Drill: Write each expression as a sum of powers of x. (No variables in the denominator!)

Drill: Write each expression as a sum of powers of x. (No variables in the denominator!)

Lesson 3.3: Derivatives Day #1 Homework page 124: 1-13, 15-22 (do not show graphically)

Derivative of a Constant Function • If is the function with the constant value c, then df/dx = d/dx (c) = 0 • Example: y = 4, find dy/dx • dy/dx= 0

Power Rule for Positive Integer Powers of x • If n is a positive integer, then d/dx (xn) = nxn-1 • Examples: • y = x3 • dy/dx= 3x2 • y = 4x4 • dy/dx = 16x3

The Sum and Difference Rule • If u and v are differentiable functions of x, then their sum and differences are differentiable at every point where u and v are differentiable: • d/dx ( u ± v) = du/dx ± dv/dx • Example: • y = x3 + ½ x2 – 3x + 2 • dy/dx= 3x2 + x - 3

Finding Horizontal Tangents • Does the curve y = x4 – 2x2 + 2 have any horizontal tangents? If so, where? • Calculate dy/dx • 4x3 – 4x • Solve the equation dy/dx = 0 for x. • 4x3 – 4x = 0 • 4x (x2 – 1) = 0 • 4x (x – 1) (x + 1) = 0 • 4x = 0; (x – 1) = 0; (x + 1) = 0 • x = 0, 1, -1

The Product Rule • d/dx (uv) = u (dv/dx) + v(du/dx) • Find f’(x) if f(x) = (x2 + 1) (x3 + 3) • Let u = x2 + 1 and v = x3 + 3 • dv/dx = 3x2 and du/dx = 2x • u (dv/dx) + v(du/dx) = (x2 + 1 )(3x2) + (x3 + 3)(2x) • 3x4 + 3x2 + 2x4 + 6x • 5x4 + 3x2 + 6x

The Quotient Rule

Drill: Find the derivative of each of the following functions. 2x6– 5x4+ 2x (x3+ x) (2x2+ 3)

How to support derivatives graphically • Let y1 = f’(x) and y2 = NDER f(x). They should coincide.

Working with Numerical Values • Let y = uv be the product of the functions u and v. Find y’(2) if • u(2) = 3, u’(2) = -4, v(2) = 1 and v’(2) = 2 • So, from the product rule: • u(2)v’(2) + v(2)u’(2) • 3(2) + 1(-4) = 2

Negative Integer Powers of x • If n is a negative integer and x ≠ 0, then • d/dx (xn) = nxn-1 • Example: f(x) = -2x-3 • f’(x) = 6x-4

Using the Power Rule • Find an equation for the line tangent to the curve y = (x2 + 3)/2x at the point (1, 2) • While we could find the derivative by the Quotient Rule, but it is easier to first simplify into a sum of two powers: (x2/2x) + 3/2x = ( ½ ) x + (3/2)x-1 • dy/dx = ½ - (3/2)x-2 • The slope at x = 1: • ½ - (3/2) (1)-2 = ½ - 3/2 = -1 • Using the slope of -1 and the point (1, 2): • y – 2 = -1 (x – 1 )  y = -1x + 3

Second and High Order Derivatives • dy/dx: first derivative • f’(x) = y’ • dy’/dx = second derivative • f’’(x) = y’’ = d2y/dx2 • dy’’/dx = third derivative • f’’’(x) = y’’’ = d3y/dx3 • etc….

Find the first four derivatives of y =x3 – 5x2 + 2 • y’: 3x2 – 10x • y’’ = 6x – 10 • y’’’ = 6 • y’’’’= 0

Finding Instantaneous Rates of Change • (page 123, example 5) • An orange farmer currently has 200 trees yielding an average of 15 bushels of oranges per tree. She is expanding her farm at the rate of 15 trees per year, while improved husbandry is improving her average annual yield by 1.2 bushels by tree. What is the current (instantaneous) rate of increase of her total annual production of oranges?

Solution • Let t(x) = the number of trees x years from now • Let y(x) = yield per tree x years from now • p(x) = t(x)y(x) is the total production of oranges in year x. • We know t(0) = 200; y (0) = 15; t’(0) = 15; y’(0) = 1.2 • We need to find p’(x), so we can use the product rule: • t(0)y’(0) + y(0)t’(0) • 200(1.2) + 15(15) = 465 bushels per year.

Homework • page 124-125:23-37, 51

Drill: Write each expression as a sum of powers of x. (No variables in the denominator!)