Definition of Derivative if the limit exists as one real number.

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Presentation transcript:

Definition of Derivative if the limit exists as one real number.

If g(t) = t n then g'(t) = n t (n-1) If g(t) = t 4 + 3t + 1, find g’(t) If f(t) = t n then f ' (t) = n t (n-1) A. 4t B. 4t + 3 C. 4t D. 4t

If g(t) = t n then g'(t) = n t (n-1) If g(t) = t 4 + 3t + 1, find g’(t) A. 4t B. 4t + 3 C. 4t D. 4t

If g’(t) = 4t 3 + 3, find g’’(t) A. 4t B. 12t C. 12t 2 D. 4t If f(t) = t n then f ' (t) = n t (n-1)

If g’(t) = 4t 3 + 3, find g’’(t) A. 4t B. 12t C. 12t 2 D. 4t If f(t) = t n then f ' (t) = n t (n-1)

If h(x) =, find h’(x) If h(x) = x n then h' (x) = n x (n-1) If h(x) = x n then h' (x) = n x (n-1) h(x) = 4x -2 h(x) = 4x -2

h(x) = =4x -2 h’(x)= A. -8x x -11 B. 8x x -11 C. -8x x -9 D. 8x x -9 If h(x) = x n then h' (x) = n x (n-1)

If h(x) =, find h’(x) A. -8x x -11 B. 8x x -11 C. -8x x -9 D. 8x x -9 If h(x) = x n then h' (x) = n x (n-1)

Dist trvl by X-2 racing car t seconds after braking.#59 x(t) = 120 t – 15 t 2. x(t) = 120 t – 15 t 2. Find the velocity for any t. Find the velocity for any t. Find the velocity when brakes applied. Find the velocity when brakes applied. When did it stop? When did it stop?

Dist trvl by X-2 racing car t seconds after braking. 59. x(t) = 120 t – 15 t 2. x(t) = 120 t – 15 t 2. Find the velocity for any t. Find the velocity for any t. x’(t) = t x’(t) = t

Dist trvl by X-2 racing car t seconds after braking. 59. x(t) = 120 t – 15 t 2. x(t) = 120 t – 15 t 2. x’(t) = t x’(t) = t Find the velocity when brakes applied. Find the velocity when brakes applied. x’(0) = 120 ft/sec x’(0) = 120 ft/sec

Dist trvl by X-2 racing car t seconds after braking. 59. x(t) = 120 t – 15 t 2. x(t) = 120 t – 15 t 2. x’(t) = t x’(t) = t Find the velocity when t = 2. Find the velocity when t = 2. x’(2) = 120 – 30(2) = 60 ft/sec x’(2) = 120 – 30(2) = 60 ft/sec

Dist trvl by X-2 racing car t seconds after braking. 59. x(t) = 120 t – 15 t 2. x(t) = 120 t – 15 t 2. x’(t) = t x’(t) = t Find the velocity when t = 2. Find the velocity when t = 2. x’(2) = 120 – 30(2) = 60 ft/sec x’(2) = 120 – 30(2) = 60 ft/sec What does positive 60 mean? What does positive 60 mean?

Dist trvl by X-2 racing car t seconds after braking. 59. x(t) = 120 t – 15 t 2. x(t) = 120 t – 15 t 2. x’(t) = t x’(t) = t Find the velocity when t = 2. Find the velocity when t = 2. x’(2) = 120 – 30(2) = 60 ft/sec x’(2) = 120 – 30(2) = 60 ft/sec What does positive 60 mean? What does positive 60 mean? Car is increasing its distance from home. Car is increasing its distance from home.

Dist trvl by X-2 racing car t seconds after braking. 59. x(t) = 120 t – 15 t 2. x(t) = 120 t – 15 t 2. x’(t) = t x’(t) = t When did it stop? When did it stop?

Dist trvl by X-2 racing car t seconds after braking. 59. x(t) = 120 t – 15 t 2. x(t) = 120 t – 15 t 2. x’(t) = t x’(t) = t When did it stop? When did it stop? When the velocity is zero. When the velocity is zero.

Dist trvl by X-2 racing car t seconds after braking. 59. x(t) = 120 t – 15 t 2. x(t) = 120 t – 15 t 2. x’(t) = t x’(t) = t When did it stop? When did it stop? x’(t) = t = 0 x’(t) = t = 0

Dist trvl by X-2 racing car t seconds after braking. 59. x(t) = 120 t – 15 t 2. x(t) = 120 t – 15 t 2. x’(t) = t x’(t) = t When did it stop? When did it stop? x’(t) = t = 0 x’(t) = t = = 30 t 120 = 30 t 4 = t 4 = t

Dist trvl by X-2 racing car t seconds after braking. 59. x(t) = 120 t – 15 t 2. x(t) = 120 t – 15 t 2. x’(t) = t x’(t) = t When did it stop? When did it stop? x’(t) = t = 0 x’(t) = t = = 30 t 120 = 30 t 4 = t 4 = t This changes the domain of x to This changes the domain of x to

Dist trvl by X-2 racing car t seconds after braking. 59. x(t) = 120 t – 15 t 2. x(t) = 120 t – 15 t 2. x’(t) = t x’(t) = t When did it stop? When did it stop? x’(t) = t = 0 x’(t) = t = = 30 t 120 = 30 t 4 = t 4 = t This changes the domain of x to [0,4]. This changes the domain of x to [0,4].

Dist trvl by X-2 racing car t seconds after braking. 59. x(t) = 120 t – 15 t 2 defined on [0,4]. x(t) = 120 t – 15 t 2 defined on [0,4]. x’(t) = t x’(t) = t How far did it travel after hitting the brakes? How far did it travel after hitting the brakes?

Dist trvl by X-2 racing car t seconds after braking. 59. x(t) = 120 t – 15 t 2 defined on [0,4]. x(t) = 120 t – 15 t 2 defined on [0,4]. x’(t) = t x’(t) = t How far did it travel after hitting the brakes? How far did it travel after hitting the brakes? x(4) = 480 – 15*16 = 240 feet x(4) = 480 – 15*16 = 240 feet

Dist trvl by X-2 racing car t seconds after braking. 59. x(t) = 120 t – 15 t 2. x(t) = 120 t – 15 t 2. x’(t) = t x’(t) = t Find the acceleration, x’’(t). Find the acceleration, x’’(t).

Dist trvl by X-2 racing car t seconds after braking. 59. x(t) = 120 t – 15 t 2. x(t) = 120 t – 15 t 2. x’(t) = t x’(t) = t Find the acceleration, x’’(t). Find the acceleration, x’’(t). x’’(t) = -30 x’’(t) = -30

Dist trvl by X-2 racing car t seconds after braking. 59. x(t) = 120 t – 15 t 2. x(t) = 120 t – 15 t 2. x’(t) = t x’(t) = t Acceleration, x’’(t) = -30. Acceleration, x’’(t) = -30. What does the negative sign mean? What does the negative sign mean?

Dist trvl by X-2 racing car t seconds after braking. 59. x(t) = 120 t – 15 t 2. x(t) = 120 t – 15 t 2. x’(t) = t x’(t) = t Acceleration, x’’(t) = -30. Acceleration, x’’(t) = -30. What does the negative sign mean? What does the negative sign mean? Your foot is on the brakes. Your foot is on the brakes.

Dist trvl by X-2 racing car t seconds after braking. 59. x(t) = 120 t – 15 t 2. x(t) = 120 t – 15 t 2. x’(t) = t x’(t) = t What is the range on [0,4]? What is the range on [0,4]?

Dist trvl by X-2 racing car t seconds after braking. 59. x(t) = 120 t – 15 t 2. x(t) = 120 t – 15 t 2. x’(t) = t x’(t) = t What is the range on [0,4]? What is the range on [0,4]? [0, 240] [0, 240]

Definition of Derivative if the limit exists as one real number.

Theorem 1 The product rule [f(x) g(x)]’ = copy diff. copy diff. f(x) g’(x) + g(x) f ’(x)

The product rule If k(x) = f(x) g(x), then k’(x) = f(x) g’(x) + g(x) f ’(x) Suppose k(x) = (3x 2 -2x) (4x 4 +1) k’(x) = (3x 2 -2x) (16x 3 ) + (4x 4 +1)(6x-2)

Theorem 1 The product rule [f(x) g(x)]’ = f(x) g’(x) + g(x) f ’(x)

The product rule [f(x) g(x)]’=f(x)g’(x) + g(x)f ’(x) If y =(x 2 -3x+1)(1/x+3x) find dy/dx dy/dx = (x 2 -3x+1)(-x -2 +3) + (1/x+3x) (2x-3)

If f(x) = x n then f ' (x) = n x (n-1) Ifthen

The product rule [f(x) g(x)]’ = f(x) g’(x) + g(x) f ’(x) [ (3x-1)]’ = (3) +(3x-1)

If y = (x-1)(3x 2 +2) find dy/dx or y’ A. 6x B. (x-1) 6x + (3x 2 +2) C. (x-1) 6x + (3x 2 +2)x D. 3x 2 +2

If y = (x-1)(3x 2 +2) find dy/dx or y’ A. 6x B. (x-1) 6x + (3x 2 +2) C. (x-1) 6x + (3x 2 +2)x D. 3x 2 +2

If y = (x 2 -x)(-3x 2 +2) find dy/dx or y’ A. (2x-1) (-6x) B. (x 2 -x)(-3x+2) + (-3x 2 +2)(2x-1) C. (x 2 -x)(-6x) + (-3x 2 +2)(2x-1)

If y = (x 2 -x)(-3x 2 +2) find dy/dx or y’ A. (2x-1) (-6x) B. (x 2 -x)(-3x+2) + (-3x 2 +2)(2x-1) C. (x 2 -x)(-6x) + (-3x 2 +2)(2x-1)

If y = (x 2 -x)(x 3 -3x 2 +2) find dy/dx or y’ A. (x 2 -x) (3x 2 - 6x) + (x 3 - 3x 2 +2) (2x-1) B. (x 2 -x) (2x-1) + (x 3 - 3x 2 +2 ) (3x 2 - 6x) C. (2x-1) (3x 2 - 6x) D. (x 2 -x) (3x 2 - 6x) - (x 3 - 3x 2 +2) (2x-1)

If y = (x 2 -x)(x 3 -3x 2 +2) find dy/dx or y’ A. (x 2 -x) (3x 2 - 6x) + (x 3 - 3x 2 +2) (2x-1) B. (x 2 -x) (2x-1) + (x 3 - 3x 2 +2 ) (3x 2 - 6x) C. (2x-1) (3x 2 - 6x) D. (x 2 -x) (3x 2 - 6x) - (x 3 - 3x 2 +2) (2x-1)

Theorem 2 The quotient rule

Never cancel across the minus sign

The quotient rule

If f(x)= find f ’(x) A. [(3x-4)2 – (2x-1)3] /(3x-4) 2 B. (3x-4)2 – (2x -1)(3) C. [(2x-1)3 – (3x-4)2] /(3x-4) 2

If f(x)= find f ’(x) A. [(3x-4)2 – (2x-1)3] /(3x-4) 2 B. (3x-4)2 – (2x -1)(3) C. [(2x-1)3 – (3x-4)2] /(3x-4) 2

If f(x)= find f ’(1) A. -11 B. -5 C. C.

If f(x)= find f ’(1) A. -11 B. -5 C. C.

If f(x)= f ’(1) = -5 Write the equation of the line tangent to f when x = 1.

If f(x)= f ’(1) = -5 Write the equation of the line tangent to f when x = 1. f(1) =

If f(x)= f ’(1) = -5 Write the equation of the line tangent to f when x = 1. f(1) = 2/(-1) = -2

If f(x)= & f ’(1) = -5 Write the eq. of the tan. line A. y + 1 = -5(x-1) B. y – 1 = -5(x-1) C. y = -5(x-1) D. y = 5(x-1) - 1 when x = 1.

If f(x)= & f ’(1) = -5 Write the eq. of the tan. line A. y + 1 = -5(x-1) B. y – 1 = -5(x-1) C. y = -5(x-1) D. y = 5(x-1) - 1 when x = 1.

If f(x)= where is the horizontal asymptote? A. d. n. e. B. y = 0 C. y = 2/3 D. y = 1

If f(x)= where is the horizontal asymptote? A. d. n. e. B. y = 0 C. y = 2/3 D. y = 1

If f(x)= where is the vertical asymptote? A. d. n. e. B. x = 1 C. x = 4/3 D. x = 3/2

If f(x)= where is the vertical asymptote? A. d. n. e. B. x = 1 C. x = 4/3 D. x = 3/2

Proof of the product rule k(x) = f(x) g(x)

Proving the product rule

The product rule

k’(x) = f(x) g’(x) + g(x) f ’(x)