Chapter 3 Shortcuts to Differentiation Section 3.1 Powers and Polynomials Section 3.2 The Exponential Function Section 3.3 The Product and Quotient Rules Section 3.4 The Chain Rule Section 3.5 The Trigonometric Functions Section 3.6 The Chain Rule and Inverse Functions Section 3.7 Implicit Functions Section 3.8 Hyperbolic Functions Section 3.9 Linear Approximation and the Derivative Section 3.10 Theorems About Differentiable Functions Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Powers and Polynomials Section 3.1 Powers and Polynomials Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Derivative of a Constant Times a Function Figure 3.1 A function and its multiples: Derivative of multiple is multiple of derivative Theorem 3.1: Derivative of a Constant Multiple If f is differentiable and c is a constant Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Derivative of Sum and Difference If f and g are differentiable, then Theorem 3.2: Derivative of Sum and Difference If f and g are differentiable, then Proof using the definition of the derivative: Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

The Power Rule For any constant real number n, Example 1 Use the power rule to differentiate (a) 1/x3 , (b) x1/2 , (c) Solution (a) For n = − 3: (b) For n = 1/2: (c) For n = − 1/3: Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Example 6 If the position of a body, in meters, is given as a function of time t, in seconds, by s = − 4.9t2 + 5t + 6, find the velocity and acceleration of the body at time t. Solution The velocity, v, is the derivative of the position: v = ds/dt = − 9.8t +5 and the acceleration, a, is the derivative of the velocity: a = dv/dt = − 9.8 Note that v is in meters/second and a is in meters/second2. Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Finding and Visualizing Derivatives of Polynomials Exercise 56 The graph of y(x) = x3 − 9x2 − 16x + 1 has a slope of 5 at two points. Find the coordinates of the points. Solution y’(x) = 3x2 − 18x − 16. Setting 3x2 − 18x − 16 = 5, 3(x2 − 6x − 7) = 0 . So x = 7 or -1. At red points, the slope of the blue graph, y(x), is 5. Parabola y’(x) (-1,7) x Cubic y(x) (7,-209) Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Section 3.2 The Exponential Function Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Derivatives of Exponential Functions We start by calculating the derivative of g(x) = 2x, which is given by Table 3.2 lists values of (2h- 1)/h as h → 0. h (2h- 1)/h -0.1 0.6697 -0.01 0.6908 -0.001 0.6929 0.001 0.6934 0.01 0.6956 0.1 0.7177 We might conclude that g’(x) ≈ (0.693) 2x . Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Exploring the Derivative of ax The derivative of 2x is proportional to 2x with constant of proportionality 0.693. A similar calculation shows that the derivative of f(x) = ax is And the value of this limit is explored in Table 3.3. Is there a value of “a” between 2 and 3, such that this limit will be 1? Table 3.3 a 2 0.693 3 1.099 4 1.386 5 1.609 6 1.792 7 1.946 Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

The Natural Number e Table 3.4 If , then for small values of h, (ah -1)/h ≈ 1 or ah ≈ 1 + h , a ≈ (1 + h)1/h We define the value of a that makes our constant of proportionality 1 as . This limit is explored in the table on the right. h (1+h)1/h −0.001 2.719642 −0.0001 2.718418 −0.00001 2.718295 0.00001 2.718268 0.0001 2.718146 0.001 2.716924 Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Formula for the Derivative of ax Based on Table 3.4, the number e ≈ 2.718… and this irrational number is the base of the natural logarithms. Using this fact, a = eln a and Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Exercise 41 In 2009, the population of Mexico was 111 million and growing 1.13% annually, while the population of the US was 307 million and growing 0.975% annually. If we measure growth rates in people per year, which population was growing faster in 2009? Solution Country Mexico United States Population Function 111 (1.0113)t 307 (1.00975)t Population Rate of Change 111 ln(1.0113) (1.0113)t 307 ln(1.00975) (1.00975)t Rate of Growth in 2009 (t = 0) 1.24727 million/year 2.97875 million/year Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Section 3.3 The Product and Quotient Rules Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Difference of a Product f(x + h)g(x + h) − f(x)g(x) = (Area of whole rectangle) − (Unshaded area) = Area of the three shaded rectangles = Δ f ・ g(x) + f(x) ・ Δ g + Δ f ・Δ g. Figure 3.13: Illustration for the product rule (with Δf, Δg positive) Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Theorem 3.3: The Product Rule If u = f(x) and v = g(x) are differentiable, then (f g)′ = f′ g + f g′. The product rule can also be written In words: The derivative of a product is the derivative of the first times the second plus the first times the derivative of the second. Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Exercise 6 Exercise 54 Find the derivative of y = (t2 + 3) et Solution dy/dt = (2t) et + (t2 + 3) et = (t2 + 2t +3) et Exercise 54 Let f(3) = 6, g(3) = 12, f′(3) = 1/2 , and g′(3) = 4/3 . Evaluate the following when x = 3. (f(x)g(x))′ − (g(x) − 4f′(x)) Solution At x = 3, (fg)’= f’(3)g(3) + f(3)g’(3) = (1/2)(12)+6(4/3) = 6+8 =14 So at x = 3, (f(x)g(x))′ − (g(x) − 4f′(x)) = 14 – (12-4·(1/2)) = 14 – 10 = 4 Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Theorem 3.4: The Quotient Rule If u = f(x) and v = g(x) are differentiable, then (f /g)′ = (f′ g - f g′)/g2 or equivalently, In words: The derivative of a quotient is the derivative of the numerator times the denominator minus the numerator times the derivative of the denominator, all over the denominator squared. Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Example 2a Differentiate 5x2 /(x3 + 1) Solution Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Section 3.4 The Chain Rule Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Intuition Behind the Chain Rule Imagine we are moving straight upward in a hot air balloon. Let y be our distance from the ground. The temperature, H, is changing as a function of altitude, so H = f(y). How does our temperature change with time? Since temperature is a function of height, H = f(y), and height is a function of time, y = g(t), we can think of temperature as a composite function of time, H = f(g(t)), with f as the outside function and g as the inside function. The example suggests the following result, which turns out to be true: Rate of change of composite function Rate of change of outside function Rate of change of inside function = × The Derivative of a Composition of Functions Symbolically, for our H = f(g(t)): Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Theorem 3.5: The Chain Rule If f and g are differentiable functions, then In words: The derivative of a composite function is the product of the derivatives of the outside and inside functions. The derivative of the outside function must be evaluated at the inside function. Example 2a Find the derivative of (x2 + 1)100 . Solution Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Example 4 Differentiate Solution The chain rule is needed four times: Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Section 3.5 The Trigonometric Functions Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

The Sine Function and Its Derivative Graphical Perspective Figure 3.22: The sine function First we might ask where the derivative is zero. Then ask where the derivative is positive and where it is negative. In exploring these answers, we get something like the following graph. Figure 3.23: Derivative of f(x) = sin x The graph of the derivative in Figure 3.23 looks suspiciously like the graph of the cosine function. This might lead us to conjecture, quite correctly, that the derivative of the sine is the cosine. Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

The Cosine Function and Its Derivative Graphical Perspective Example 2 Starting with the graph of the cosine function, sketch a graph of its derivative. Solution Looking at the graph of g(x) = cos x its derivative is 0 at x = 0, ±π, ± 2π, … It’s derivative (slope) is positive for (- π,0), (π, 2π), (3π, 4π), … It’s derivative (slope) is negative for (- 2π, π), (0, π), (2π, 3π), … Figure 3.24: g(x) = cos x and its derivative, g′(x) The derivative of the cosine in Figure 3.24(b) looks exactly like the graph of sine, except reflected across the x-axis. Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Calculus, 6th edition, Hughes-Hallett et. al Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Examples Solution Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Derivative of the Tangent Function Since tan x = sin x/cos x, we differentiate tan x using the quotient rule. Writing (sin x)′ for d(sin x)/dx, we have: For x in radians, Exercise 30 Find the derivative of g(z) = tan ez . Solution g’(z) = ez/(cos2 ez) Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Section 3.6 The Chain Rule and Inverse Functions Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Derivative of ln x We use the chain rule to differentiate an identity involving ln x. Since eln x = x, on the one hand we have However, we can also use the chain rule to get Equating these to one another, we have our desired result. Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Derivative of ax Revisited In Section 3.2, we saw that the derivative of ax is proportional to ax. Now we see another way of calculating the constant of proportionality. We use the identity ln(ax) = x ln a. Differentiating both sides, using the chain rule, and remembering that ln a is a constant, we obtain: So we have the result from Section 3.2. Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Derivatives of Inverse Trigonometric Functions To find d/dx (arctan x) we use the identity tan(arctan x) = x. Differentiating both sides and using the chain rule gives So Using the identity 1 + tan2θ = 1/cos2θ, and replacing θ by arctan x, Thus we have Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Derivative of the Arcsine and Examples By a similar argument, we obtain the result: Example 2 Differentiate (a) arctan(t2) (b) arcsin(tan θ). Solution Use the chain rule: (a) (b) Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Derivative of a General Inverse Function In general, if a function f has a differentiable inverse, f−1, we find its derivative by differentiating f(f−1(x)) = x by the chain rule, yielding the following result: Exercise 65 Use the table and the fact that f(x) is invertible and differentiable everywhere to find (f−1)′(3). x f(x) f′(x) 3 1 7 6 2 10 9 5 Solution Begin by observing that f−1(3)=9, since f(9) = 3 Then (f−1)′(3) = 1/f’(9) = 1/5 Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Section 3.7 Implicit Functions Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

What Is an Implicit Function? In earlier chapters, most functions were written in the form y = f(x); here y is said to be an explicit function of x. An equation such as x2 + y2 = 4 is said to give y as an implicit function of x. Its graph is the circle below. Since there are x-values which correspond to two y-values, y is not a function of x on the whole circle. Figure 3.35: Graph of x2 + y2 = 4 Note that y is a function of x on the top half, and y is a different function of x on the bottom half. Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

d/dx (x2) + d/dx (y2) = d/dx (4). Differentiating Implicitly Let us consider the circle as a whole. The equation does represent a curve which has a tangent line at each point. The slope of this tangent can be found by differentiating the equation of the circle with respect to x: d/dx (x2) + d/dx (y2) = d/dx (4). If we think of y as a function of x and use the chain rule, we get 2x + 2y dy/dx = 0. Solving gives dy/dx = − x/y. The derivative here depends on both x and y (instead of just on x). Differentiating the equation of the circle has given us the slope of the curve at all points except (2, 0) and (−2, 0), where the tangent is vertical. In general, this process of implicit differentiation leads to a derivative whenever the expression for the derivative does not have a zero in the denominator. Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

· Example 2 y3 − x y = −6 Find all points where the tangent line to y3 − x y = −6 is either horizontal or vertical. Solution Differentiating implicitly, 3y2·dy/dx–y–x·dy/dx=0 So dy/dx = y/(3y2 − x). The tangent is horizontal when the numerator of dy/dx equals 0, so y = 0. Since we also must satisfy y3 − x y = −6, we get 0=-6, which is impossible. We conclude that there are no points on the curve where the tangent line is horizontal. The tangent is vertical when the denominator of dy/dx is 0, giving 3y2−x = 0. Thus, x = 3y2 at any point with a vertical tangent line. Again, we must also satisfy y3 − x y = −6, so y3=3. Solving for x, with this value of y, we conclude there is a vertical tangent at the point (6.240,1.442). The figure below verifies graphically what was determined analytically. y3 − x y = −6 · (6.240,1.442) Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Section 3.8 Hyperbolic Functions Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Hyperbolic Functions Graphs of Hyperbolic Cosine and Sine Figure 3.37: Graph of y = cosh x Figure 3.38: Graph of y = sinh x Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Properties of Hyperbolic Functions cosh 0 = 1 sinh 0 = 0 cosh(−x) = cosh x sinh(−x) = −sinh x Example 2 Describe and explain the behavior of cosh x as x → ∞ and as x → −∞. Solution From Figure 3.37, it appears that as x → ∞, the graph of cosh x resembles the graph of ½ ex. Similarly, as x → −∞, the graph of cosh x resembles the graph of ½ e-x. This behavior is explained by using the formula for cosh x and the facts that e−x → 0 as x → ∞ and ex → 0 as x → −∞: Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Identity Involving cosh x and sinh x cosh2 x − sinh2 x = 1 This identity shows us how the hyperbolic functions got their name. Suppose (x, y) is a point in the plane and x = cosh t and y = sinh t for some t. Then the point (x, y) lies on the hyperbola x2 − y2 = 1. Extending the analogy to the trigonometric functions, we define the hyperbolic tangent. y The Hyperbolic Tangent y =tanh x x Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Derivatives of Hyperbolic Functions We calculate the derivatives using the fact that d/dx (ex) = ex. The results are again reminiscent of the trigonometric functions. Example 3 Compute the derivative of tanh x. Solution Using the quotient rule gives Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Section 3.9 Linear Approximation and the Derivative Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

The Tangent Line Approximation Suppose f is differentiable at a. Then, for values of x near a, the tangent line approximation to f(x) is f(x) ≈ f(a) + f ′(a)(x − a). The expression f(a)+ f ′(a)(x −a) is called the local linearization of f near x = a. We are thinking of a as fixed, so that f(a) and f ′(a) are constant. The error, E(x), in the approximation is defined by E(x) = f(x) − f(a) − f ′(a)(x − a). Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Visualization of the Tangent Line Approximation Figure 3.40: The tangent line approximation and its error It can be shown that the tangent line approximation is the best linear approximation to f near a. Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Example 1 What is the tangent line approximation for f(x) = sin x near x = 0? Solution The tangent line approximation of f near x = 0 is f(x) ≈ f(0) + f ′(0)(x − 0). If f(x) = sin x, then f ′(x) = cos x, so f(0) = sin 0 = 0 and f ′(0) = cos 0 = 1, and the approximation is sin x ≈ x. This means that, near x = 0, the function f(x) = sin x is well approximated by the function y = x. If we zoom in on the graphs of the functions sin x and x near the origin, we won’t be able to tell them apart. Figure 3.41: Tangent line approximation to y = sin x Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

E(x) = f(x) − f(a) − f ′(a)(x − a). Estimating the Error in Linear Approximation Theorem 3.6: Differentiability and Local Linearity Suppose f is differentiable at x = a and E(x) is the error in the tangent line approximation, that is: E(x) = f(x) − f(a) − f ′(a)(x − a). Then Proof Using the definition of E(x), we have Taking the limit as x → a and using the definition of the derivative, we see that An error estimate that will be developed later is Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

E(x) = True value − Approximation = (x3 − 5x + 3) − (1 + 7(x − 2)). Example 3 Let E(x) be the error in the tangent line approximation to f(x) = x3 − 5x + 3 for x near 2. What does a table of values for E(x)/(x − 2) suggest about the limit of this ratio as x→2? Make another table to see that E(x) ≈ k(x − 2) 2. Estimate the value of k. Check that a possible value is k = f ′′(2)/2. Solution Since f(x) = x3 − 5x + 3 , we have f ′(x) = 3x2 − 5, and f′′(x) = 6x. Thus, f(2) = 1 and f ′(2) = 3·22 − 5 = 7, so the tangent line approximation for x near 2 is f(x) ≈ f(2) + f ′(2)(x − 2) ≈ 1 + 7(x − 2). E(x) = True value − Approximation = (x3 − 5x + 3) − (1 + 7(x − 2)). x E(x)/(x − 2) E(x)/(x − 2)2 2.1 0.61 6.1 2.01 0.0601 6.01 2.001 0.006001 6.001 2.0001 0.0006 6.0001 These values suggest that E(x)/(x − 2) → 0 as x → 2 and E(x)/(x − 2)2 → 6 as x → 2. So E(x) ≈ 6(x − 2) 2 Also note that f ′′(2)/2 = 12/2=6 Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Section 3.10 Theorems About Differentiable Functions Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

A Relationship Between Local and Global: The Mean Value Theorem Theorem 3.7: The Mean Value Theorem If f is continuous on a ≤ x ≤ b and differentiable on a < x < b, then there exists a number c, with a < c < b, such that f ′(c) = (f(b) − f(a))/(b − a). In other words, f(b) − f(a) = f ′(c)(b − a). To understand this theorem geometrically, look at Figure 3.44. Join the points on the curve where x = a and x = b with a secant line and observe that the slope of secant line = (f(b) − f(a))/(b − a). There appears to be at least one point between a and b where the slope of the tangent line to the curve is precisely the same as the slope of the secant line. Figure 3.44 Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Theorem 3.8: The Increasing Function Theorem Suppose that f is continuous on a ≤ x ≤ b and differentiable on a < x < b. • If f ′(x) > 0 on a < x < b, then f is increasing on a ≤ x ≤ b. • If f ′(x) ≥ 0 on a < x < b, then f is nondecreasing on a ≤ x ≤ b. Theorem 3.9: The Constant Function Theorem Suppose that f is continuous on a ≤ x ≤ b and differentiable on a < x < b. If f ′(x) = 0 on a < x < b, then f is constant on a ≤ x ≤ b. Theorem 3.10: The Racetrack Principle Suppose that g and h are continuous on a ≤ x ≤ b and differentiable on a < x < b, and that g′(x) ≤ h′(x) for a < x < b. • If g(a) = h(a), then g(x) ≤ h(x) for a ≤ x ≤ b. • If g(b) = h(b), then g(x) ≥ h(x) for a ≤ x ≤ b. Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved