√ Definition of Derivatives, Types, Formulas, & Example Problems

August 15, 2023
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The discussion of derivatives needs to be studied. By using the limit concept that you have learned, you will easily learn the following derivative material.

Definition of Derivative

The derivative is a calculation of changes in function values due to changes in input values (variables).

The derivative can also be called differential and the process of determining the derivative of a function is called differentiation.

Using the limit concept that has been studied, the derivative can be defined as

the derivative is defined as the limit of the average change of the value of the function to variable x.

In the following, an example of the implementation of inheritance will be explained.

Derived Application

Here are some derived implementations.

The derivative can be applied to calculate the gradient of the tangent to a curve.
The derivative can be used to determine the interval over which a function increases or decreases.
Derivatives can be applied to determine the stationary value of a function.

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Derivatives can be applied in solving problems related to the equation of motion.
Derivatives can be used to solve maximum-minimum problems.

The following will explain the derivative formula.

Derivative Formulas

Here are some basic formulas for determining the derivative.

f(x) = c, where c is a constant

The derivative of this function is f'(x) = 0.

f(x) = x

The derivative of this function is f'(x) = 1.

f(x) = axⁿ

The derivative of this function is f'(x) = anx^n–1

Function addition: h(x) = f(x) + g(x)

The derivative of this function is h'(x) = f'(x) + g'(x).

Subtraction function: h (x) = f (x) – g (x)

The derivative of this function is h'(x) = f'(x) – g'(x)

Constant multiplication by a function (kf)(x).

The derivative of this function is k. f'(x).

In the following, we will explain the derivative function.

Function Derivation

Suppose there is a function f (x) = axⁿ. The derivative of this function is f'(x) = anx^n–1.

Examples are:

f(x) = 3x³

the derivative of the function i.e

f'(x) = 3 (3) x^{3 – 1} = 9x².

Another example is for example g (x) = -5y^-3.

The derivative of this function is g'(y) = -5 (-3) y^{-3 – 1} = 15y^-4.

The following will explain the derivative of algebraic functions.

Derivative of Algebraic Functions

The discussion on the derivatives of algebraic functions in this section includes derivatives in the form of multiplication and derivatives in the distribution of algebraic functions.

The derivative of the algebraic function in multiplication form is as follows.

Suppose there is a multiplication of functions: h (x) = u (x). v(x).

The derivative of this function is h'(x) = u'(x). v(x) + u(x). v'(x).

Information:

h(x): function in multiplication form.
h'(x): the derivative of the multiplication form function
u(x), v(x): functions with variable x
u'(x), v'(x): derivative of functions with variable x

The derivative of the algebraic function in division form is:

Suppose there is a multiplication function: h (x) = u (x)/v (x). The derivative of this function is

h'(x) = (u'(x). v(x) – u(x). v'(x))/v²(x).

Information:

h(x): function in multiplication form.
h'(x): the derivative of the multiplication form function
u(x), v(x): functions with variable x
u'(x), v'(x): derivative of functions with variable x

The following will explain about root derivatives.

Root Derivatives

Suppose there is a root function as follows

To determine the derivative of this function, we first change it to the exponential function. The exponential form of the function is f (x) = x^a/b.

The derivative of this function is f'(x) = a/b. x^{(a/b) – 1}.

What if the function looks like this?

To determine the derivative of the function above, it must first be changed to exponential form.

f(x) = g(x)^z/b

The derivative of this function is f'(x) = a/b. g(x)^{(a/b) – 1}. g'(x).

The following will explain about partial derivatives.

Partial Derivative

What is a partial derivative? The partial derivative is a derivative of the function of many variables with respect to a variable, while the other variables are maintained.

Suppose there is a function: f (x, y) = 2xy, the partial derivative of the function with respect to variable x is f_x’(x, y) = 2y.

The partial derivative of the variable y is f_y’(x, y) = -6xy.

The following will explain about implicit derivatives.

Implicit Derivative

The implicit derivative is determined based on the variables contained in the function.

A function with variable x, its derivative: x d/dx.

A function with variable y, its derivative: y d/dy. dy/dx.

A function with variables x and y, derivative: xy d/dx + xy d/dy. dy/dx.

Another example is, there is a function g (x, y) = -3xy²

To better understand derivatives, try doing the following questions and then check your answers using the discussion in the section below.

Examples of Derivative Questions

1. Find the derivative of the following function.

f(x) = 8
g(x) = 3x + 5
h(x) = 6x³
k(x) = 3x^5/3
m(x) = (3x² + 3)⁴

Discussion

f'(x) = 0
g'(x) = 3
h'(x) = 6 (3) x^{3 – 1}= 18x²
k'(x) = 3 (5/3) x^{(5/3) – 1} = 5x^2/3
m'(x) = 4. (3x² + 3)^{4 – 1}. 6x = 24x. (3x² + 3)³
2. Find the derivative of the following function.
f(x) = (3x + 2). (2x² – 1)

Discussion

For example: u (x) = 3x + 2 and v (x) = 2x² – 1

f'(x) = u'(x). v(x) + u(x). v'(x)

f'(x) = 3. (2x² – 1) + (3x + 2). (4x)

f'(x) = 6x² – 3 + 12x² + 8x = 18x² +8x – 3

3. Given a function of order 2 as below

Determine the value of f (0) + 3f'(1)

Discussion

To do this problem, we can enter the value 0 into the function.

After you, get the value of f(0). We can work on the derivative of the quotient function using any of the derived properties.

To use the formula, we can use the example and its derivatives as below.

U = x2 + 3; U' = 2x

V = 2x + 1; V' = 2

Then, we can enter this example into the previous derivative formula and we can directly enter f'x (1).

So, the result f (0) + 3f'(1) = 3 + 3(0) = 3

4. Find the derivative f (x) = (x² + 2x + 3)(3x + 2)

Discussion

Just like the previous problem, to work on the derivative problem in multiplication form, we can use the formula for the derived property and use an example in the function as below.

F'(x) = u'v + uv'

U = x² +2x +3; U' = 2x + 3

V = 3x + 2; V' = 3

F'(x) = u'v + uv'

F'(x) = (2x+3)(3x + 2) + (x² + 2x + 3)(3)

F'(x) = 6x²+13x +6 +3x²+6x+9

F'(x) = 9x² +19x +15

So the final form F'(x) is 9x² +19x +15

5. If there is f (x) = (2x-1)²(x+2). What is the value of f'x (2)
Discussion
To do this problem, we can use the derivative property of the function f'(x) = u'v + v'u to get the final result. So we can do the separation again.

F'(x) = u'v + uv'

U= (2x-1)² = 4x²– 4x + 1; U' = 8x – 4

V = x + 2; V' = 1

F'(x) = u'v + uv'

F'(x) = (8x – 4)(x + 2) + (4x²– 4x + 1)(1); we can enter the value 2 as in the problem

F'(2) = ((8(2) – 4)(2 + 2)) + ((4(2)²– 4(2) + 1)(1))

F'(2) = ((16-4)(4)) + ((16-8+1)(1))

F'(2) = 96 + 9 = 105

So that the final value of F'(2) is 105

6. Find a tangent to the curve y= -2x² + 6x + 7 which is perpendicular to the line x – 2y +13 = 0

Discussion

It is stated in the problem that there are 2 lines that are perpendicular to each other, so we can assume that the two lines have a certain slope. We can determine the value of m₁ and M₂ from both lines.

m₁is the slope of the line y= -2x² +6x+7. To find the value of m₁, can be done by deriving the function y= -2x² +6x+7.

m₁= y'(x) = -4x + 6

m₂is the slope of x – 2y +13. To find the value of m₂, we have to change the function to function y.

x – 2y +13 = 0

x + 13 = 2y

y = 0.5x + 6.5

m₂= y'(x) = 0.5

Because the two lines are perpendicular to each other, the value of m₁x m₂= -1.

m₁x m₂= -1

(-4x + 6)0.5 = -1

-2x + 3 = -1

-2x = -4

X = 2

We plug it into the equation m₁so that the value of m is obtained₁ = -2. After finding the value of x, we enter that value into the y function so that we get the value y = 11.

To make a tangent line, the formula used is (y-y₁) = m₁(x – x₁).

(y – 11) = -2 (x – 2)

Y – 11 = -2x +4

Y = -2x + 15

The tangent is y+2x-15 = 0

7. There is a box without a lid with a square base having an area of 512 cm². What is the length of the edge so that the volume has a maximum value
Discussion
In this question, it is explained that the box does not have a lid. Thus, the box consists of 4 sides and 1 base. Assume the side of the base is s and the height of the side is t. We can write the box equation as below.

512 = area of the base + 4 sides of the box

512 = s.s + 4.s.t
512 = s² + 4st
512 – s² = 4th

After getting t, we can find the volume of the box

V = s³ = s². t

To get the maximum volume, we can derive the volume equation above

V'(s) = 0

S² = 170.67 cm²

S = 13.07 cm

Thus, the length s required for the maximum volume is 13.07 cm.

The derivative is a calculation of changes in function values due to changes in input values (variables).

There are several kinds of derivatives, namely algebraic derivatives, root derivatives, partial derivatives, implicit derivatives, and others.

That's the discussion about inheritance. Hopefully it can help you in learning about derivatives. Thank You.

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