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maths Individual assignment on differentiation

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maths Individual assignment on differentiation

  1. 1. Name: xxxxxxxxxxxxxxx ID: xxxxxxxxxxx Class: xxxx INDIVIDUAL ASSIGNMENT 1. The balance of a bank account after 3 years with respect to the interest rate r% is given by A= 1000(1+r/1000)36 Find the rate of change of the balance with respect to r when r = 6 SOLUTION A=1000(1+6/1000) =1000(1+0.006) 36 36 = 1000(1.24) =1240.3 36= 3 YEARS 12= 1 YEAR A=1000(1+6/1000) 12 =1074.42 AVERAGE RATE OF CHANGE IS GIVEN BY: Y3-Y1/X3-X1 1240.3-1074.42/36-12 ARC = 6.912 2- Find f (a) for given value of a. f(x) = 5x2- 4x + 7; and a = 4: SOLUTION 2 F(x) =5x -4x+7 F’(x) =10x-4
  2. 2. F’ (4) =10(4)-4 F’ (a) =36 3- Find f (g(x)) and g (f(x)) for f(x) = 4x + 3; and g(x) = -2x + 1: SOLUTION F (G(X)) = f (-2x+1) = 4(-2x+1) +3 = -8x+7 G (f(x)) = g (4x+3) = -2(4x+3) +1 =-8x-5 4- Let s be a distance given by S (t) = 3t3 + 4t2 + 8t: Find the acceleration. SOLUTION 3 2 S (t) = 3t +4t +8t To find the acceleration we derive the function twice 2 S’ (t) = 9t +8t S’’ (t) = 18t+8 5- Find f0(x) for f(x) = 10/x Solution 6 F (x) = 10/x =10x -6 6
  3. 3. F’ (x) = -60x -7 6- For the function f(x) = 5x2 + 4x. Find f(x + h) - f(x)/h For: a. x = 2; h = 0:1. b. x = 1; h = 0:5. c. x = 3; h = 1. Solution 2 a- f (2) =5(2) +4(2) =28 2 F (2+0.1) = 5(2+0.1) +4(2+0.1) =30.09 30.09-28/0.1=20.9 b- F (1) =5(1) +4(1) =9 2 F (1+0.5) =f (1.5) +4(1.5) =17.25 17.25-9/0.5=16.5 2 c- F (3) =5(3) +4(3) =57 2 F (3+1) =5(3+1) +4(3+1) =96 96-57/1=39 7- Find the limit if it exists lim 10x+3 X 3 Lim 10x+3 = lim 10x+3 = 33 + x 3 x 3 3 8- Differentiate the function y = 5x (2x - 7x)
  4. 4. SOLUTION). USING CHAIN RULE: F(X).G(X) = F’(X).G(X) + G’(X).F(X) F(X) =5X F’(X) =5 3 2 G’(X) = 6X -7 G(X) =2X -7X 3 2 Y’= 5. (2x -7x)+ (6x -7)5x 3 Y’= 4x -7x 2 9- Differentiate the function y= 2x/x +4 F(x) =2x f’(x) =2 2 G(x) =x +4 g’(x) = 2x Y’ = g(x).f’(x)-f(x).g’(x)/g(x) 2 2 2 Y’ = (x +4).2 – 2x (2x) / (x +4) 2 2 Y’ = -2(x +8) / (x +4) 2 2 10- Find an expression for dy/dx: y = 3u2 and u = 3x + 1: Dy/du= 6u, and du/dx= 3 Dy/dx= dy/du.U + du/dx.y 2 = 6u. (3x + 1) + 3. (3u ) = 18ux + 6u + 9u 2 2 = 3u + 6ux + 2u 3 2 12- Find the absolute maximum and minimum of the function f(x) = 2x - 3x + 4 on the interval [-1; 2]. SOLUTION
  5. 5. 2 F’(x) = 6x -6x F’(x) =0 2 6x -6x =0 6x(x-1) =0 6x=0 or x-1=0 X=0, X=1 {-1, 0, 1, 2} F (-1) =-1 F (0) =4 F (1) =3 F (2) =8 The absolute maximum is 8= f (2) The absolute minimum is -1= f (-1) 3 13- Find dy for given values of x and dx: y = x ; Dy/dx=3x 2 x = 4 and dx = 0:1 2 dy= 3x .dx Dy= 3(4). (0.1) Dy=4.8 3 14- Determine where the function is concave up and down: f(x) = x - 3x + 2: Concavity can be found using the second derivative. 2 F’ (x) = 3x -3 F’’(x) = 6x -to find where the function is concave up, we find where the second derivative is positive 6x>0 x>0: the function is concave up for all x- values > 0. Similarly, the function is concave down for all x-values < 0.
  6. 6. 3 2 16- The cost of producing x items is given by: C(x) = x -34x + 5600x: Find the production level that minimizes the average cost. 3 C(x)/x 2 2 x -34x +5600x/x = x -34x+5600 3 2 3 17- Find dy=dx by implicit differentiation, given that x + xy + y = 3: 3 2 3 D/dx(x +xy +y ) =d/dx (3) 3 2 3 D/dx(x ) +d/dx (xy ) +d/dx (y ) =0 2 2 2 3x +x.2y.dy/dx+y +3y d/dx=0 2 2 2 2 2 3y d/dx+2yxdy/dx=-3x -y 2 (3y +2yx) dy/dx=-3x -y 2 2 2 Dy/dx= -3x -y /3y +2yx Find the slope of the tangent line at point (1; 1). y - y0 = m(x - x0) 18- The cost and the revenue with respect to the number of items is given by C(x) = 200x + 2000 and 2 R(x) = 5x + 100x + 80: Find the marginal profit Solution P(X) = R(x) – C(x) 2 P(x) = (5x +100x+80) – (200x +200) 2 P(x) = 5x – 100x -120 P’(x) = 10x-100

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