9 The Derivative Limits One-sided Limits and Continuity The Derivative Basic Rules of Differentiation The Product and Quotient Rules: Higher-Order Derivatives The Chain Rule Differentiation of Exponential and Logarithmic Functions Marginal Functions in Economics

Basic Rules of Differentiation 9.4 Basic Rules of Differentiation

Four Basic Rules We’ve learned that to find the rule for the derivative f ′of a function f, we first find the difference quotient But this method is tedious and time consuming, even for relatively simple functions. This chapter we will develop rules that will simplify the process of finding the derivative of a function.

Rule 1: Derivative of a Constant We will use the notation to mean “the derivative of f with respect to x at x.” Rule 1: Derivative of a constant The derivative of a constant function is equal to zero.

Rule 1: Derivative of a Constant We can see geometrically why the derivative of a constant must be zero. The graph of a constant function is a straight line parallel to the x axis. Such a line has a slope that is constant with a value of zero. Thus, the derivative of a constant must be zero as well. y f(x) = c x

Rule 1: Derivative of a Constant We can use the definition of the derivative to demonstrate this:

Rule 2: The Power Rule Rule 2: The Power Rule If n is any real number, then

Rule 2: The Power Rule Lets verify this rule for the special case of n = 2. If f(x) = x2, then

Rule 2: The Power Rule Practice Examples: If f(x) = x, then

Rule 2: The Power Rule Practice Examples: Find the derivative of

Rule 2: The Power Rule Practice Examples: Find the derivative of

Rule 3: Derivative of a Constant Multiple Function If c is any constant real number, then

Rule 3: Derivative of a Constant Multiple Function Practice Examples: Find the derivative of

Rule 3: Derivative of a Constant Multiple Function Practice Examples: Find the derivative of

Rule 4: The Sum Rule Rule 4: The Sum Rule

Rule 4: The Sum Rule Practice Examples: Find the derivative of

Rule 4: The Sum Rule Practice Examples: Find the derivative of

Applied Example: Conservation of a Species A group of marine biologists at the Neptune Institute of Oceanography recommended that a series of conservation measures be carried out over the next decade to save a certain species of whale from extinction. After implementing the conservation measure, the population of this species is expected to be where N(t) denotes the population at the end of year t. Find the rate of growth of the whale population when t = 2 and t = 6. How large will the whale population be 8 years after implementing the conservation measures?

Applied Example: Conservation of a Species Solution The rate of growth of the whale population at any time t is given by In particular, for t = 2, we have And for t = 6, we have Thus, the whale population’s rate of growth will be 34 whales per year after 2 years and 338 per year after 6 years.

Applied Example: Conservation of a Species Solution The whale population at the end of the eighth year will be

The Product and Quotient Rules 9.5 The Product and Quotient Rules

Rule 5: The Product Rule The derivative of the product of two differentiable functions is given by

Rule 5: The Product Rule Practice Examples: Find the derivative of

Rule 5: The Product Rule Practice Examples: Find the derivative of

Rule 6: The Quotient Rule The derivative of the quotient of two differentiable functions is given by

Rule 6: The Quotient Rule Practice Examples: Find the derivative of

Rule 6: The Quotient Rule Practice Examples: Find the derivative of

Applied Example: Rate of Change of DVD Sales The sales ( in millions of dollars) of DVDs of a hit movie t years from the date of release is given by Find the rate at which the sales are changing at time t. How fast are the sales changing at: The time the DVDs are released (t = 0)? And two years from the date of release (t = 2)?

Applied Example: Rate of Change of DVD Sales Solution The rate of change at which the sales are changing at time t is given by

Applied Example: Rate of Change of DVD Sales Solution The rate of change at which the sales are changing when the DVDs are released (t = 0) is That is, sales are increasing by $5 million per year.

Applied Example: Rate of Change of DVD Sales Solution The rate of change two years after the DVDs are released (t = 2) is That is, sales are decreasing by $600,000 per year.

Higher-Order Derivatives The derivative f ′ of a function f is also a function. As such, f ′ may also be differentiated. Thus, the function f ′ has a derivative f ″ at a point x in the domain of f if the limit of the quotient exists as h approaches zero. The function f ″ obtained in this manner is called the second derivative of the function f, just as the derivative f ′ of f is often called the first derivative of f. By the same token, you may consider the third, fourth, fifth, etc. derivatives of a function f.

Higher-Order Derivatives Practice Examples: Find the third derivative of the function f(x) = x2/3 and determine its domain. Solution We have and So the required derivative is The domain of the third derivative is the set of all real numbers except x = 0.

Higher-Order Derivatives Practice Examples: Find the second derivative of the function f(x) = (2x2 +3)3/2 Solution Using the general power rule we get the first derivative:

Higher-Order Derivatives Practice Examples: Find the second derivative of the function f(x) = (2x2 +3)3/2 Solution Using the product rule we get the second derivative:

Applied Example: Acceleration of a Maglev The distance s (in feet) covered by a maglev moving along a straight track t seconds after starting from rest is given by the function s = 4t2 (0  t  10) What is the maglev’s acceleration after 30 seconds? Solution The velocity of the maglev t seconds from rest is given by The acceleration of the maglev t seconds from rest is given by the rate of change of the velocity of t, given by or 8 feet per second per second (ft/sec2).

9.6 The Chain Rule

Deriving Composite Functions Consider the function To compute h′(x), we can first expand h(x) and then derive the resulting polynomial But how should we derive a function like H(x)?

Deriving Composite Functions Note that is a composite function: H(x) is composed of two simpler functions So that We can use this to find the derivative of H(x).

Deriving Composite Functions To find the derivative of the composite function H(x): We let u = f(x) = x2 + x + 1 and y = g(u) = u100. Then we find the derivatives of each of these functions The ratios of these derivatives suggest that Substituting x2 + x + 1 for u we get

Rule 7: The Chain Rule If h(x) = g[f(x)], then Equivalently, if we write y = h(x) = g(u), where u = f(x), then

The Chain Rule for Power Functions Many composite functions have the special form h(x) = g[f(x)] where g is defined by the rule g(x) = xn (n, a real number) so that h(x) = [f(x)]n In other words, the function h is given by the power of a function f. Examples:

The General Power Rule If the function f is differentiable and h(x) = [f(x)]n (n, a real number), then

The General Power Rule Practice Examples: Find the derivative of Solution Rewrite as a power function: Apply the general power rule:

The General Power Rule Practice Examples: Find the derivative of Solution Apply the product rule and the general power rule:

The General Power Rule Practice Examples: Find the derivative of Solution Rewrite as a power function: Apply the general power rule:

The General Power Rule Practice Examples: Find the derivative of Solution Apply the general power rule and the quotient rule:

Applied Problem: Arteriosclerosis Arteriosclerosis begins during childhood when plaque forms in the arterial walls, blocking the flow of blood through the arteries and leading to heart attacks, stroke and gangrene.

Applied Problem: Arteriosclerosis Suppose the idealized cross section of the aorta is circular with radius a cm and by year t the thickness of the plaque is h = g(t) cm then the area of the opening is given by A = p (a – h)2 cm2 Further suppose the radius of an individual’s artery is 1 cm (a = 1) and the thickness of the plaque in year t is given by h = g(t) = 1 – 0.01(10,000 – t2)1/2 cm

Applied Problem: Arteriosclerosis Then we can use these functions for h and A h = g(t) = 1 – 0.01(10,000 – t2)1/2 A = f(h) = p (1 – h)2 to find a function that gives us the rate at which A is changing with respect to time by applying the chain rule:

Applied Problem: Arteriosclerosis For example, at age 50 (t = 50), So that That is, the area of the arterial opening is decreasing at the rate of 0.03 cm2 per year for a typical 50 year old.

Differentiation of the Exponential and Logarithmic Functions 9.7 Differentiation of the Exponential and Logarithmic Functions

Rule 8 Derivative of the Exponential Function The derivative of the exponential function with base e is equal to the function itself:

Examples Find the derivative of the function Solution Using the product rule gives

Examples Find the derivative of the function Solution Using the general power rule gives

Rule 9 Chain Rule for Exponential Functions If f(x) is a differentiable function, then

Examples Find the derivative of the function Solution

Examples Find the derivative of the function Solution

Examples Find the derivative of the function Solution

Examples Find the derivative of the function Solution

Examples Find the derivative of the function Solution

Rule 10 Derivative of the Natural Logarithm The derivative of ln x is

Examples Find the derivative of the function Solution

Examples Find the derivative of the function Solution

Rule 11 Chain Rule for Logarithmic Functions If f(x) is a differentiable function, then

Examples Find the derivative of the function Solution

Examples Find the derivative of the function Solution

Marginal Functions in Economics 9.8 Marginal Functions in Economics MC(251) = C(251) – C(250) = [8000 + 200(251) – 0.2(251)2] – [8000 + 200(250) – 0.2(250)2] = 45,599.8 – 45,500 = 99.80

Marginal Analysis Marginal analysis is the study of the rate of change of economic quantities. These may have to do with the behavior of costs, revenues, profit, output, demand, etc. In this section we will discuss the marginal analysis of various functions related to: Cost Average Cost Revenue Profit

Applied Example: Rate of Change of Cost Functions Suppose the total cost in dollars incurred each week by Polaraire for manufacturing x refrigerators is given by the total cost function C(x) = 8000 + 200x – 0.2x2 (0  x  400) What is the actual cost incurred for manufacturing the 251st refrigerator? Find the rate of change of the total cost function with respect to x when x = 250. Compare the results obtained in parts (a) and (b).

Applied Example: Rate of Change of Cost Functions Solution The cost incurred in producing the 251st refrigerator is C(251) – C(250) = [8000 + 200(251) – 0.2(251)2] – [8000 + 200(250) – 0.2(250)2] = 45,599.8 – 45,500 = 99.80 or $99.80.

Applied Example: Rate of Change of Cost Functions Solution The rate of change of the total cost function C(x) = 8000 + 200x – 0.2x2 with respect to x is given by C´(x) = 200 – 0.4x So, when production is 250 refrigerators, the rate of change of the total cost with respect to x is C´(x) = 200 – 0.4(250) = 100 or $100.

Applied Example: Rate of Change of Cost Functions Solution Comparing the results from (a) and (b) we can see they are very similar: $99.80 versus $100. This is because (a) measures the average rate of change over the interval [250, 251], while (b) measures the instantaneous rate of change at exactly x = 250. The smaller the interval used, the closer the average rate of change becomes to the instantaneous rate of change.

Applied Example: Rate of Change of Cost Functions Solution The actual cost incurred in producing an additional unit of a good is called the marginal cost. As we just saw, the marginal cost is approximated by the rate of change of the total cost function. For this reason, economists define the marginal cost function as the derivative of the total cost function.

Applied Example: Marginal Cost Functions A subsidiary of Elektra Electronics manufactures a portable music player. Management determined that the daily total cost of producing these players (in dollars) is C(x) = 0.0001x3 – 0.08x2 + 40x + 5000 where x stands for the number of players produced. Find the marginal cost function. Find the marginal cost for x = 200, 300, 400, and 600. Interpret your results.

Applied Example: Marginal Cost Functions Solution If the total cost function is: C(x) = 0.0001x3 – 0.08x2 + 40x + 5000 then, its derivative is the marginal cost function: C´(x) = 0.0003x2 – 0.16x + 40

Applied Example: Marginal Cost Functions Solution The marginal cost for x = 200, 300, 400, and 600 is: C´(200) = 0.0003(200)2 – 0.16(200) + 40 = 20 C´(300) = 0.0003(300)2 – 0.16(300) + 40 = 19 C´(400) = 0.0003(400)2 – 0.16(400) + 40 = 24 C´(600) = 0.0003(600)2 – 0.16(600) + 40 = 52 or $20/unit, $19/unit, $24/unit, and $52/unit, respectively.

Applied Example: Marginal Cost Functions Solution From part (b) we learn that at first the marginal cost is decreasing, but as output increases, the marginal cost increases as well. This is a common phenomenon that occurs because of several factors, such as excessive costs due to overtime and high maintenance costs for keeping the plant running at such a fast rate.

Applied Example: Marginal Revenue Functions Suppose the relationship between the unit price p in dollars and the quantity demanded x of the Acrosonic model F loudspeaker system is given by the equation p = – 0.02x + 400 (0  x  20,000) Find the revenue function R. Find the marginal revenue function R′. Compute R′(2000) and interpret your result.

Applied Example: Marginal Revenue Functions Solution The revenue function is given by R(x) = px = (– 0.02x + 400)x = – 0.02x2 + 400x (0  x  20,000)

Applied Example: Marginal Revenue Functions Solution Given the revenue function R(x) = – 0.02x2 + 400x We find its derivative to obtain the marginal revenue function: R′(x) = – 0.04x + 400

Applied Example: Marginal Revenue Functions Solution When quantity demanded is 2000, the marginal revenue will be: R′(2000) = – 0.04(2000) + 400 = 320 Thus, the actual revenue realized from the sale of the 2001st loudspeaker system is approximately $320.

Applied Example: Marginal Profit Function Continuing with the last example, suppose the total cost (in dollars) of producing x units of the Acrosonic model F loudspeaker system is C(x) = 100x + 200,000 Find the profit function P. Find the marginal profit function P′. Compute P′ (2000) and interpret the result.

Applied Example: Marginal Profit Function Solution From last example we know that the revenue function is R(x) = – 0.02x2 + 400x Profit is the difference between total revenue and total cost, so the profit function is P(x) = R(x) – C(x) = (– 0.02x2 + 400x) – (100x + 200,000) = – 0.02x2 + 300x – 200,000

Applied Example: Marginal Profit Function Solution Given the profit function P(x) = – 0.02x2 + 300x – 200,000 we find its derivative to obtain the marginal profit function: P′(x) = – 0.04x + 300

Applied Example: Marginal Profit Function Solution When producing x = 2000, the marginal profit is P′(2000) = – 0.04(2000) + 300 = 220 Thus, the profit to be made from producing the 2001st loudspeaker is $220.

End of Chapter