PREVIEW VERSION—NOT FINAL 3 DIFFERENTIATION ifferential calculus is the study of the derivative, and differentiation is the process of D computing derivatives. What is a derivative? There are three equally important an- swers: Aderivative is a rate of change, it is the slope of a tangent line, and (more formally), it is the limit of a difference quotient, as we will explain shortly. In this chapter, we explore all three facets of the derivative and develop the basic rules of differentiation. When you master these techniques, you will possess one of the most useful and flexible tools that mathematics has to offer. 3.1 Definition of the Derivative We begin with two questions: What is the precise definition of a tangent line?And how can we compute its slope? To answer these questions, let’s return to the relationship between Calculus is the foundation for all of our tangent and secant lines first mentioned in Section 2.1. understanding of motion, including the The secant line through distinct points P = (a, f (a)) and Q = (x, f (x)) on the graph aerodynamic principles that made supersonic of a function f (x) has slope [Figure 1(A)] flight possible. − f = f (x) f (a) x x − a where f = f (x) − f (a) and x = x − a f (x) − f (a) The expression is called the difference quotient. REMINDER A secant line is any line x − a through two points on a curve or graph. y y Q = (x, f(x)) P = (a, f(a)) Δf = f(x) − f(a) Tangent P = (a, f(a)) x = x − a FIGURE 1 The secant line has slope x x f/x. Our goal is to compute the slope xa a of the tangent line at (a, f (a)). (A) (B) Now observe what happens as Q approaches P or, equivalently, as x approaches a. Figure 2 suggests that the secant lines get progressively closer to the tangent line. If we imagine Q moving toward P , then the secant line appears to rotate into the tangent line as in (D). Therefore, we may expect the slopes of the secant lines to approach the slope of the tangent line. Based on this intuition, we define the derivative f (a) (which is read “f prime of a”) as the limit f (x) − f (a) f (a) = lim x→a x − a Limit of slopes of secant lines 120 October 11, 2010 PRINTER FILE PREVIEW VERSION—NOT FINAL SECTION 3.1 Definition of the Derivative 121 y y y y Q Q Q Q P P P P x x x x xa xa xa a x (A) (B) (C) (D) FIGURE 2 The secant lines approach the tangent line as Q approaches P . There is another way of writing the difference quotient using a new variable h: h = x − a y We have x = a + h and, for x = a (Figure 3), Q − + − f (x) f (a) = f (a h) f (a) − f(a + h) − f(a) x a h P The variable h approaches 0 as x → a, so we can rewrite the derivative as h f (a + h) − f (a) f (a) = lim x h→0 h a x = a + h FIGURE 3 The difference quotient can be Each way of writing the derivative is useful. The version using h is often more convenient written in terms of h. in computations. DEFINITION The Derivative The derivative of f (x) at x = a is the limit of the dif- ference quotients (if it exists): f (a + h) − f (a) f (a) = lim 1 h→0 h When the limit exists, we say that f is differentiable at x = a.An equivalent definition of the derivative is f (x) − f (a) f (a) = lim 2 x→a x − a We can now define the tangent line in a precise way, as the line of slope f (a) through P = (a, f (a)). = REMINDER The equation of the line DEFINITION Tangent Line Assume that f (x) is differentiable at x a. The tangent = = through P = (a, b) of slope m in line to the graph of y f (x) at P (a, f (a)) is the line through P of slope f (a). point-slope form: The equation of the tangent line in point-slope form is − = − y b m(x a) y − f (a) = f (a)(x − a) 3 October 11, 2010 PRINTER FILE PREVIEW VERSION—NOT FINAL 122 CHAPTER 3 DIFFERENTIATION y EXAMPLE 1 Equation of a Tangent Line Find an equation of the tangent line to the 75 y = x2 graph of f (x) = x2 at x = 5. 50 y = 10x − 25 Solution First, we must compute f (5). We are free to use either Eq. (1) or Eq. (2). Using Eq. (2), we have 25 (5, 25) 2 f (x) − f(5) x − 25 (x − 5)(x + 5) f (5) = lim = lim = lim x x→5 x − 5 x→5 x − 5 x→5 x − 5 357 = 2 = = lim (x + 5) = 10 FIGURE 4 Tangent line to y x at x 5. x→5 Next, we apply Eq. (3) with a = 5. Because f(5) = 25, an equation of the tangent line is y − 25 = 10(x − 5), or, in slope-intercept form: y = 10x − 25 (Figure 4). The next two examples illustrate differentiation (the process of computing the deriva- Isaac Newton referred to calculus as the tive) using Eq. (1). For clarity, we break up the computations into three steps. “method of fluxions” (from the Latin word for “flow”), but the term “differential EXAMPLE 2 Compute f (3), where f (x) = x2 − 8x. calculus”, introduced in its Latin form “calculus differentialis” by Gottfried Solution Using Eq. (1), we write the difference quotient at a = 3 as Wilhelm Leibniz, eventually won out and f (a + h) − f (a) f(3 + h) − f(3) was adopted universally. = (h = 0) h h Step 1. Write out the numerator of the difference quotient. f(3 + h) − f(3) = (3 + h)2 − 8(3 + h) − 32 − 8(3) = (9 + 6h + h2) − (24 + 8h) − (9 − 24) = h2 − 2h Step 2. Divide by h and simplify. f(3 + h) − f(3) h2 − 2h h(h − 2) = = = h − 2 h h h Cancel h Step 3. Compute the limit. f(3 + h) − f(3) f (3) = lim = lim (h − 2) =−2 h→0 h h→0 y 1 EXAMPLE 3 Sketch the graph of f (x) = and the tangent line at x = 2. x 3 (a) Based on the sketch, do you expect f (2) to be positive or negative? 2 (b) Find an equation of the tangent line at x = 2. Solution The graph and tangent line at x = 2 are shown in Figure 5. 1 (a) We see that the tangent line has negative slope, so f (2) must be negative. x 54321 (b) We compute f (2) in three steps as before. = 1 FIGURE 5 Graph of f (x) x . The tangent Step 1. Write out the numerator of the difference quotient. line at x = 2 has equation y =−1 x + 1. 4 1 1 2 2 + h h f(2 + h) − f(2) = − = − =− 2 + h 2 2(2 + h) 2(2 + h) 2(2 + h) Step 2. Divide by h and simplify. + − f(2 h) f(2) = 1 · − h =− 1 h h 2(2 + h) 2(2 + h) October 11, 2010 PRINTER FILE PREVIEW VERSION—NOT FINAL SECTION 3.1 Definition of the Derivative 123 Step 3. Compute the limit. f(2 + h) − f(2) −1 1 f (2) = lim = lim =− h→0 h h→0 2(2 + h) 4 = 1 1 The function value is f(2) 2 , so the tangent line passes through 2, 2 and has equation 1 1 y − =− (x − 2) y 2 4 4 f x = mx + b ( ) =−1 + In slope-intercept form, y 4 x 1. 2 The graph of a linear function f (x) = mx + b (where m and b are constants) is a line of slope m. The tangent line at any point coincides with the line itself (Figure 6), so we should expect that f (a) = m for all a. Let’s check this by computing the derivative: x −1 2 4 f (a + h) − f (a) (m(a + h) + b) − (ma + b) FIGURE 6 The derivative of = = f (a) lim lim f (x) = mx + b is f (a) = m for all a. h→0 h h→0 h = mh = = y lim lim m m h→0 h h→0 4 f(x) = b If m = 0, then f (x) = b is constant and f (a) = 0 (Figure 7). In summary, 2 THEOREM 1 Derivative of Linear and Constant Functions • If f (x) = mx + b is a linear function, then f (a) = m for all a. x • 2 4 If f (x) = b is a constant function, then f (a) = 0 for all a. FIGURE 7 The derivative of a constant function f (x) = b is f (a) = 0 for all a. EXAMPLE 4 Find the derivative of f (x) = 9x − 5 at x = 2 and x = 5. Solution We have f (a) = 9 for all a. Hence, f (2) = f (5) = 9. Estimating the Derivative Approximations to the derivative are useful in situations where we cannot evaluate f (a) y exactly. Since the derivative is the limit of difference quotients, the difference quotient should give a good numerical approximation when h is sufficiently small: Secant f (a + h) − f (a) f (a) ≈ if h is small Q Tangent h P Graphically, this says that for small h, the slope of the secant line is nearly equal to the slope of the tangent line (Figure 8).