Lecture 1 - Derivative of a Polynomial Functions

L1 – Derivative of a Polynomial Functions MCV4U Jensen Unit 1 In advanced functions, you should have been introduced to the idea that the instantaneous rate of change is represented by the slope of the tangent at a point on the curve. You also learned that you can determine this value by taking the derivative of the function using the Newton Quotient. Newton Quotient Example: Newton’s Quotient: a) Find the equation of the derivative of 𝑓(𝑥) = 3𝑥 2 + 2𝑥 + 4 𝑓 ′ (𝑥) = lim ℎ→0 𝑓(𝑥 + ℎ) − 𝑓(𝑥) ℎ 3(𝑥 + ℎ)2 + 2(𝑥 + ℎ) + 4 − (3𝑥 2 + 2𝑥 + 4) ℎ→0 ℎ 𝑓 ′ (𝑥) = lim 3(𝑥 2 + 2𝑥ℎ + ℎ2 ) + 2𝑥 + 2ℎ + 4 − 3𝑥 2 − 2𝑥 − 4 ℎ→0 ℎ 𝑓 ′ (𝑥) = lim 3𝑥 2 + 6𝑥ℎ + 3ℎ2 + 2𝑥 + 2ℎ + 4 − 3𝑥 2 − 2𝑥 − 4 ℎ→0 ℎ 𝑓 ′ (𝑥) = lim 6𝑥ℎ + 3ℎ2 + 2ℎ ℎ→0 ℎ 𝑓 ′ (𝑥) = lim 𝑓 ′ (𝑥) = lim ℎ→0 ℎ(6𝑥 + 3ℎ + 2) ℎ 𝑓 ′ (𝑥) = 6𝑥 + 3(0) + 2 𝑓 ′ (𝑥) = 6𝑥 + 2 f(x) b) Calculate 𝑓′(5). What does it represent? 𝑓 ′ (5 ) = 6 (5 ) + 2 𝑓 ′ (5) = 32 This tells us that the instantaneous rate of change of the original function when 𝑥 = 5 is 32. Graphically speaking, this means the slope of the tangent line drawn on the original function at (5, 𝑓(5)) is 32. (5, f(5)) Mathematicians have derived a set of rules for calculating derivatives that make this process more efficient. Rule Constant Rule Derivative If 𝑓(𝑥 ) = 𝑐 where 𝑐 is a constant Power Rule 𝑛 If 𝑓(𝑥) = 𝑥 Constant Multiple Rule 𝑓 ′ (𝑥) = 0 Example 𝑓(𝑥) = 87 𝑓 ′ (𝑥) = 0 𝑓 ′ (𝑥) = 𝑛𝑥 𝑛−1 𝑓(𝑥) = 𝑥 5 𝑓 ′ (𝑥) = 5𝑥 4 𝑓 ′ (𝑥) = 𝑐 ∙ 𝑔′(𝑥) If 𝑓(𝑥) = 𝑐 ∙ 𝑔(𝑥) where 𝑐 is a constant Sum Rule If ℎ(𝑥) = 𝑓(𝑥) + 𝑔(𝑥) Difference Rule 𝑓(𝑥) = 3𝑥 5 𝑓 = 3 ∙ 5𝑥 4 𝑓 ′ (𝑥) = 15𝑥 4 ′ (𝑥) ℎ′ (𝑥) = 𝑓 ′ (𝑥) + 𝑔′(𝑥) ℎ(𝑥) = 𝑥 5 + 𝑥 4 ℎ′ (𝑥) = 5𝑥 4 + 4𝑥 3 ℎ′ (𝑥) = 𝑓 ′ (𝑥) − 𝑔′(𝑥) ℎ (𝑥 ) = 𝑥 5 − 𝑥 4 ℎ′ (𝑥) = 5𝑥 4 − 4𝑥 3 If ℎ(𝑥) = 𝑓(𝑥) − 𝑔(𝑥) Proof of Power Rule: (𝑥 + ℎ)𝑛 − 𝑥 𝑛 𝑑 𝑛 𝑥 = lim ℎ→0 𝑑𝑥 ℎ 𝑛 Use Binomial Theorem: 𝑛 𝑛 𝑛 ( 0 )𝑥 𝑛 ℎ0 + ( 1 )𝑥 𝑛−1 ℎ1 + ( 2 )𝑥 𝑛−2 ℎ2 + ⋯ + (𝑛)𝑥 0 ℎ𝑛 − 𝑥 𝑛 𝑑 𝑛 𝑥 = lim ℎ→0 𝑑𝑥 ℎ 𝑛 𝑛 𝑛 𝑛 𝑛 ( 1 )𝑥 𝑛−1 ℎ1 + ( 2 )𝑥 𝑛−2 ℎ2 + ⋯ + (𝑛)𝑥 0 ℎ𝑛 𝑑 𝑛 𝑥 = lim ℎ→0 𝑑𝑥 ℎ 𝑛 ℎ[( 1 )𝑥 𝑛−1 + ( 2 )𝑥 𝑛−2 ℎ1 + ⋯ + (𝑛)𝑥 0 ℎ𝑛−1 ] 𝑑 𝑛 𝑥 = lim ℎ→0 𝑑𝑥 ℎ 𝑑 𝑛 𝑛 𝑛 𝑛 𝑥 = lim ( ) 𝑥 𝑛−1 + ( ) 𝑥 𝑛−2 ℎ1 + ⋯ + ( ) 𝑥 0 ℎ𝑛−1 ℎ→0 1 𝑑𝑥 2 𝑛 𝑑 𝑛 𝑛 𝑛 𝑛 𝑥 = ( ) 𝑥 𝑛−1 + ( ) 𝑥 𝑛−2 01 + ⋯ + ( ) 𝑥 0 0𝑛−1 𝑑𝑥 1 2 𝑛 𝑑 𝑛 𝑥 = 𝑛𝑥 𝑛−1 𝑑𝑥 𝑛 𝑡𝑟+1 = ( ) 𝑎𝑛−𝑟 𝑏𝑟 𝑟 Example 1: Determine the equation of the derivative of each of the following functions: a) 𝑓(𝑥) = 3𝑥 5 𝑓 ′ (𝑥) = 15𝑥 4 b) 𝑓(𝑥) = 71 c) 𝑓(𝑥) = √𝑥 𝑓 ′ (𝑥) = 0 𝑓(𝑥) = 𝑥 2 1 1 1 𝑓 ′ (𝑥) = 2 𝑥 2−1 1 1 𝑓 ′ (𝑥) = 2 𝑥 −2 𝑓 ′ (𝑥) = 1 d) 𝑦 = 3√𝑥 1 1 𝑦′ = 𝑦′ = f) 𝑦 = − 𝑥 5 𝑦 = 𝑥 −1 𝑦 = −1𝑥 −5 𝑦 ′ = −1𝑥 −2 1 𝑦 ′ = 3 𝑥 3−1 𝑦′ = 1 −2 𝑥 3 3 −1 𝑥2 𝑑𝑦 𝑑𝑥 = 5𝑥 −6 𝑑𝑦 𝑑𝑥 = 𝑥6 5 1 2 3𝑥 3 Example 2: Differentiate each function a) 𝑦 = 5𝑥 6 − 4𝑥 3 + 6 b) 𝑓(𝑥) = −3𝑥 5 + 8√𝑥 − 9.3 1 𝑦 ′ = 30𝑥 5 − 12𝑥 2 + 0 𝑓(𝑥) = −3𝑥 5 + 8𝑥 2 − 9.3 𝑦 ′ = 30𝑥 5 − 12𝑥 2 𝑓 ′ (𝑥) = −15𝑥 4 + 4𝑥 −2 − 0 1 𝑓 ′ (𝑥) = −15𝑥 4 + c) 𝑔(𝑥) = (2𝑥 − 3)(𝑥 + 1) 𝑔(𝑥) = 2𝑥 2 −𝑥−3 𝑔′ (𝑥) = 4𝑥 − 1 2√𝑥 1 e) 𝑦 = 𝑥 𝑦 = 𝑥3 1 d) ℎ(𝑥) = ℎ(𝑥) = 4 √𝑥 −8𝑥 6 +8𝑥 2 4𝑥 5 −8𝑥 6 4𝑥 5 8𝑥 2 + 4𝑥 5 ℎ(𝑥) = −2𝑥 + 2𝑥 −3 ℎ′ (𝑥) = −2 − 6𝑥−4 ℎ′ (𝑥) = −2 − 6 𝑥4 Example 3: Determine an equation for the tangent to the curve 𝑓(𝑥) = 4𝑥 3 + 3𝑥 2 − 5 at 𝑥 = −1. Point on the tangent line: Slope of tangent line: 𝑓(−1) = 4(−1)3 + 3(−1)2 − 5 𝑓 𝑓(−1) = −6 𝑓 ′ (−1) = 12(−1)2 + 6(−1) (−1, −6) 𝑓 ′ (−1) = 6 ′ (𝑥) 2 = 12𝑥 + 6𝑥 Remember: The equation of the derivative tells you the slope of the tangent to the original function. Slope of the tangent is 6 Equation of tangent line: 𝑦 = 𝑚𝑥 + 𝑏 −6 = 6(−1) + 𝑏 𝑏=0 𝑦 = 6𝑥 Example 4: Determine the point(s) on the graph of 𝑦 = 𝑥 2 (𝑥 + 3) where the slope of the tangent is 24. 𝑦 = 𝑥 3 + 3𝑥 2 𝑑𝑦 = 3𝑥 2 + 6𝑥 𝑑𝑥 24 = 3𝑥 2 + 6𝑥 0 = 3𝑥 2 + 6𝑥 − 24 0 = 3(𝑥 2 + 2𝑥 − 8) 0 = (𝑥 + 4)(𝑥 − 2) 𝑥1 = −4 𝑥2 = 2 Now find 𝑦-coordinates of points: Point 1: Point 2: 𝑦 = (−4)3 + 3(−4)2 𝑦 = (2)3 + 3(2)2 𝑦 = −16 𝑦 = 20 (−4, 16) (2, 20)