Chapter 2: Differentiation Spring 2018 Department of Mathematics Hong Kong Baptist University 1 / 82
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
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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
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
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
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(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
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
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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 = 0. 12 / 82
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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
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 0 1 27 + 6h = 1 27. 15 / 82
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
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
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) = 0. 18 / 82
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 1. 19 / 82
Table 1: Some elementary functions and their derivatives f(x) f (x) c 0 x 1 x 2 1 1 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
Leibniz Notation 21 / 82
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
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 = 48. 23 / 82
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
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
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
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
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
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
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
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
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 1. 32 / 82
Example 7 Find the derivatives of (a) y = x 2 + x + 1 x 3 ; and (b) y = x 2 x + 1. Solution: 33 / 82
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
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) 2. 35 / 82
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
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
Example 8 Verify the Chain rule using the function y = 1/(x 2 4). Proof: 38 / 82
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. 39 / 82
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
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
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 3 + 1 42 / 82
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
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 3. 44 / 82
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 2.3. 45 / 82
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
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
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
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
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
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
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 2 6. 52 / 82
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
Example 13 Calculate the first 3 derivatives of f (x) = x 2 + 1. 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
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
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
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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
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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 > 0. 61 / 82
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
Increasing and Decreasing Functions 63 / 82
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
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
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Example 16 On what intervals is the function f (x) = what intervals is it decreasing? Solution: The derivative of f is x x 2 + 1 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
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
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
1 (3/5, 4/5) 0.5 0 0.5 1 (3/5, 4/5) 1.5 1 0.5 0 0.5 1 1.5 70 / 82
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 ( 5 + 4 ) 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 = 0. 71 / 82
3 Isolate dy/dx to get dy dx = x y dy = dx = x=3/5 3 5 4 5 = 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=3/5 3 5 4 5 = + 3 4. Thus, the dependence on y in the implicit derivative reflects the fact that the equation defines more than one function. 72 / 82
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
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 2 + 1. dy dx = y 2 + 1 1 2xy. 74 / 82
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 5. 75 / 82
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
Example 20 Verify that F (x) = x 3 + 5x is an antiderivative of f (x) = 3x 2 + 5. Solution: F (x) is an antiderivative of f (x) if F (x) = f (x). Differentiating F (x) we get F (x) = 3x 2 + 5 = 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
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
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
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
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
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. 2 82 / 82