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Lesson 2-8

Lesson 2-8. Introduction to Derivatives. 2-7 Review Problem. For a particle whose position at time t is f ( t ) = 6 t 2 - 4 t +1 ft.: b. Find the instantaneous velocity of the particle at t = 1 sec.  . f(1+h) – f(1) lim --------------------- =

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Lesson 2-8

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  1. Lesson 2-8 Introduction to Derivatives

  2. 2-7 Review Problem For a particle whose position at time t is f(t) = 6t2 - 4t +1 ft.: b. Find the instantaneous velocity of the particle at t = 1 sec.   f(1+h) – f(1) lim --------------------- = h0 h f(a+h) – f(a) lim --------------------- = h0 h (6h² + 8h + 3) – (3) lim --------------------------- = h0 h f(1) = 6 - 4 + 1 = 3 f(1+h) = 6(1+h)² - 4(1+h) + 1 = 6h² + 12h + 6 – 4h – 4 + 1 = 6h² + 8h + 3 6h² + 8h lim ------------- = h0 h lim 6h + 8 = 8 ft h0

  3. Objectives • Understand the derivative as the slope of the tangent and as a rate of change

  4. Vocabulary • Derivative – the instantaneous rate of change of a function. It is also a function, denoted by f’(x). At a point it is the slope of the tangent line at a point on the curve (lim ∆y/∆x (∆x→0) ) f(a+h) – f(a) f’(a) = lim --------------------- h0 h

  5. Example A Use the definition to find f’(1) if f(x) = 1/x f(1+h) – f(1) lim --------------------- = h0 h f(a+h) – f(a) lim --------------------- = h0 h (1/(h+1)) – (1) lim --------------------- = h0 h f(1) = 1/(1) = 1 f(1+h) = 1/(1+h) 1 h + 1 - h ------ - --------- = ----------- h + 1 h + 1 h + 1 - h / (h + 1) lim ---------------- = h0 h - 1 lim ---------- = - 1 h0 h + 1

  6. Example A Generalized Use the definition to find f’(a) if f(x) = 1/x 1/(h+a) – (1/a) lim --------------------- = h0 h f(a+h) – f(a) lim --------------------- = h0 h f(a) = 1/(a) f(a+h) = 1/(h+a) -h lim --------------------- = h0 a(h + a)h 1 1 ------ - ---- = h + a a a(1) - (h + a) - h ---------------------- = ---------------- a(h+a) a(h+a) - 1 -1 lim ---------------- = ------- h0 ah + a² a²

  7. Example B Use the definition to find f’(x) if f(x) = 2x³ - x² + 3x - 1 f(a+h) – f(a) lim --------------------- = h0 h f(a+h) = 2(a+h)³ - (a+h)² + 3(a+h) - 1 = 2a³ + 6a²h + 6ah² + 2h³ - a² - 2ah - h² + 3a + 3h - 1 f(a) = 2a³ - a² + 3a - 1 ---------------------------------------------------------------------------------- f(a+h) – f(a) = 6a²h + 6ah² + 2h³ – 2ah – h² + 3h 6a²h + 6ah² + 2h³ – 2ah – h² + 3h lim -------------------------------------------------- = h0 h lim 6a² + 6ah + 2h² – 2a – h + 3 = 6a² - 2a + 3 h0

  8. Example C 2(a+h)² - 3(a+h) + 1 f(a+h) = ---------------------------- a+h 2x² - 3x + 1 f(x) = ---------------- x 2a² - 3a + 1 f(a) = ---------------- a Use the definition to find f’(x) if f(x) = (2x² - 3x + 1) / x 2(a+h)² - 3(a+h) + 1 (2a² - 3a + 1) f(a+h) – f(a) = ---------------------------- - ------------------ a+h a 2a(a+h)² - 3a(a+h) + a – (2a²(a+h) – 3a(a+h) + (a+h)) = ----------------------------------------------------------------------- a(a+h) 2a³ + 4a²h + 2ah² – 3a² – 3ah + a – 2a³ – 2a²h + 3a² + 3ah – a – h = ----------------------------------------------------------------------------------------- a² + ah 2a²h + 2ah² - h = ------------------------- a² + ah f(a+h) – f(a)2a²h + 2ah² - h 2a² + 2ah - 1 2a² - 1 Lim ---------------- = Lim ------------------------- = Lim ------------------- = --------- h (a² + ah)h (a² + ah) a²

  9. Alternative Form of Derivative f(x) – f(a) f’(a) = lim ------------------- xa x – a Using the alternate form, find f’(3) for f(x) = x2 + 10x f(x) = x² + 10x f(3) = 3² + 10(3) = 39 f(x) – f(3) = x² + 10x – 39 = (x + 13) (x – 3) f(x) – f(3) (x + 13) (x – 3) f’(3) = lim ------------------- = lim ---------------------- x3 x – 3 x3 x – 3 = lim (x + 13) = 16 x3

  10. Graphical Relationships Below is a graph of the function f(x), place the following quantities in order from lowest to highest. 5 f’(a) f’(b) slope of the secant line PQ slope of the secant line QR (f(c) – f(a)) / (c – a) 1 4 2 3

  11. Derivative Definition Reversed Each of the following is a derivative, but of what function (f(x)=?) and at what point (a=?)? (4+ ∆x)² – 16 a. lim --------------------- ∆x0 ∆x a = 4 f(x) = x² 2(5+ ∆x)³ – 2(5)³ b. lim ----------------------- ∆x0 ∆x a = 5 f(x) = 2x³ cos (π/4+ ∆x) – 2/2 c. lim -------------------------------- ∆x0 ∆x a = π/4 f(x) = cos x

  12. Derivative Definition Reversed Each of the following is a derivative, but of what function (f(x)=?) and at what point (a=?)? x² – 4 d. lim --------------- x2 x – 2 a = 2 f(x) = x² x³ + x – 30 e. lim ----------------------- x3 x – 3 a = 3 f(x) = x³ + x (3+ ∆x)² + 2(3 + ∆x) – 15 f. lim ------------------------------------ ∆x0 ∆x a = 3 f(x) = x² + 2x (2/x) – (2/3) g. lim -------------------- x3 x – 3 a = 3 f(x) = 2/x

  13. Summary & Homework • Summary: • Slope of the tangent line to a curve at a point is the derivative of the function evaluated at that point • Homework: pg 163 - 164: 4, 13, 19, 29

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