Lesson 2-8 Introduction to Derivatives

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

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

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

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

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

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 6a²h + 6ah² + 2h³ – 2ah – h² + 3h 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 f(a+h) – f(a) = 6a²h + 6ah² + 2h³ – 2ah – h² + 3h lim 6a² + 6ah + 2h² – 2a – h + 3 = 6a² - 2a + 3 h  0

Example C Use the definition to find f’(x) if f(x) = (2x² - 3x + 1) / x 2x² - 3x + 1 f(x) = x 2a² - 3a + 1 f(a) = a 2(a+h)² - 3(a+h) + 1 f(a+h) = a+h 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

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) = x x f(x) – f(3) (x + 13) (x – 3) f’(3) = lim = lim x  3 x – 3 x  3 x – 3 f(x) = x² + 10x f(3) = 3² + 10(3) = 39 f(x) – f(3) = x² + 10x – 39 = (x + 13) (x – 3) = lim (x + 13) = 16 x  3

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

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 cos (π/4+ ∆x) –  2/2 c. lim ∆x  0 ∆x a = 5 f(x) = 2x³ a = π/4 f(x) = cos x

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 (3+ ∆x)² + 2(3 + ∆x) – 15 f. lim ∆x  0 ∆x a = 3 f(x) = x³ + x a = 3 f(x) = x² + 2x (2/x) – (2/3) g. lim x  3 x – 3 a = 3 f(x) = 2/x

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 : 4, 13, 19, 29