log a (x) + log a (y) log a ( x y ) = log a(x) log a (y) log a (x r ) = r log a (x).

a x a y = a x+y a x a = y ax y (a x ) r = a rx and log a (xy) = log a (x) + log a (y) log a ( x y ) = log a(x) log a (y) log a (x r ) = r log a (x). You should prepare the following topics for our final exam. () Pre-calculus. (2) Inverses. (3) Algebra of Limits. (4) Derivative Formulas and Rules. (5) Graphing Techniques. (6) Optimization (Maxima and

More information

Learning Objectives for Math 165

Learning Objectives for Math 165 Learning Objectives for Math 165 Chapter 2 Limits Section 2.1: Average Rate of Change. State the definition of average rate of change Describe what the rate of change does and does not tell us in a given

More information

Formulas to remember

Formulas to remember Complex numbers Let z = x + iy be a complex number The conjugate z = x iy Formulas to remember The real part Re(z) = x = z+z The imaginary part Im(z) = y = z z i The norm z = zz = x + y The reciprocal

More information

Inverse Trig Functions

Inverse Trig Functions 6.6i Inverse Trigonometric Functions Inverse Sine Function Does g(x) = sin(x) have an inverse? What restriction would we need to make so that at least a piece of this function has an inverse? Given f (x)

More information

MAT137 Calculus! Lecture 6

MAT137 Calculus! Lecture 6 MAT137 Calculus! Lecture 6 Today: 3.2 Differentiation Rules; 3.3 Derivatives of higher order. 3.4 Related rates 3.5 Chain Rule 3.6 Derivative of Trig. Functions Next: 3.7 Implicit Differentiation 4.10

More information

Calculating the Derivative Using Derivative Rules Implicit Functions Higher-Order Derivatives

Calculating the Derivative Using Derivative Rules Implicit Functions Higher-Order Derivatives Topic 4 Outline 1 Derivative Rules Calculating the Derivative Using Derivative Rules Implicit Functions Higher-Order Derivatives D. Kalajdzievska (University of Manitoba) Math 1500 Fall 2015 1 / 32 Topic

More information

Tangent Lines Sec. 2.1, 2.7, & 2.8 (continued)

Tangent Lines Sec. 2.1, 2.7, & 2.8 (continued) Tangent Lines Sec. 2.1, 2.7, & 2.8 (continued) Prove this Result How Can a Derivative Not Exist? Remember that the derivative at a point (or slope of a tangent line) is a LIMIT, so it doesn t exist whenever

More information

Blue Pelican Calculus First Semester

Blue Pelican Calculus First Semester Blue Pelican Calculus First Semester Student Version 1.01 Copyright 2011-2013 by Charles E. Cook; Refugio, Tx Edited by Jacob Cobb (All rights reserved) Calculus AP Syllabus (First Semester) Unit 1: Function

More information

February 21 Math 1190 sec. 63 Spring 2017

February 21 Math 1190 sec. 63 Spring 2017 February 21 Math 1190 sec. 63 Spring 2017 Chapter 2: Derivatives Let s recall the efinitions an erivative rules we have so far: Let s assume that y = f (x) is a function with c in it s omain. The erivative

More information

Calculus I Review Solutions

Calculus I Review Solutions Calculus I Review Solutions. Compare and contrast the three Value Theorems of the course. When you would typically use each. The three value theorems are the Intermediate, Mean and Extreme value theorems.

More information

Calculus & Analytic Geometry I

Calculus & Analytic Geometry I TQS 124 Autumn 2008 Quinn Calculus & Analytic Geometry I The Derivative: Analytic Viewpoint Derivative of a Constant Function. For c a constant, the derivative of f(x) = c equals f (x) = Derivative of

More information

Chapter 3 Differentiation Rules

Chapter 3 Differentiation Rules Chapter 3 Differentiation Rules Derivative constant function if c is any real number, then Example: The Power Rule: If n is a positive integer, then Example: Extended Power Rule: If r is any real number,

More information

b n x n + b n 1 x n b 1 x + b 0

b n x n + b n 1 x n b 1 x + b 0 Math Partial Fractions Stewart 7.4 Integrating basic rational functions. For a function f(x), we have examined several algebraic methods for finding its indefinite integral (antiderivative) F (x) = f(x)

More information

One of the powerful themes in trigonometry is that the entire subject emanates from a very simple idea: locating a point on the unit circle.

One of the powerful themes in trigonometry is that the entire subject emanates from a very simple idea: locating a point on the unit circle. 2.24 Tanz and the Reciprocals Derivatives of Other Trigonometric Functions One of the powerful themes in trigonometry is that the entire subject emanates from a very simple idea: locating a point on the

More information

Chapter 12: Differentiation. SSMth2: Basic Calculus Science and Technology, Engineering and Mathematics (STEM) Strands Mr. Migo M.

Chapter 12: Differentiation. SSMth2: Basic Calculus Science and Technology, Engineering and Mathematics (STEM) Strands Mr. Migo M. Chapter 12: Differentiation SSMth2: Basic Calculus Science and Technology, Engineering and Mathematics (STEM) Strands Mr. Migo M. Mendoza Chapter 12: Differentiation Lecture 12.1: The Derivative Lecture

More information

Find the indicated derivative. 1) Find y(4) if y = 3 sin x. A) y(4) = 3 cos x B) y(4) = 3 sin x C) y(4) = - 3 cos x D) y(4) = - 3 sin x

Find the indicated derivative. 1) Find y(4) if y = 3 sin x. A) y(4) = 3 cos x B) y(4) = 3 sin x C) y(4) = - 3 cos x D) y(4) = - 3 sin x Assignment 5 Name Find the indicated derivative. ) Find y(4) if y = sin x. ) A) y(4) = cos x B) y(4) = sin x y(4) = - cos x y(4) = - sin x ) y = (csc x + cot x)(csc x - cot x) ) A) y = 0 B) y = y = - csc

More information

Sec 4.1 Limits, Informally. When we calculated f (x), we first started with the difference quotient. f(x + h) f(x) h

Sec 4.1 Limits, Informally. When we calculated f (x), we first started with the difference quotient. f(x + h) f(x) h 1 Sec 4.1 Limits, Informally When we calculated f (x), we first started with the difference quotient f(x + h) f(x) h and made h small. In other words, f (x) is the number f(x+h) f(x) approaches as h gets

More information

Math 229 Mock Final Exam Solution

Math 229 Mock Final Exam Solution Name: Math 229 Mock Final Exam Solution Disclaimer: This mock exam is for practice purposes only. No graphing calulators TI-89 is allowed on this test. Be sure that all of your work is shown and that it

More information

Calculus I Announcements

Calculus I Announcements Slie 1 Calculus I Announcements Office Hours: Amos Eaton 309, Monays 12:50-2:50 Exam 2 is Thursay, October 22n. The stuy guie is now on the course web page. Start stuying now, an make a plan to succee.

More information

2.12: Derivatives of Exp/Log (cont d) and 2.15: Antiderivatives and Initial Value Problems

2.12: Derivatives of Exp/Log (cont d) and 2.15: Antiderivatives and Initial Value Problems 2.12: Derivatives of Exp/Log (cont d) and 2.15: Antiderivatives and Initial Value Problems Mathematics 3 Lecture 14 Dartmouth College February 03, 2010 Derivatives of the Exponential and Logarithmic Functions

More information

WORKBOOK. MATH 31. CALCULUS AND ANALYTIC GEOMETRY I.

WORKBOOK. MATH 31. CALCULUS AND ANALYTIC GEOMETRY I. WORKBOOK. MATH 31. CALCULUS AND ANALYTIC GEOMETRY I. DEPARTMENT OF MATHEMATICS AND COMPUTER SCIENCE Contributors: U. N. Iyer and P. Laul. (Many problems have been directly taken from Single Variable Calculus,

More information

Using this definition, it is possible to define an angle of any (positive or negative) measurement by recognizing how its terminal side is obtained.

Using this definition, it is possible to define an angle of any (positive or negative) measurement by recognizing how its terminal side is obtained. Angle in Standard Position With the Cartesian plane, we define an angle in Standard Position if it has its vertex on the origin and one of its sides ( called the initial side ) is always on the positive

More information

3.1 Day 1: The Derivative of a Function

3.1 Day 1: The Derivative of a Function A P Calculus 3.1 Day 1: The Derivative of a Function I CAN DEFINE A DERIVATIVE AND UNDERSTAND ITS NOTATION. Last chapter we learned to find the slope of a tangent line to a point on a graph by using a

More information

Math 250 Skills Assessment Test

Math 250 Skills Assessment Test Math 5 Skills Assessment Test Page Math 5 Skills Assessment Test The purpose of this test is purely diagnostic (before beginning your review, it will be helpful to assess both strengths and weaknesses).

More information

Math Camp II. Calculus. Yiqing Xu. August 27, 2014 MIT

Math Camp II. Calculus. Yiqing Xu. August 27, 2014 MIT Math Camp II Calculus Yiqing Xu MIT August 27, 2014 1 Sequence and Limit 2 Derivatives 3 OLS Asymptotics 4 Integrals Sequence Definition A sequence {y n } = {y 1, y 2, y 3,..., y n } is an ordered set

More information

2. Theory of the Derivative

2. Theory of the Derivative 2. Theory of the Derivative 2.1 Tangent Lines 2.2 Definition of Derivative 2.3 Rates of Change 2.4 Derivative Rules 2.5 Higher Order Derivatives 2.6 Implicit Differentiation 2.7 L Hôpital s Rule 2.8 Some

More information

Topics and Concepts. 1. Limits

Topics and Concepts. 1. Limits Topics and Concepts 1. Limits (a) Evaluating its (Know: it exists if and only if the it from the left is the same as the it from the right) (b) Infinite its (give rise to vertical asymptotes) (c) Limits

More information

There are some trigonometric identities given on the last page.

There are some trigonometric identities given on the last page. MA 114 Calculus II Fall 2015 Exam 4 December 15, 2015 Name: Section: Last 4 digits of student ID #: No books or notes may be used. Turn off all your electronic devices and do not wear ear-plugs during

More information

AP CALCULUS SUMMER WORKSHEET

AP CALCULUS SUMMER WORKSHEET AP CALCULUS SUMMER WORKSHEET DUE: First Day of School, 2011 Complete this assignment at your leisure during the summer. I strongly recommend you complete a little each week. It is designed to help you

More information

Unit IV Derivatives 20 Hours Finish by Christmas

Unit IV Derivatives 20 Hours Finish by Christmas Unit IV Derivatives 20 Hours Finish by Christmas Calculus There two main streams of Calculus: Differentiation Integration Differentiation is used to find the rate of change of variables relative to one

More information

Unit IV Derivatives 20 Hours Finish by Christmas

Unit IV Derivatives 20 Hours Finish by Christmas Unit IV Derivatives 20 Hours Finish by Christmas Calculus There two main streams of Calculus: Differentiation Integration Differentiation is used to find the rate of change of variables relative to one

More information

CALCULUS. Berkant Ustaoğlu CRYPTOLOUNGE.NET

CALCULUS. Berkant Ustaoğlu CRYPTOLOUNGE.NET CALCULUS Berkant Ustaoğlu CRYPTOLOUNGE.NET Secant 1 Definition Let f be defined over an interval I containing u. If x u and x I then f (x) f (u) Q = x u is the difference quotient. Also if h 0, such that

More information

Mth Review Problems for Test 2 Stewart 8e Chapter 3. For Test #2 study these problems, the examples in your notes, and the homework.

Mth Review Problems for Test 2 Stewart 8e Chapter 3. For Test #2 study these problems, the examples in your notes, and the homework. For Test # study these problems, the examples in your notes, and the homework. Derivative Rules D [u n ] = nu n 1 du D [ln u] = du u D [log b u] = du u ln b D [e u ] = e u du D [a u ] = a u ln a du D [sin

More information

Copyright c 2007 Jason Underdown Some rights reserved. quadratic formula. absolute value. properties of absolute values

Copyright c 2007 Jason Underdown Some rights reserved. quadratic formula. absolute value. properties of absolute values Copyright & License Formula Copyright c 2007 Jason Underdown Some rights reserved. quadratic formula absolute value properties of absolute values equation of a line in various forms equation of a circle

More information

f(x 0 + h) f(x 0 ) h slope of secant line = m sec

f(x 0 + h) f(x 0 ) h slope of secant line = m sec Derivatives Using limits, we can define the slope of a tangent line to a function. When given a function f(x), and given a point P (x 0, f(x 0 )) on f, if we want to find the slope of the tangent line

More information

3 Algebraic Methods. we can differentiate both sides implicitly to obtain a differential equation involving x and y:

3 Algebraic Methods. we can differentiate both sides implicitly to obtain a differential equation involving x and y: 3 Algebraic Methods b The first appearance of the equation E Mc 2 in Einstein s handwritten notes. So far, the only general class of differential equations that we know how to solve are directly integrable

More information

Section 4.4. Using the Fundamental Theorem. Difference Equations to Differential Equations

Section 4.4. Using the Fundamental Theorem. Difference Equations to Differential Equations Difference Equations to Differential Equations Section 4.4 Using the Fundamental Theorem As we saw in Section 4.3, using the Fundamental Theorem of Integral Calculus reduces the problem of evaluating a

More information

MATH1013 Calculus I. Revision 1

MATH1013 Calculus I. Revision 1 MATH1013 Calculus I Revision 1 Edmund Y. M. Chiang Department of Mathematics Hong Kong University of Science & Technology November 27, 2014 2013 1 Based on Briggs, Cochran and Gillett: Calculus for Scientists

More information

AP CALCULUS SUMMER WORKSHEET

AP CALCULUS SUMMER WORKSHEET AP CALCULUS SUMMER WORKSHEET DUE: First Day of School Aug. 19, 2010 Complete this assignment at your leisure during the summer. It is designed to help you become more comfortable with your graphing calculator,

More information

Chapter 3: Transcendental Functions

Chapter 3: Transcendental Functions Chapter 3: Transcendental Functions Spring 2018 Department of Mathematics Hong Kong Baptist University 1 / 32 Except for the power functions, the other basic elementary functions are also called the transcendental

More information

Math 106 Calculus 1 Topics for first exam

Math 106 Calculus 1 Topics for first exam Math 06 Calculus Topics for first exam Precalculus = what comes before its. Lines and their slopes: slope= rise over run = (change in y-value)/(corresponding change in x value) y y 0 slope-intercept: y

More information

Section 4.8 Anti Derivative and Indefinite Integrals 2 Lectures. Dr. Abdulla Eid. College of Science. MATHS 101: Calculus I

Section 4.8 Anti Derivative and Indefinite Integrals 2 Lectures. Dr. Abdulla Eid. College of Science. MATHS 101: Calculus I Section 4.8 Anti Derivative and Indefinite Integrals 2 Lectures College of Science MATHS 101: Calculus I (University of Bahrain) 1 / 28 Indefinite Integral Given a function f, if F is a function such that

More information

UNIT 3: DERIVATIVES STUDY GUIDE

UNIT 3: DERIVATIVES STUDY GUIDE Calculus I UNIT 3: Derivatives REVIEW Name: Date: UNIT 3: DERIVATIVES STUDY GUIDE Section 1: Section 2: Limit Definition (Derivative as the Slope of the Tangent Line) Calculating Rates of Change (Average

More information

f(g(x)) g (x) dx = f(u) du.

f(g(x)) g (x) dx = f(u) du. 1. Techniques of Integration Section 8-IT 1.1. Basic integration formulas. Integration is more difficult than derivation. The derivative of every rational function or trigonometric function is another

More information

(e) 2 (f) 2. (c) + (d). Limits at Infinity. 2.5) 9-14,25-34,41-43,46-47,56-57, (c) (d) 2

(e) 2 (f) 2. (c) + (d). Limits at Infinity. 2.5) 9-14,25-34,41-43,46-47,56-57, (c) (d) 2 Math 150A. Final Review Answers, Spring 2018. Limits. 2.2) 7-10, 21-24, 28-1, 6-8, 4-44. 1. Find the values, or state they do not exist. (a) (b) 1 (c) DNE (d) 1 (e) 2 (f) 2 (g) 2 (h) 4 2. lim f(x) = 2,

More information

Math 180, Final Exam, Fall 2012 Problem 1 Solution

Math 180, Final Exam, Fall 2012 Problem 1 Solution Math 80, Final Exam, Fall 0 Problem Solution. Find the derivatives of the following functions: (a) ln(ln(x)) (b) x 6 + sin(x) e x (c) tan(x ) + cot(x ) (a) We evaluate the derivative using the Chain Rule.

More information

Core Mathematics 3 Differentiation

Core Mathematics 3 Differentiation http://kumarmaths.weebly.com/ Core Mathematics Differentiation C differentiation Page Differentiation C Specifications. By the end of this unit you should be able to : Use chain rule to find the derivative

More information

Chapter 2 Derivatives

Chapter 2 Derivatives Contents Chapter 2 Derivatives Motivation to Chapter 2 2 1 Derivatives and Rates of Change 3 1.1 VIDEO - Definitions................................................... 3 1.2 VIDEO - Examples and Applications

More information

f(x) f(a) Limit definition of the at a point in slope notation.

f(x) f(a) Limit definition of the at a point in slope notation. Lesson 9: Orinary Derivatives Review Hanout Reference: Brigg s Calculus: Early Transcenentals, Secon Eition Topics: Chapter 3: Derivatives, p. 126-235 Definition. Limit Definition of Derivatives at a point

More information

ECM Calculus and Geometry. Revision Notes

ECM Calculus and Geometry. Revision Notes ECM1702 - Calculus and Geometry Revision Notes Joshua Byrne Autumn 2011 Contents 1 The Real Numbers 1 1.1 Notation.................................................. 1 1.2 Set Notation...............................................

More information

TRIGONOMETRY OUTCOMES

TRIGONOMETRY OUTCOMES TRIGONOMETRY OUTCOMES C10. Solve problems involving limits of trigonometric functions. C11. Apply derivatives of trigonometric functions. C12. Solve problems involving inverse trigonometric functions.

More information

2009 A-level Maths Tutor All Rights Reserved

2009 A-level Maths Tutor All Rights Reserved 2 This book is under copyright to A-level Maths Tutor. However, it may be distributed freely provided it is not sold for profit. Contents the derivative formula 3 tangents & normals 7 maxima & minima 10

More information

Math 180, Exam 2, Practice Fall 2009 Problem 1 Solution. f(x) = arcsin(2x + 1) = sin 1 (3x + 1), lnx

Math 180, Exam 2, Practice Fall 2009 Problem 1 Solution. f(x) = arcsin(2x + 1) = sin 1 (3x + 1), lnx Math 80, Exam, Practice Fall 009 Problem Solution. Differentiate the functions: (do not simplify) f(x) = x ln(x + ), f(x) = xe x f(x) = arcsin(x + ) = sin (3x + ), f(x) = e3x lnx Solution: For the first

More information

Chapter 2: Functions, Limits and Continuity

Chapter 2: Functions, Limits and Continuity Chapter 2: Functions, Limits and Continuity Functions Limits Continuity Chapter 2: Functions, Limits and Continuity 1 Functions Functions are the major tools for describing the real world in mathematical

More information

Chapter II.B. The Chain Rule

Chapter II.B. The Chain Rule Chapter IIB The Chain Rule x x Preface: To find the derivative of f (x) = [sin(x)] and g (x) = exp(x) = e = [e ] you could x x view these functions as the products, sin(x) sin(x) or e e With this view

More information

Practice Differentiation Math 120 Calculus I Fall 2015

Practice Differentiation Math 120 Calculus I Fall 2015 . x. Hint.. (4x 9) 4x + 9. Hint. Practice Differentiation Math 0 Calculus I Fall 0 The rules of differentiation are straightforward, but knowing when to use them and in what order takes practice. Although

More information

Chapter 6. Techniques of Integration. 6.1 Differential notation

Chapter 6. Techniques of Integration. 6.1 Differential notation Chapter 6 Techniques of Integration In this chapter, we expand our repertoire for antiderivatives beyond the elementary functions discussed so far. A review of the table of elementary antiderivatives (found

More information

11.5. The Chain Rule. Introduction. Prerequisites. Learning Outcomes

11.5. The Chain Rule. Introduction. Prerequisites. Learning Outcomes The Chain Rule 11.5 Introduction In this Section we will see how to obtain the derivative of a composite function (often referred to as a function of a function ). To do this we use the chain rule. This

More information

Your signature: (1) (Pre-calculus Review Set Problems 80 and 124.)

Your signature: (1) (Pre-calculus Review Set Problems 80 and 124.) (1) (Pre-calculus Review Set Problems 80 an 14.) (a) Determine if each of the following statements is True or False. If it is true, explain why. If it is false, give a counterexample. (i) If a an b are

More information

Analysis/Calculus Review Day 2

Analysis/Calculus Review Day 2 Analysis/Calculus Review Day 2 Arvind Saibaba arvindks@stanford.edu Institute of Computational and Mathematical Engineering Stanford University September 14, 2010 Limit Definition Let A R, f : A R and

More information

7.1. Calculus of inverse functions. Text Section 7.1 Exercise:

7.1. Calculus of inverse functions. Text Section 7.1 Exercise: Contents 7. Inverse functions 1 7.1. Calculus of inverse functions 2 7.2. Derivatives of exponential function 4 7.3. Logarithmic function 6 7.4. Derivatives of logarithmic functions 7 7.5. Exponential

More information

Chapter 3a Topics in differentiation. Problems in differentiation. Problems in differentiation. LC Abueg: mathematical economics

Chapter 3a Topics in differentiation. Problems in differentiation. Problems in differentiation. LC Abueg: mathematical economics Chapter 3a Topics in differentiation Lectures in Mathematical Economics L Cagandahan Abueg De La Salle University School of Economics Problems in differentiation Problems in differentiation Problem 1.

More information

Slide 1. Slide 2. Slide 3 Remark is a new function derived from called derivative. 2.2 The derivative as a Function

Slide 1. Slide 2. Slide 3 Remark is a new function derived from called derivative. 2.2 The derivative as a Function Slide 1 2.2 The derivative as a Function Slide 2 Recall: The derivative of a function number : at a fixed Definition (Derivative of ) For any number, the derivative of is Slide 3 Remark is a new function

More information

Unit #3 : Differentiability, Computing Derivatives

Unit #3 : Differentiability, Computing Derivatives Unit #3 : Differentiability, Computing Derivatives Goals: Determine when a function is differentiable at a point Relate the derivative graph to the the graph of an original function Compute derivative

More information

CALCULUS ASSESSMENT REVIEW

CALCULUS ASSESSMENT REVIEW CALCULUS ASSESSMENT REVIEW DEPARTMENT OF MATHEMATICS CHRISTOPHER NEWPORT UNIVERSITY 1. Introduction and Topics The purpose of these notes is to give an idea of what to expect on the Calculus Readiness

More information

Chapter 6. Techniques of Integration. 6.1 Differential notation

Chapter 6. Techniques of Integration. 6.1 Differential notation Chapter 6 Techniques of Integration In this chapter, we expand our repertoire for antiderivatives beyond the elementary functions discussed so far. A review of the table of elementary antiderivatives (found

More information

1.1 Definition of a Limit. 1.2 Computing Basic Limits. 1.3 Continuity. 1.4 Squeeze Theorem

1.1 Definition of a Limit. 1.2 Computing Basic Limits. 1.3 Continuity. 1.4 Squeeze Theorem 1. Limits 1.1 Definition of a Limit 1.2 Computing Basic Limits 1.3 Continuity 1.4 Squeeze Theorem 1.1 Definition of a Limit The limit is the central object of calculus. It is a tool from which other fundamental

More information

Math 210 Midterm #1 Review

Math 210 Midterm #1 Review Math 20 Miterm # Review This ocument is intene to be a rough outline of what you are expecte to have learne an retaine from this course to be prepare for the first miterm. : Functions Definition: A function

More information

June 9 Math 1113 sec 002 Summer 2014

June 9 Math 1113 sec 002 Summer 2014 June 9 Math 1113 sec 002 Summer 2014 Section 6.5: Inverse Trigonometric Functions Definition: (Inverse Sine) For x in the interval [ 1, 1] the inverse sine of x is denoted by either and is defined by the

More information

Unit #3 : Differentiability, Computing Derivatives, Trig Review

Unit #3 : Differentiability, Computing Derivatives, Trig Review Unit #3 : Differentiability, Computing Derivatives, Trig Review Goals: Determine when a function is differentiable at a point Relate the derivative graph to the the graph of an original function Compute

More information

MAT137 Calculus! Lecture 5

MAT137 Calculus! Lecture 5 MAT137 Calculus! Lecture 5 Today: 2.5 The Pinching Theorem; 2.5 Trigonometric Limits. 2.6 Two Basic Theorems. 3.1 The Derivative Next: 3.2-3.6 DIfferentiation Rules Deadline to notify us if you have a

More information

Table of Contents. Module 1

Table of Contents. Module 1 Table of Contents Module Order of operations 6 Signed Numbers Factorization of Integers 7 Further Signed Numbers 3 Fractions 8 Power Laws 4 Fractions and Decimals 9 Introduction to Algebra 5 Percentages

More information

Calculus: Early Transcendental Functions Lecture Notes for Calculus 101. Feras Awad Mahmoud

Calculus: Early Transcendental Functions Lecture Notes for Calculus 101. Feras Awad Mahmoud Calculus: Early Transcendental Functions Lecture Notes for Calculus 101 Feras Awad Mahmoud Last Updated: August 2, 2012 1 2 Feras Awad Mahmoud Department of Basic Sciences Philadelphia University JORDAN

More information

The Derivative of a Function Measuring Rates of Change of a function. Secant line. f(x) f(x 0 ) Average rate of change of with respect to over,

The Derivative of a Function Measuring Rates of Change of a function. Secant line. f(x) f(x 0 ) Average rate of change of with respect to over, The Derivative of a Function Measuring Rates of Change of a function y f(x) f(x 0 ) P Q Secant line x 0 x x Average rate of change of with respect to over, " " " " - Slope of secant line through, and,

More information

Chapter 1: Limits and Continuity

Chapter 1: Limits and Continuity Chapter 1: Limits and Continuity Winter 2015 Department of Mathematics Hong Kong Baptist University 1/69 1.1 Examples where limits arise Calculus has two basic procedures: differentiation and integration.

More information

Practice Problems: Integration by Parts

Practice Problems: Integration by Parts Practice Problems: Integration by Parts Answers. (a) Neither term will get simpler through differentiation, so let s try some choice for u and dv, and see how it works out (we can always go back and try

More information

Core A-level mathematics reproduced from the QCA s Subject criteria for Mathematics document

Core A-level mathematics reproduced from the QCA s Subject criteria for Mathematics document Core A-level mathematics reproduced from the QCA s Subject criteria for Mathematics document Background knowledge: (a) The arithmetic of integers (including HCFs and LCMs), of fractions, and of real numbers.

More information

Calculus I: Practice Midterm II

Calculus I: Practice Midterm II Calculus I: Practice Mierm II April 3, 2015 Name: Write your solutions in the space provided. Continue on the back for more space. Show your work unless asked otherwise. Partial credit will be given for

More information

Chapter 1. Functions 1.3. Trigonometric Functions

Chapter 1. Functions 1.3. Trigonometric Functions 1.3 Trigonometric Functions 1 Chapter 1. Functions 1.3. Trigonometric Functions Definition. The number of radians in the central angle A CB within a circle of radius r is defined as the number of radius

More information

II. The Calculus of The Derivative

II. The Calculus of The Derivative II The Calculus of The Derivative In Chapter I we learned that derivative was the mathematical concept that captured the common features of the tangent problem, instantaneous velocity of a moving object,

More information

Chapter 7: Techniques of Integration

Chapter 7: Techniques of Integration Chapter 7: Techniques of Integration MATH 206-01: Calculus II Department of Mathematics University of Louisville last corrected September 14, 2013 1 / 43 Chapter 7: Techniques of Integration 7.1. Integration

More information

Lecture Notes for Math 1000

Lecture Notes for Math 1000 Lecture Notes for Math 1000 Dr. Xiang-Sheng Wang Memorial University of Newfoundland Office: HH-2016, Phone: 864-4321 Office hours: 13:00-15:00 Wednesday, 12:00-13:00 Friday Email: xswang@mun.ca Course

More information

Indefinite Integration

Indefinite Integration Indefinite Integration 1 An antiderivative of a function y = f(x) defined on some interval (a, b) is called any function F(x) whose derivative at any point of this interval is equal to f(x): F'(x) = f(x)

More information

3. On the grid below, sketch and label graphs of the following functions: y = sin x, y = cos x, and y = sin(x π/2). π/2 π 3π/2 2π 5π/2

3. On the grid below, sketch and label graphs of the following functions: y = sin x, y = cos x, and y = sin(x π/2). π/2 π 3π/2 2π 5π/2 AP Physics C Calculus C.1 Name Trigonometric Functions 1. Consider the right triangle to the right. In terms of a, b, and c, write the expressions for the following: c a sin θ = cos θ = tan θ =. Using

More information

Math Final Exam Review

Math Final Exam Review Math - Final Exam Review. Find dx x + 6x +. Name: Solution: We complete the square to see if this function has a nice form. Note we have: x + 6x + (x + + dx x + 6x + dx (x + + Note that this looks a lot

More information

Grade: The remainder of this page has been left blank for your workings. VERSION E. Midterm E: Page 1 of 12

Grade: The remainder of this page has been left blank for your workings. VERSION E. Midterm E: Page 1 of 12 First Name: Student-No: Last Name: Section: Grade: The remainder of this page has been left blank for your workings. Midterm E: Page of Indefinite Integrals. 9 marks Each part is worth 3 marks. Please

More information

In general, if we start with a function f and want to reverse the differentiation process, then we are finding an antiderivative of f.

In general, if we start with a function f and want to reverse the differentiation process, then we are finding an antiderivative of f. Math 1410 Worksheet #27: Section 4.9 Name: Our final application of derivatives is a prelude to what will come in later chapters. In many situations, it will be necessary to find a way to reverse the differentiation

More information

1 The Derivative and Differrentiability

1 The Derivative and Differrentiability 1 The Derivative and Differrentiability 1.1 Derivatives and rate of change Exercise 1 Find the equation of the tangent line to f (x) = x 2 at the point (1, 1). Exercise 2 Suppose that a ball is dropped

More information

Calculus. Contents. Paul Sutcliffe. Office: CM212a.

Calculus. Contents. Paul Sutcliffe. Office: CM212a. Calculus Paul Sutcliffe Office: CM212a. www.maths.dur.ac.uk/~dma0pms/calc/calc.html Books One and several variables calculus, Salas, Hille & Etgen. Calculus, Spivak. Mathematical methods in the physical

More information

MAT265 Calculus for Engineering I

MAT265 Calculus for Engineering I MAT65 Calculus for Engineering I Joe Wells April 8, 06 Contents 0 Prerequisites 3 0. Review of Functions..................................... 3 0.. The Definition..................................... 3

More information

PART ONE: Solve algebraically and check. Be sure to show all work.

PART ONE: Solve algebraically and check. Be sure to show all work. NAME AP CALCULUS BC SUMMER ASSIGNMENT 2017 DIRECTIONS: Each part must be completed separately on looseleaf. All work should be shown and done in a neat and precise manner. Any questions pertaining to the

More information

Functions and Graphs. Chapter Numbers (1.2.1, 1.2.4)

Functions and Graphs. Chapter Numbers (1.2.1, 1.2.4) Chapter 1 Functions and Graphs 1.1 Numbers (1.2.1, 1.2.4) The most fundamental type of number are those we use to count with: 0,1,2,... These are called the natural numbers: the set of all natural numbers

More information