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10.4: The Derivative

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The average rate of change is the ratio of the change in y to the change in x The instantaneous rate of change of f at a is the limit of the average rates of change of f over shorter and shorter intervals around a Another name for the instantaneous rate of change is THE DERIVATIVE Note: Sometimes the word “instantaneous” is omitted

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Think of the odometer and the speedometer in a car. The odometer measure the distance (in miles) that the car travels and the speedometer measure how fast (miles per hour) the car travels at a particular time. The speedometer will give you the derivative at a certain point.

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Ex: Before I headed to Virginia, I looked at the odometer: 50,155. The trip started at 2:00 PM and ended at 8:30 PM. When I got to my parent-in-law’s house in VA, I checked my odometer: 50,517. Find the average speed? (50,517 - 50,155) / (8.5-2) = 362 / 6.5 = 55.7 mph The average speed is 55.7 mph; however, during my trip, I didn’t travel 55.7 mph all the time. Sometimes I went faster and sometimes I went slower. If I calculate the average speed at a smaller time interval (for example, during 1 second interval or smaller than that), I will have instantaneous average speed which is the derivative.

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Given y = f (x), the average rate of change from x = a to x = a + h is The above expression is also called a difference quotient. It can be interpreted as the slope of a secant. f (a + h) – f (a) h

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Application: The revenue generated by producing and selling calculators is given by R(x) = x (75 – 3x) for 0 x 20. What is the change in revenue if production changes from 9 to 12? R(12) – R(9) = $468 – $432 = $36. Increasing production from 9 to 12 will increase revenue by $36.

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The revenue is R(x) = x (75 – 3x) for 0 x 20. What is the average rate of change in revenue (per unit change in x) if production changes from 9 to 12? To find the average rate of change we divide the change in revenue by the change in production: Thus the average change in revenue is $12 when production is increased from 9 to 12.

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Example 1: given f(x)=x 2 A)Find the slope of the secant line for a =2 and h =1 Two ways to do this: *** Use slope formula We have (2, 4) and (3, 9) giving the above information *** Use different quotient TRY THE SAME PROBLEM WITH a=2, h=2. Do you get 6 for the answer?

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Example 1- continue: given f(x)=x 2 B)Find and simplify the slope of the secant line for a = 2 and h is any nonzero number. *** Use different quotient

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Example 1- continue: given f(x)=x 2 C)Find the limit of the expression in part B. D)Find the slope of the graph and the slope of the tangent line at a=2 The slope obtained from the limit of slopes of secant lines in part C is call the slope of the graph. Therefore, the slope of the graph and the slope of the tangent line at a=2 is 4 E)Find the equation of the tangent line at 2 Slope m=4, (2,4) y = mx + b 4 = 4(2) + b -4 = b Therefore the equation of the tangent line is y = 4x - 4

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The Derivative For y = f (x), we define the derivative of f at x, denoted f ’ (x), to be if the limit exists. If f ’(a) exists, we call f differentiable at a. If f ’(x) exist for each x in the open interval (a, b), then f is said to be differentiable over (a, b).

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Interpretations of the Derivative If f is a function, then f ’ is a new function with the following interpretations: ■ For each x in the domain of f ’, f ’ (x) is the slope of the line tangent to the graph of f at the point (x, f (x)). ■ For each x in the domain of f ’, f ’ (x) is the instantaneous rate of change of y = f (x) with respect to x. ■ If f (x) is the position of a moving object at time x, then v = f ’ (x) is the velocity of the object at that time.

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Finding the Derivative To find f ‘ (x), we use a four-step process: Step 1. Find f (x + h) Step 2. Find f (x + h) – f (x) Step 3. Find Step 4. Find

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Find the derivative of f (x) = x 2 – 3x Step 1. f (x + h) = (x + h) 2 – 3(x + h) = x 2 + 2xh + h 2 – 3x – 3h Step 2. Find f (x + h) – f (x) = x 2 + 2xh + h 2 – 3x – 3h – (x 2 – 3x) = 2xh + h 2 – 3h Step 3. Find Step 4. Find Example

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continue Find the slope of the tangent to the graph of f (x) = x 2 – 3x at x = 0, x = 2, and x = 3. Solution: In example 2 we found the derivative of this function at x to be f ’ (x) = 2x – 3 Hence f ’ (0) = -3 f ’ (2) = 1, and f ’ (3) = 3 Graphing Calculator: Press: y= Type in the equation Graph Press: 2 nd then Trace Press 6 (dy/dx) Type in the number Enter

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Practice: Find the derivatives

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Example: application The total sales of a company (in millions of dollars) t months from now are given by Find S(12) and S’(12), and interpret. Use these results to estimate the total sales after 13 months and after 14 months. Answer: Use G.C: S’(12) = 0.125 In 12 months, the sales will be 4 million dollars In 12 months, the rate is increasing at 0.125 million dollars ($125,000) per month. After 13 months, the sales will be 4.125 (4+0.125) million dollars After 14 months, the sales will be 4.250 (4.125+0.125) million dollars

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Nonexistence of the Derivative Some of the reasons why the derivative of a function may not exist at x = a are ■ The graph of f has a hole or break at x = a, or ■ The graph of f has a sharp corner at x = a, or ■ The graph of f has a vertical tangent at x = a.

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The Derivative. Objectives Students will be able to Use the “Newton’s Quotient and limits” process to calculate the derivative of a function. Determine.

The Derivative. Objectives Students will be able to Use the “Newton’s Quotient and limits” process to calculate the derivative of a function. Determine.

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