Why Does the Derivative Change for Different Values of X?

• racer
In summary, the derivative function gives the instantaneous rate of change at a particular point, while the slope formula gives the average rate of change between two points. This is an important distinction to make when working with derivatives and understanding their applications.

racer

Hello guys

I've been recently thinking about derivative and it's applications.

Let's say Y=3x

When X = 1, Y=3
wHEN X = 2, Y = 6
When X = 3, Y = 9

Dy/Dx is the rate of change in Y with respect X, it is 6-3/2-1 =3/1 = 3
and the derivative is 3.

Let's Y = (X)^2

When X = 1, Y=1
When X = 2, Y = 4
When X = 3 , Y=9
When X = 4 , Y = 16

Dy/dx = 4-1/2-1 = 3/1 = 3
Dy/dx = 9-4/3-2 =5/1 = 5
Dy Dx = 16-9/4-3=7/1 = 7

So the rate of Change is not constant.

When you derive you get 2x

When you plug values in the function, you get different values from the the rate of change, Where is my mistake?

Thanks.

The function is different than its derivative. Of course, the values at any given x can be different.

Hi racer!

You have to choose the half-way value …

3 5 and 7 correspond to 1,2 2,3 and 3,4.

half-way is 1.5 2.5 and 3.5, and 2x of that is 3 5 and 7 !

Happier?

You should expect dy/dx to be different than (y(x2)-y(x1)) / (x2-x1). The former is exactly the rate of change, while the latter is just an approximation to the rate of change.

Much what the others are saying but:

[f(x1)- f(x0)]/(x1- x0) is the average rate of change of f between x0 and x1. The derivative of f (at a point x= a) is the instantaneous rate of change at x= a. If a function is linear (as your example f(x)= 3x) then the average rate of change over interval, starting at any point is a constant (the slope of the line) so the two are exactly the same. If a function is not linear (as f(x)= x2), the average rate of change depends upon both the starting point and the interval and the "average rate of change" and derivative are different.

You have to choose the half-way value …

3 5 and 7 correspond to 1,2 2,3 and 3,4.

half-way is 1.5 2.5 and 3.5, and 2x of that is 3 5 and 7 !

Happier?

That was amazing, I'll see if this applies to other functions.

Much what the others are saying but:

[f(x1)- f(x0)]/(x1- x0) is the average rate of change of f between x0 and x1. The derivative of f (at a point x= a) is the instantaneous rate of change at x= a. If a function is linear (as your example f(x)= 3x) then the average rate of change over interval, starting at any point is a constant (the slope of the line) so the two are exactly the same. If a function is not linear (as f(x)= x2), the average rate of change depends upon both the starting point and the interval and the "average rate of change" and derivative are different

Is there a realistic proof or example that contains releastic things?

It is logical, I would like to see a realistic example and I know that it is hard to find a realistic example that proves derivatives but there has to be an example proves derivatives realistically specially of functions that are second degree and above.

You should expect dy/dx to be different than (y(x2)-y(x1)) / (x2-x1). The former is exactly the rate of change, while the latter is just an approximation to the rate of change
The function is different than its derivative. Of course, the values at any given x can be different.

Yeah, I made a mistake because I read somewhere that the tan of the angle between the line that a function makes and the x Axis is the first derivative.

Thanks guys.

racer said:
That was amazing, I'll see if this applies to other functions.

The slope formula will give you the rate of change between two points.

The derivative function will give you the rate of change at one particular point.

That's a big difference! Hopefully you will take some time to think that through.

So say you have any function f(x). Pick two points $$(x_1, y_1)$$ and $$(x_2, y_2)$$

Now from what I have said above, you should know tha the slope formula will allow you to calculate the rate of change between these two points. Where as the derivative, f'(x), give you the rate of change at one particular point.

One interesting result, if you know the rate of change between two points (call it m), you are not guaranteed that the rate of change at either of those points is m! However, you are guaranteed that there is a point between those two points where the rate of change is m! In the example of f(x) = x^2, it just so happens that the point where this occurs is half way between any two points. For other functions it probably won't be exactly halfway.

Last edited:

1. What is a derivative?

A derivative is a mathematical concept that represents the rate of change or slope of a function at a specific point. It is calculated by finding the limit of the ratio of change in the function's output to the change in its input as the change in input approaches zero.

2. Why are derivatives important?

Derivatives are important because they allow us to analyze the behavior of functions and make predictions about their values. They are used in many areas of mathematics, science, and engineering, including calculus, physics, economics, and statistics.

3. What are some real-world applications of derivatives?

Some common real-world applications of derivatives include finding maximum and minimum values, calculating rates of change in physics and engineering problems, and predicting the behavior of stock prices in finance.

4. How do I find the derivative of a function?

The process of finding a derivative is called differentiation. It involves using specific rules and formulas to calculate the rate of change at a given point. The most common method is using the power rule, where the derivative of a power function is found by multiplying the coefficient by the exponent and subtracting one from the exponent.

5. Can derivatives be negative?

Yes, derivatives can be negative. This means that the function is decreasing at that point. A negative derivative can also indicate a local maximum, where the function is increasing before and after that point, but decreasing at that specific point.