TURUNAN Definisi : Turunan pertama dari fungsi y = f (x) didefinisikan sebagai berikut : f ‘ (x) = y’ = p)x(f)px(flimdxdy0p
RUMUS-RUMUS TURUNAN
1. Jika y = c ( konstanta ) , maka y’ = 0
2. Jika y = x n , maka y’ = n.x n-1
3. Jika y = sin x , maka y’ = cos x
4. Jika y = cos x , maka y’ = –sin x
5. Jika y = tan x , maka y’ = sec2x
6. Jika y = cot x maka y’ = – csc2 x
7. Jika y = sec x maka y’ = secx tan x
8. Jika y = cscx maka y’ = – csc x.cot x
9. Jika y = ln x , maka y’ = x1
10. Jika y = ex , maka y’ = ex
SIFAT-SIFAT TURUNAN
1. Jika y = u ± v , maka y’ = u’ ± v’
2. Jika y = u . v , maka y’ = u’.v + u.v’
3. Jika y = vu , maka y’ = 2v'v.uv'.u
4. Jika y = u n , maka y’ = n. u n-1 . u’
5. Jika y = f ( u ) , maka y’ = f ’ ( u ) . u’
6. Jika y = f ( t ) dan t = g (x) , maka dxdt.dtdydxdy
PENGGUNAAN TURUNAN
1. f ’ (x ) = 0  didapat titik kritis
2. f ’ (x) > 0  f (x) naik
3. f ‘ (x) < 0  f (x) turun
4. f ‘ (x) = 0 dan f “ (x) < 0  didapat titik ekstrim maksimum
5. f ‘ (x) = 0 dan f ” (x) > 0  didapat titik ekstrim minimum

mathematic

LINEAR PROGRAM Programing: The allocation of limited resources to meet specific objectives. Linear Programming: Programming that involves issues in which the relationship between variables-variables are all linear. Some mathematical understanding will be found in linear programming problems, among others: 1. Constraints, namely the terms of conditions associated with the source. 2. The objective function or objective function or objective function, which is a function of the form Z = C1x1 + C2x2 + C3x3 + ... ... ... .... + Cnxn where xi  0 for each i = 1, 2, 3 ... n. .. C1, C2, C3, ... .... Cn is usually called the coefficient of cost. 3. Answer Feasible, ie users who meet the requirements given. 4. Answer infeasible, ie users who do not meet the requirements given. The purpose of the program is to maximize or minimize a linear objective function with linear-shaped linear terms. In general mathematical model of the form of linear programming in two dimensions (the field) are: Maximize or minimize the objective function Z = C1x1 + Cx2dengan terms: 2 K1 a1x1 + ax   22 d1 K2   bx2 b1x1 + d2 2 x1  0 and x2  0 Extreme Point Extreme point is a point which lies in the answer such that the objective function will reach the extreme price at that point. Example: 1. Maximize the objective function of the form Z = 5x1 + 3x2dengan condition: 3x + 5x K1   15 12 K2   10 5x1 + 2x2 x1  0 and x  0 2 Irvan Dedy Dwiwarna high school Tutoring Answer: We answer the first set (feasible region) in the field of XOY (field built by x1 and x2), then the answer is OABC and the dashed line is the line of the objective function. Visible lines built by the objective function and has the highest position is the line through point B (1920.1945). This means that Z = 5x + 3x reach the maximum price at point B. The result obtained x1x2AC255x1 12 035 + 2x2 = 10  K23x1 + 5x2 = 15  K1 B37, 121945.3 1920.5Zmaks    2. Maximize the objective function Z = 2.5 x + y with the following requirements: K1   15 3x + 5Y K  5x + 2y  10 2 x  0 and y  0035 x1x2AC255x1 + 2x2 = 10  K23x1 + 5x2 = 15  K1 B Answer: It turned out that the objective function z = 2.5 x + y coincides with the line 5x + 2y = 10, result 5  maksZ 3. Maximize z = 2x + 2y with the terms: 21  1  4 yx K1 K2 K1    x  y 1 K2    x  2y 4 x  0, y  0 From the picture above we get x and y   ~ ~. In this case the infinite responsibility. In other words the target function has no maximum price. 4. Determine the maximum price of the objective function z = 3x - 2y with the following requirements: K1   x + y 1 x y 01 2 2 1 K2 K1 K2   4 2x + 2y x  0 and y  0 Answer: In this persoalaan we find that there are no areas that meet the requirements given. As a result there is no price x and y that satisfy the objective function. 5. A fruit peddler yag using carts, selling apples and bananas. Purchase price of apples USD. 1000.00 USD per kg, and banana. 400.00 per kg. Capital is only Rp. 250,000.00 and wagon capacity of not more than 400 kg. If the profit per kg of apples per kg two times the profit of bananas, so as to benefit as much as possible, these traders have to buy how many kg of apples and how many pounds of bananas. Answer: Suppose that the number of apples that must be purchased x kg, and banana y kg. So the model mathematics are: Objective function: z = px + ½ p y, p  profit per kg of apples Provided that: K1  1000 x + 400 y  250 000 K2  x + y  400 x  0, y  0 Irvan Dedy Dwiwarna high school Tutoring Irvan Dedy Dwiwarna high school Tutoring AB C400250 O400625 K1 K2 x y z = p (x + ½ y) O 0 0 0 A 250 0 250 p B 150 250 275 p C 0 400 250 p Seen from above table that p. Thus the number of apples to be purchased is 150 kg and 250 kg of banana. 275Zmaks  6. An aircraft having a seating capacity of not more than 48 people, divided in the main class and economy class. Also capable of carrying baggage weighing up to 1440 kg. Each main class passengers to carry luggage no more than 60 kg for economy class while a maximum of 20 kg. If the cost (price kasrcis) for first class and economy class respectively, are USD. 100,000.00 and USD. 50000.00 per person, determine the number of passengers of each class for the biggest ticket sales. Answer: Suppose the number of passengers x men first class and economy class y people, the importance of the mathematical model as follows: AB C4824048K1K272 Objective function: z = 100,000 x + 50,000 y Term limits: K1  x + y  48 K2   60x + 20y 1440 x  0, y  0 x y z O 0 0 0 A 24 0 2.400.000 B 12 36 3.000.000 C 0 48 2.400.000 For ticket sales revenue reaches the largest number of passengers then the number of primary classes have 12 people while 36 people in economy class.