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Power Rule

Power Rule. k(x) = g n (x) = [g(x)] n k’(x) = n [g (x)] n-1 g’(x). Power Rule. k(x) = g n (x) = [g(x)] n k’(x) = n [g (x)] n-1 g’(x). k(x) = (x 2 - 4) 4 k’(x)=. 4 (x 2 - 4) 3 (2x) 4 (2x) 3 4 (x 2 - 4) 3 (2x - 4). t(x) = t’(x)=.

nelle-guerra
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Power Rule

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  1. Power Rule k(x) = gn(x) = [g(x)]n k’(x) = n [g(x)] n-1g’(x)

  2. Power Rule k(x) = gn(x) = [g(x)]n k’(x) = n [g(x)] n-1g’(x)

  3. k(x) = (x2 - 4)4k’(x)= • 4 (x2 - 4)3 (2x) • 4 (2x)3 • 4 (x2 - 4)3 (2x - 4)

  4. t(x) = t’(x)= • . • .

  5. #69-The assets of socially responsible funds in bill. f(t) = 23.7(0.2t+1)1.32 on [0,11] 0 is code for 1991 • What were the assets in 1991? • What were the assets in 2000?

  6. f(t) = 23.7(0.2t+1)1.32on [0,11] 0 is code for 1991 What were the assets in 1991? 23.7 * ( 1) = $23.7 billion What were the assets in 2000? =23.7*(1.8+1)^1.32=$92.26 billion

  7. f(t) = 23.7(0.2t+1)1.32on [0,11] 0 is code for 1991 How fast are the assets changing? f’(t) = 1.32(23.7)[0.2t+1] 0.32 [0.2]

  8. f’(t) = 1.32(23.7)[0.2t+1] 0.32 [0.2] How fast are the assets changing in 1991? f’(0) = 1.32(23.7)[1] 0.32 [0.2] = $6.26 billion per year How fast in 2000? f’(9) = 1.32(23.7)[0.2(9)+1] 0.32 [0.2] = $8.7 billion per year

  9. N(t)=.002(t+5)2.5 on [0,15] where 0 means 2000 Total number of US jobs (in millions) projected to leave the US by year t. a) How many left by 2008?

  10. N(t)=.002(t+5)2.5How many millions left by 2008? • 1.22 • 0.1

  11. N(t)=.002(t+5)2.5 on [0,15] where 0 means 2000 Total number of US jobs (in millions) projected to leave the US by year t. • How many left by 2008? N(8) = 0.002(8+5)2.5 = 0.002(13)2.5 = 1.22 million jobs.

  12. Power Rule k(x) = [x2 + x] 3 k’(x) = 3 [x2 + x] 2 (2x+1)

  13. k(x) = ( x2 + x)3 Evaluate k’(x) • 3( x2 + x)2(2x+1) • 3(2x+1) 2 • 3( x2 + x)(2x+1)

  14. N(t)=.002(t+5)2.5 on [0,15] where 0 means 2000 How fast is the graph of N growing after t years? Find the rate of change of N after t years? Find N’(t).

  15. N(t)=.002(t+5)2.5 Find the rate of change of N. • N’(t)=.05(t+5) 2 • N’(t)=.005(t+5) 1.5 • N’(t)=.005(t+5) 2.5

  16. Theorem 1 The chain rule If k(x) =f (g(x)), then k’(x) = f ’ ( g(x) ) g’(x)

  17. If k(x) =f (g(x)), thenk’(x) = f ’(g(x))g’(x) k(x) = ( x2 + x)3 k’(x) = 3 ( x2 + x)2(2x +1)

  18. The chain rule If y = (u)3 and u(x) = x2 + x then dy/dx = dy/du du/dx dy/du = 3(u)2 du/dx = 2x + 1 dy/dx = 3(u)2(2x + 1) = 3(x2 + x)2(2x + 1)

  19. R is daily revenue ($K) and r (%) is occupancy rate r(t) = 10/81 t3 – 10/3 t2 +200/9 t +60 on [0,12] 0=12 means January 1st R(r) = -3/150000 r3 +9/1500 r2 on [0, 100]

  20. r(t) = 10/81 t3 – 10/3 t2 +200/9 t +60 What is the average occupancy rate on the ides of March? r(2.5) = 10/81 2.53 - 62.5/3 + 500/9 + 60 = 96.65%

  21. R(r) = -3/150000 r3+9/1500 r2 What is the revenue on the ides of March when the occupancy rate is 96.65%? R(96.65)=$38 thousand

  22. dy/dx = dy/du du/dx Find the rate of change of revenue with respect to t. dR/dt = dR/dr dr/dt $1000/day =($1000/%point)(%point/day)

  23. R is Revenue r in % dR/dt = dR/dr dr/dt Find the rate of change of revenue with respect to t. ($1000/day) R(r) = -3/150000 r3 +9/1500 r2 dR/dr = -9/150000 r2 + 18/1500 r r(t) = 10/81 t3 – 10/3 t2 +200/9 t +60 dr/dt = 10/27 t2 – 20/3 t + 200/9

  24. dR/dt = dR/dr dr/dt Find the rate of change of revenue with respect to r on the ides of March dR/dr = -9/150000 r2 + 18/1500 r dR/dr (96.65)= -9/150000 (96.65)2 + 18/1500 (96.65) = 0.60 =$600/point

  25. dR/dt = dR/dr dr/dt Find dr/dt on the ides of March r(t) = 10/81 t3 – 10/3 t2 +200/9 t +60 dr/dt = 10/27 t2 – 20/3 t + 200/9 dr/dt (2.5) = 10/27(2.5)2 – 50/3 + 200/9 = 7.87 points per day

  26. dR/dt = dR/dr dr/dt Find dR/dt on the ides of March dR/dt = 0.60 * 7.87 = 4.722 The revenue is increasing at the rate of $4722 per day

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