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**The Derivative Section 2.1**

Calculus for Business, Economics, the Social and Life Sciences The Derivative Section 2.1 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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Learning Objectives Examine slopes of tangent lines and rates of change Define the derivative, and study its basic properties Compute and interpret a variety of derivatives using the definition Study the relationship between differentiability and continuity

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Average Rate of Change The average rate of change of function f on the interval [a,b] is given by Note that this is the “slope” of a function between two points.

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**Let , then the average rate of change of f between x=-1 and x=2 is**

Quick example: Let , then the average rate of change of f between x=-1 and x=2 is Note that because f is linear, this is just the slope of the line, which equals -2.

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**EXAMPLE 1 Average Rate of Change in Cost**

The cost to produce k quarts of kettle corn is dollars. What is the average rate of change in cost when changing from producing 10 to 20 quarts? Include units.

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**EXAMPLE 1 Average Rate of Change in Cost**

SOLUTION The average rate of change from k = 10 to k = 20 is Then we have and

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**EXAMPLE 1 Average Rate of Change in Cost**

SOLUTION So the average rate is The units on the average are in units of the output (cost in dollars) divided by input (quarts produced). Thus the average rate of change is 0.2 dollars per quart.

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**EXAMPLE 2 Average Rate of Change on a Small Interval**

Compute the average rate of change of ... … on the interval [1,2]. ... on the interval [1,1.1]. … “at” x = 1.

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**EXAMPLE 2 Average Rate of Change on a Small Interval**

SOLUTION … on the interval [1,2].

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**EXAMPLE 2 Average Rate of Change on a Small Interval**

SOLUTION (b) ... on the interval [1,1.1].

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**EXAMPLE 2 Average Rate of Change on a Small Interval**

SOLUTION (c) … “at” x = 1. Here we first compute the average rate of change for an arbitrary distance, h, from x = 1:

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**EXAMPLE 2 Average Rate of Change on a Small Interval**

SOLUTION If we observe the trend in this expression as h approaches zero, we should get a sense for the slope “at” the value x = 1:

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**EXAMPLE 2 Average Rate of Change on a Small Interval**

SOLUTION The graph of along with its secant line on The graph of along with its tangent line at x = 1. the interval [1,2].

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The Derivative The derivative of function f is given by At a point c, is the slope of the tangent line to f at c. To compute the derivative of a function is to differentiate.

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**Let , the derivative of f is**

Quick example: Let , the derivative of f is

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**Like the average rate of change computed earlier, this is merely the slope of the line.**

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**EXAMPLE 3 Instantaneous Change in Cost**

The cost in dollars to produce k quarts of kettle corn is Compute the derivative of the cost function. At what instantaneous rate is cost changing when 10 quarts are produced?

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**EXAMPLE 3 Instantaneous Change in Cost**

SOLUTION We compute the derivative by evaluating the limit

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**EXAMPLE 3 Instantaneous Change in Cost**

SOLUTION

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**EXAMPLE 3 Instantaneous Change in Cost**

SOLUTION Thus Then we find the instantaneous rate of change at , that is, the derivative at 10: When producing 10 quarts of kettle corn, cost is increasing at a rate of 0.3 dollars per additional quart produced.

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**EXAMPLE 4 Change in U.S. Demand for Rice**

The demand for rice in the U.S. in 2009 approximately followed where p is the price per ton and D is the demand in millions of tons of rice. Find and interpret Find the equation of the tangent line to D at p = 500.

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**EXAMPLE 4 Change in U.S. Demand for Rice**

SOLUTION Using the definition of the derivative, we get We can remove the fractions by multiplying by

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**EXAMPLE 4 Change in U.S. Demand for Rice**

SOLUTION The fraction is still undefined at h=0, so we apply an algebra trick, multiplying by the conjugate

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**EXAMPLE 4 Change in U.S. Demand for Rice**

SOLUTION

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**EXAMPLE 4 Change in U.S. Demand for Rice**

SOLUTION

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**EXAMPLE 4 Change in U.S. Demand for Rice**

SOLUTION so The units on demand are millions of tons of rice, and the units on price are dollars per ton. Because the derivative is negative, at a unit price of $500 per ton, demand is decreasing by about 4,470 tons per $1 increase in unit price.

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**EXAMPLE 4 Change in U.S. Demand for Rice**

SOLUTION (b) Find the equation of the tangent line to D at p = 500. The tangent line to a function f is defined to be the line passing through the point and having slope equal to the derivative at that point.

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**EXAMPLE 4 Change in U.S. Demand for Rice**

SOLUTION First, we find the value of D at p=500: So we know that the tangent line passes through the point

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**EXAMPLE 4 Change in U.S. Demand for Rice**

SOLUTION Next, we use the derivative of D for the slope of the tangent line: Finally, we use the point-slope formula and simplify:

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**EXAMPLE 4 Change in U.S. Demand for Rice**

SOLUTION The graph of along with its tangent line at p = 500.

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**Continuity vs. Differentiability**

A function f is differentiable at if is defined. If a function is differentiable at a point, then it is continuous at that point. Note that being continuous at a point does not guarantee that the function is differentiable there.

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**and so the derivative of f is**

Quick example: Let , and recall that and so the derivative of f is Note that the derivative cannot exist at because the derivatives on each side disagree.

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