Chapter 3 Limits and the Derivative Section 5 Basic Differentiation Properties (Part 1)

Objectives for Section 3.5 Power Rule and Differentiation Properties The student will be able to: Calculate the derivative of a constant function. Apply the power rule. Apply the constant multiple and sum and difference properties. Barnett/Ziegler/Byleen Business Calculus 12e

The Derivative Recall from the previous lesson: f(x) is a “slope machine”. 𝑓 ′ (𝑎) will tell you the slope of the line tangent to the graph of f(x) at x=a (if it exists). f(x) is also the instantaneous rate of change of f(x). f(x) is also the instantaneous velocity of an object. In this lesson, you will learn an easier way to find 𝑓 ′ 𝑥 . Barnett/Ziegler/Byleen Business Calculus 12e

Derivative Notation In the preceding section we defined the derivative of a function. There are several widely used symbols to represent the derivative. Given y = f (x), the derivative of f at x may be represented by any of the following: f (x) y 𝑑𝑦 𝑑𝑥 Later on, you will see how each of these symbols has its particular advantage in certain situations. Barnett/Ziegler/Byleen Business Calculus 12e

Derivative Rules In the next several slides, you will learn rules for finding derivatives of: Constant functions Power functions Functions multiplied by constants These rules will enable you to find 𝑓′(𝑥) easier compared to using a limit. Note that you will be tested on both methods. Barnett/Ziegler/Byleen Business Calculus 12e

Derivative of a Constant What is the slope of a constant function? The graph of f (x) = C is a horizontal line with slope 0, so we would expect f (x) = 0. Theorem 1. Let y = f (x) = C be a constant function, then y = f (x) = 0. Barnett/Ziegler/Byleen Business Calculus 12e

Example 1 Derivatives of constant functions: If f(x) = 8 f (x)=0 If y = -4 y = 0 If y =  𝑑𝑦 𝑑𝑥 =0 𝑑 𝑑𝑥 12=0 Barnett/Ziegler/Byleen Business Calculus 12e

Derivative of a Power Function A function of the form f (x) = xn is called a power function. (n is a real number) Theorem 2. (Power Rule) Let y = xn be a power function, then f (x) = nxn – 1. THEOREM 2 IS VERY IMPORTANT. IT WILL BE USED A LOT! Barnett/Ziegler/Byleen Business Calculus 12e

Example 2 Derivatives of power functions: If 𝑓 𝑥 = 𝑥 5 If y=𝑥 If 𝑓 𝑥 = 𝑥 5 4 If 𝑦= 3 𝑥 If 𝑓 𝑥 = 1 𝑥 2 then 𝑓 ′ 𝑥 =5 𝑥 5−1 =5 𝑥 4 then 𝑑𝑦 𝑑𝑥 = 1𝑥 1−1 = 𝑥 0 =1 then 𝑓 ′ 𝑥 = 5 4 𝑥 1 4 then 𝑦 ′ = 1 3 𝑥 −2 3 𝑦= 𝑥 1 3 then 𝑓 ′ (𝑥)= −2𝑥 −3 𝑓(𝑥)= 𝑥 −2 Barnett/Ziegler/Byleen Business Calculus 12e

Constant Multiple Property Theorem 3. Let y = k u(x) where k is a constant. Then y = k  u (x) In words: The derivative of a constant times a function is the constant times the derivative of the function. Barnett/Ziegler/Byleen Business Calculus 12e

Example 3 Differentiate each function: f (x) = 7x4 𝑦=−3 𝑥 2 𝑓 𝑥 =6 𝑥 2 3 𝑦= 2 3 𝑥 6 𝑓′(𝑥)=28 𝑥 3 𝑦 ′ =−6𝑥 𝑓 ′ 𝑥 =4 𝑥 − 1 3 = 2 𝑥 −6 3 𝑑𝑦 𝑑𝑥 =−4 𝑥 −7 Barnett/Ziegler/Byleen Business Calculus 12e

Sum and Difference Properties Theorem 5. If y = f (x) = u(x) + v(x), then y = f (x) = u(x) + v(x). (this is also true for subtraction) The derivative of the sum of two differentiable functions is the sum of the derivatives. The derivative of the difference of two differentiable functions is the difference of the derivatives. Barnett/Ziegler/Byleen Business Calculus 12e

Example 5 Differentiate f (x) = 3x5 + x4 – 2x3 + 5x2 – 7x + 4. Answer: f (x) = 15x4 + 4x3 – 6x2 + 10x – 7 Find 𝑑𝑦 𝑑𝑥 for 𝑦= 𝑥 −5x Answer: 𝑑𝑦 𝑑𝑥 = 1 2 𝑥 −1 2 −5 Barnett/Ziegler/Byleen Business Calculus 12e

Homework #3-5A: Pg. 185 (25-41 odd, 49) Mammoth opens on Nov. 7th! Barnett/Ziegler/Byleen Business Calculus 12e

Chapter 3 Limits and the Derivative Section 5 Basic Differentiation Properties (Part 2)

Objectives for Section 3.5 Power Rule and Differentiation Properties The student will be able to: Solve applications. Barnett/Ziegler/Byleen Business Calculus 12e

Applications Remember that the derivative has these meanings: Slope of the tangent line at a point on the graph of a function. Instantaneous velocity. Instantaneous rate of change. Barnett/Ziegler/Byleen Business Calculus 12e

Tangent Line Example Let f (x) = x4 – 6x2 + 10. (a) Find f (x) (b) Find the equation of the tangent line at x = 1 (c) Find the values of x where the tangent line is horizontal. Solution: f (x) = 4x3 - 12x Barnett/Ziegler/Byleen Business Calculus 12e

Example (continued) f (x) = x4 – 6x2 + 10. (b) Find the equation of the tangent line at x = 1 Solution: Slope: f (1) = 4(13) – 12(1) = -8. Point: If x = 1, then y = f (1) = 5 Point-slope form: y – y1 = m(x – x1) y – 5 = –8(x –1) y = –8x + 13 Barnett/Ziegler/Byleen Business Calculus 12e

Barnett/Ziegler/Byleen Business Calculus 12e

Example (continued) Let f (x) = x4 – 6x2 + 10. c) Find the values of x where the tangent line is horizontal. Solution: Tangent line is horizontal means slope is zero. So set derivative = 0 and solve for x. 4𝑥 3 −12𝑥=0 4𝑥 𝑥 2 −3 =0 𝑥=0, ± 3 Barnett/Ziegler/Byleen Business Calculus 12e

Barnett/Ziegler/Byleen Business Calculus 12e

Instantaneous Velocity An object moves along the y-axis (marked in feet) so that its position at time x (in seconds) is: 𝑓 𝑥 = 𝑥 3 −15 𝑥 2 +72𝑥 Find the instantaneous velocity function. Find the velocity at 2 and 5 seconds. Find the time(s) when the velocity is 0. Barnett/Ziegler/Byleen Business Calculus 12e

Instantaneous Velocity An object moves along the y-axis (marked in feet) so that its position at time x (in seconds) is: 𝑓 𝑥 = 𝑥 3 −15 𝑥 2 +72𝑥 Find the instantaneous velocity function. 𝑓 ′ 𝑥 =3 𝑥 2 −30𝑥+72 Barnett/Ziegler/Byleen Business Calculus 12e

Instantaneous Velocity An object moves along the y-axis (marked in feet) so that its position at time x (in seconds) is: 𝑓 𝑥 = 𝑥 3 −15 𝑥 2 +72𝑥 Find the instantaneous velocity function. Find the velocity at 2 and 5 seconds. 𝑓 ′ 𝑥 =3 𝑥 2 −30𝑥+72 𝑓 ′ 2 =24 𝑓 ′ 5 =−3 The velocity at 2 seconds is 24 ft/sec. The velocity at 5 seconds is −3 ft/sec. Barnett/Ziegler/Byleen Business Calculus 12e

Instantaneous Velocity An object moves along the y-axis (marked in feet) so that its position at time x (in seconds) is: 𝑓 𝑥 = 𝑥 3 −15 𝑥 2 +72𝑥 Find the time(s) when the velocity is 0. 𝑓 ′ 𝑥 =3 𝑥 2 −30𝑥+72 0=3 𝑥 2 −30𝑥+72 0=3( 𝑥 2 −10𝑥+24) 0=3(𝑥−6)(𝑥−4) 𝑥=4, 6 𝑇ℎ𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑖𝑠 0 𝑎𝑡 4 𝑎𝑛𝑑 6 𝑠𝑒𝑐𝑜𝑛𝑑𝑠. Barnett/Ziegler/Byleen Business Calculus 12e

Instantaneous Rate of Change If C(x) is the total cost of producing x items, then C(x) is the instantaneous rate of change of cost at a production level of x items. Barnett/Ziegler/Byleen Business Calculus 12e

Application Example The total cost (in dollars) of producing x portable radios per day is C(x) = 1000 + 100x – 0.5x2 for 0 ≤ x ≤ 100. Find 𝐶 ′ 𝑥 Solution: C(x) = 100 – x. Barnett/Ziegler/Byleen Business Calculus 12e

Example (continued) Find C(80) and 𝐶′(80) and interpret these results. Solution: 𝐶 80 =1000 + 100(80) – 0.5(80)2=5800 C(80) = 100 – 80 = 20 At a production level of 80 radios, the total cost is $5800 and is increasing at a rate of $20 per radio. Barnett/Ziegler/Byleen Business Calculus 12e

#3-5B: Pg. 185 (32-42 even, 50, 55, 56, 81, 87, 89) Homework Barnett/Ziegler/Byleen Business Calculus 12e