Derivative – Mathematica Education

Derivative function (differential) is another function of a previous function, for example the function F becomes f ‘which has an irregular value. The concept of derivatives as a major part of calculus was thought at the same time by Sir Isaac Newton Mathematics and Physics Experts of the nation and Gottifred Wilhelm Leibniz (1646 – 1716), a German mathematician. Derivatives (differential) are used as a tool to solve various problems in geometry and mechanics. Derivatives can be determined without a limit process^.For this purpose the theorem is designed about basic derivatives, derivatives of algebra operations in two functions, chain rules for composition function derivatives, and inverse functions.

The rules in the derivative function are:

1. f (x), eye f ‘(x) = 0
2. If f (x) = x, then f ‘(x) = 1
3. Rules: If f (x) = x^Nface f ‘(x) = nx ^{n – 1}
4. The rules of the multiple of the constants: (KF) (x) = k. F ‘(x)
5. Chain rules: (fog) (x) = f ‘(g (x)). g ‘(x))
6. Derivative amount, difference, product, and the results for two functions

For example the functions f and g are differentiated in interval I, then the functions f + g, f – g, fg, f/g, (g (x) ≠ 0 at i) differentiated in i with the rules:

1. (f + g) ‘(x) = f’ (x) + g ‘(x)
2. (f – g) ‘(x) = f’ (x) – g ‘(x)
3. (fg) ‘(x) = f’ (x) g (x) + g ‘(x) f (x)
4. ((f)/g) ‘(x) = (g (x) f’ (x)- f (x) g ‘(x))/((g (x)²)
5. Derivative trigonometric function
6. D/Dx (sin x) = cos x
7. D/Dx (cos x) = – sin x
8. D/Dx (tan x) = seconds²X
9. D/Dx (Cot X) = – CSC²X
10. D/Dx (sec x) = dtk x tan x
11. D/Dx (CSC x) = -CSC X Cot X
12. Derivative inverse function

(F^-1) (y) = 1/(f ‘(x), changes/dx 1/(dx/dx/dy

2.2 Formulas of Mathematical Function Derivatives

To facilitate learning, the following formula Calculates various derivative formulas.

1. Application 1: If y = cxn with constants C and n, then dy/dx = cn xn-1

Example:

y = 2 x 4 then to/dx = 4.2 × 4-1 = 8 × 3

Sometimes there are questions about wearing a small part

y = = 2 x^1/2 The derivative is (= x- =

1. Formula 2: If y = c with c is a constant then dy/dx = 0

For example if y = 6 then the derivative is equal to zero (0)

1. Formula 3: If y = f (x) + g (x) then the derivative is the same as the derivative of each function = f ‘(x) + g’ (x)

Example:

y = x³ + 2 x² Then and ‘= 3 x² + 4x

y = 2x⁵ + 6 Makay = 10x⁴ + 0 = 10x⁴

1. Formula 4: Function multiplication derivatives if yf (x) .g (x) then y ‘= f’ (x) .g (x) + g ‘(x) .f (x)

Example:

y = x² (X² + 2) For

f (x) = x²

F ‘(x) = 2x

g (x) = x² + 2

g ‘(x) = 2x

We enter the formula y ‘= f’ (x) .g (x) + g ‘(x). F (x)

y = 2x (x² + 2) + 2x.x²

y = 4x³ + 4x

The use of derivatives in everyday life

This topic is the last topic of derivative material. On this topic, we will learn how to model and solve problems in everyday life that involve descent. One of the derivative concepts that is often used is the first derivative and maximum value and minimum function. The first derivative concept of function is widely used in the issue of speed with the known position function, while the concept of maximum and minimum function is used in the area of ​​area such as land and buildings, the volume of building space, and economics.

The maximum and minimum value of a function is often referred to as extreme values. The extreme value of the function y = f (x) is obtained for x that meets the equation f ′ (x) = 0. If X = A is the value of X which meets the equation f ′ (x) = 0, then (a, f (a)) is the extreme point of the function y = f (x) and f (a) is the extreme value of the function y = f (x).
Types of extreme functions can be determined as follows.

This extreme value will be the maximum value if:

f ‘(x) = 0 and f “(x) <0.

The extreme value will be the minimum value if:

F ‘(x) = 0 and F “(x)> 0.

Example 1

A fireworks are launched into the air. Height of fireworks h = f

(in meters) Solid seconds are modeled with f

Determine the speed of the fireworks when t = 3 seconds.

Solution:

It is known that the height of the fireworks when t seconds is: f

The speed of the fireworks is obtained the first derivative of the function

height (position) of fireworks as follows:

f ‘

So, the speed of the fireworks when t = 3 seconds is 296 m/s.

Example 2

A company produces x units of goods per day at a cost x³ –

600x² + 112,500 x rupiah. How many units of goods to be produced each

the day so that the production costs become minimal?

Solution:

For example production costs per day are P (x), then the production costs will be

Minimum for the value of x that meets the equation p ′ (x) = 0 and p ′ (x)> 0.

P ′ (x) = 0

⇔3x² −1.200x+112,500 = 0

⇔x²−400x+37,500 = 0

⇔ (x – -130) (x -250) = 0

⇔x = 150 or x = 250

Because P ′ ′ (x) = 6x – 1.200 and p ′ (250) = 6 (250) −1.200 = 300> 0, then

the amount of goods that must be produced every day so that the minimum cost is

250 units.

Example 3

A tennis ball is fired up. If the height of the tennis ball (cm) from

the ground surface after t seconds are formulated with h

Determine the maximum height of the tennis ball.

Solution:

The tennis ball will reach the maximum height of the ground surface

for t ′ ′

H ′

⇔120−10t = 0

⇔10t = 120

⇔t = 12

Because H “(x) = -10 <0, the tennis ball will reach a height

maximum from the ground surface. Next by substituting

t = 12 to h

H (12) = 120 (12) −5 (12) 2 = 720.

Thus, the maximum height of the tennis ball is 720

cm.

Example 4

A kitchen equipment company produces x units of goods with

Cost (x²−70x+250) thousands of stems. If the income after all goods run out

sold is 100x thousand rupiah, then calculate the maximum profit

can be obtained by the company.

Solution:

Suppose the company’s profits are f (x), so:

f (x) = income – cost. Production

⇔F (x) = 100x− (x²−70x+250)

⇔F (x) = – x²+170x – 250

The maximum profit will be obtained for the x value

Meet f ′ (x) = 0 and f “(x) <0

f ′ (x) = 0

⇔ – 2x+170 = 0

⇔2x = 170

⇔x = 85

Because F “(x) = -2 <0, the profit gained is

maximum.

The amount of profit when x = 85 is f (85) = – 852−70 (85)+250 = 175.

So, the company’s maximum profit is 175,000 rupiah.