The first derivative is a tool in differential calculus that allows us to understand the “rate of change” of a function at a particular point. It can be thought of as the speed of change of the function. If we look at a graph of a function, the first derivative at any point is the slope of the line tangent to the graph at that point.
A derivative is a function like any function and also from it which derives a derivative. When deriving a derivative from a derivative, the first is called the first derivative and the last is called the second derivative. The chain of functions and their symbolism is as follows:
- Original function: ‘f(x)'
- First derivative: ‘f'(x)'
- Second derivative: ‘f”(x)'
The Meaning of the Second Derivative
The second derivative gives expression to the trend of changes in the slope of the source function. When the second derivative is positive, the slopes of the source function are increasing. The rate of growth in the slopes is in accordance with the result obtained. If the result is 2, then the slope increases by 2 with each additional step.
What is the first derivative used for?
- Measure the pace of change
The first derivative tells us how much the function “rises” or “falls” at each point. If the derivative is positive, the function increases by a point. If it is negative, the function decreases.
- Find Maximum and Minimum Points
When the first derivative is reset at a certain point, i.e. its value is zero, it can imply that we have reached an extreme point – a maximum (peak) or a minimum (low) at that point. To know if this is a maximum point or a minimum, we will use the second derivative.
- Identify when a function is constant or changes at a constant rate
If the first derivative is equal to zero over a certain domain, it means that the function is constant there. If it is constant at a certain value (e.g. 5), it means that the function rises at a constant rate of 5.
- Variable Rate of Change – Quadratic Function
In the case of a quadratic function such as \( y = x^2 \), the rate of change (i.e., the gradient) depends on the value of \( x \): the larger \( x \), the greater the gradient. Such functions demonstrate a change in an increasing or decreasing rate.
We will use examples – an example accompanied by Figure 3.4
- Original function: ‘f(x)=x^2'
- First derivative: ‘f'(x)=2x'
- Second derivative: ‘f”(x)=2'
This means that when ‘f”(x)=2', the slope of the original function increases by 2 units per step.
- where x=0, the slope is 0.
- where x = 1, the slope is 2.
- where x = 2, the slope is 4.
- where x = 3, the slope is 6.
Even in the domain where the function decreases (negative gradient), the slopes increase by 2 steps per step. For example:
- where x = 3, the slope is -6.
- where x = -2, the slope is -4.
- where x = -1, the slope is -2.
- where x=0, the slope is 0.
Example with Diagram 3.5
- Original function: ‘f(x)=-x^2'
- First derivative: ‘f'(x)=-2x'
- Second derivative: ‘f”(x)=-2'
This means that the slopes of the original function are reduced by 2 per step.
The information obtained from the second derivative – expansion
When the second derivative is positive at any value of x, then the original function is convex at the same value as x. Also, when the second derivative is positive along the length of any segment, then the original function is convex at the same segment.
When the second derivative is negative at any value of x, then the original function is concave at the same value as x. Also, when the second derivative is negative along a segment, then the original function is concave in the same segment.
The Practical Use of the Second Derivative
background
With the help of the second derivative, we can determine, even without the use of diagrams, which types of curves a point with a slope of 0 belongs to. This is very important. If it belongs to a convex curve, the point should be a minimum point. If it belongs to a concave curve, the point should be a maximum point. If it is in the middle of 2 types of curves, it is an inflection point. The point where the slope 0 is called the 0 point.
Examination of Affiliation, and Conclusions
If the second derivative is positive at the 2 points near the 0 point on either side, the 0 point belongs to a convex curve and is therefore a minimum point. If the second derivative is negative at the 2 points near the 0 point, then the 0 point belongs to a concave curve and is therefore a maximum point. If the second derivative at one adjacent point is positive and the other is negative, the 0 point is between 2 types of curves and is therefore a twisting point.
Illustrative Examples
Only by means of examples can the contribution of the second derivative be gradually digested.
Example 1
- The example refers to an original function with the form ‘f(x)=x^2-6x'.
- Let's assume that we don't know its route.
- Using the first derivative, we can verify whether there are points in the path of the function where the slope = 0 and where (in what values of x are they located). The form of the derivative is: ‘f'(x)=2x-6'. A slope of 0 is obtained only when x=3. That is, where 0=2x-6 and the result is 3=x. and the meaning, where x=3 the slope of the function's path is 0.
- Since we are not familiar with the route of the function, we do not know whether it is a maximum point or a minimum point or a turning point.
- To ascertain which of the 3 options is correct, we need to find the second derivative. The second derivative form is ‘f”(x)=2'. From the result, we learn that the second derivative is positive (2) along the entire length of the original function, which means that the gradients in the original function increase by 2 with each step. It is clear from here that even at the 2 points near the zero point, the second derivative is positive and equal to 2.
Suppose the adjacent dots refer to x values of 2.9 and 3.1. In these two values, the second derivative is positive, and therefore the 0 point is a minimal extreme point.
Example 2
Original function: ‘f(x)=4+5x'
First derivative: ‘f'(x)=5'
In this example, there is no point in the function where the slope is 0. Of course, we know that this is a function that represents a straight line with a slope of 5 along its entire length and therefore does not have a point with a slope of 0.
Example 3
- Original function: ‘f(x)=-3x^2-2x'
- First derivative: ‘f'(x)=-6x-2'
- Second derivative: ‘f”(x)=-6'
From the result we learn that the second derivative is negative along the entire length of the original function, which means that the slopes in the original function decrease by 6 with each step. It is clear from here that even at the 2 points near the zero point, the second derivative is negative and hence it is amaximum extreme point.
**Additional Rules of Decree**
In many cases, the function we want to derive consists of different operations between several functions – such as multiplication, division, or assembly of functions. For these cases, there are special rules of sentencing, beyond the basic rule of presumption.
1. **The Rule of Multiplication**
When we have a function that is a product of two other functions, \( f(x) = u(x) \cdot v(x) \), we will use the product rule to derive it. The Rule of the Patriarchs says:
\( f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \)
That is, the derivative of the product of two functions is the sum of the two products: the derivative of the first function multiplied by the second function, plus the first function multiplied by the derivative of the second function.
**Dogma:**
Suppose that \( f(x) = x^2 \cdot \sin(x) \). To find \( f'(x) \), we'll use the product rule:
– \( u(x) = x^2 \) and \( v(x) = \sin(x) \) is selected.
– We will derive each function separately: \( u'(x) = 2x \) and \( v'(x) = \cos(x) \).
Commissioner according to the rule of the multiplier:
\( f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) = 2x \cdot \sin(x) + x^2 \cdot \cos(x) \)
2. **Dish Rule**
When we have a function that is a quotient of two functions, \( f(x) = \frac{u(x)}{v(x)} \), we will use the quotient rule to derive it. The rule of the dish says:
\( f'(x) = \frac{u'(x) \cdot v(x) – u(x) \cdot v'(x)}{(v(x))^2} \)
Here, the derivative of a quotient of two functions is the difference of the derivative multiples (similar to the product rule), the squared parts of the function in the denominator.
**Dogma:**
Suppose that \( f(x) = \frac{x^2}{\cos(x)} \). We will use the entire dish:
– Select \( u(x) = x^2 \) and \( v(x) = \cos(x) \).
– We will derive each function separately: \( u'(x) = 2x \) and \( v'(x) = -\sin(x) \).
We will place according to the rule of the dish:
\( f'(x) = \frac{2x \cdot \cos(x) – x^2 \cdot (-\sin(x))}{\cos^2(x)} = \frac{2x \cdot \cos(x) + x^2 \cdot \sin(x)}{\cos^2(x)} \)
3. **Chain Rule**
When we have a complex function that is an assembly of functions, i.e. \( f(x) = g(h(x)) \), we will use the chain rule to derive it. The chain rule says:
\( f'(x) = g'(h(x)) \cdot h'(x) \)
That is, we will derive the external function \( g \) according to the internal value \( h\), and then multiply by the derivative of the internal function.
**Dogma:**
Suppose that \( f(x) = \sin(x^2) \). To find \( f'(x) \), we'll use the chain rule:
– \( g(x) = \sin(x) \) and \( h(x) = x^2 \ are selected.
– We will derive each function separately: \( g'(x) = \cos(x) \) and \( h'(x) = 2x \).
We will place according to the chain rule:
\( f'(x) = g'(h(x)) \cdot h'(x) = \cos(x^2) \cdot 2x = 2x \cdot \cos(x^2) \)
