Derivative using first principle definition

In summary, the first principle definition of a derivative is a mathematical formula used to find the instantaneous rate of change of a function at a specific point, also known as the limit definition of a derivative. It is calculated by finding the limit of the difference quotient as the change in x approaches 0. This definition is significant in calculus as it serves as the foundation for finding the derivative of any function and has many real-world applications. While it can be used to find the derivative of any function, it is often reserved for more complex functions and provides a deeper understanding of the concept of a derivative.
  • #1
pbonnie
92
0

Homework Statement


Calculate the derivative of f(x) = x^3 - 3x^2


Homework Equations


(f(x+h) - f(x))/h


The Attempt at a Solution


Just wondering if someone can check my solution.
(f(x+h)- f(x))/h= ((x+h)^3- 3(x+h)^2- (x^3-3x^2))/h
= (x^3+ 3x^2 h+3xh^2+ h^3-3(x^2+ 2xh+ h^2 )- x^3+3x^2)/h
= (x^3+3x^2 h+3xh^2+h^3-3x^2-6xh-3h^2-x^2+3x^2)/h
= (x^3+3x^2 h+3xh^2+h^3-6xh-3h^2-x^2)/h
= x3 + 3x2 + 3xh + h2 – 6x – 3h –x2
= x3 + 2x2 + 3xh + h2 – 6x – 3h

lim┬(h→0)⁡〖x^3+ 2x^2- 6x〗

f’(x) = x^3+ 2x^2- 6x
 
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  • #2
pbonnie said:

Homework Statement


Calculate the derivative of f(x) = x^3 - 3x^2


Homework Equations


(f(x+h) - f(x))/h


The Attempt at a Solution


Just wondering if someone can check my solution.
(f(x+h)- f(x))/h= ((x+h)^3- 3(x+h)^2- (x^3-3x^2))/h
= (x^3+ 3x^2 h+3xh^2+ h^3-3(x^2+ 2xh+ h^2 )- x^3+3x^2)/h
= (x^3+3x^2 h+3xh^2+h^3-3x^2-6xh-3h^2-x^2+3x^2)/h
= (x^3+3x^2 h+3xh^2+h^3-6xh-3h^2-x^2)/h
= x3 + 3x2 + 3xh + h2 – 6x – 3h –x2
= x3 + 2x2 + 3xh + h2 – 6x – 3h

lim┬(h→0)⁡〖x^3+ 2x^2- 6x〗

f’(x) = x^3+ 2x^2- 6x

You are doing the right general thing but the algebra starts going badly after the third line. How come the x^3 and the 3x^2 didn't just cancel? Then I completely start losing track.
 
  • #3
I thought they could only cancel out if they were of the same degree?
 
  • #4
It's 3x^2h so I think I can't cancel it out without a negative 3x^2h?
 
  • #5
Sorry, I copy and pasted it from word and didn't edit it, I see now what you mean
 
  • #6
The third line should read:
(x^3 + 3x^2h + 3xh^2 + h^3 - 3x^2 - 6xh - 3h^2 - x^2 + 3x^2)/h
 
  • #7
pbonnie said:
The third line should read:
(x^3 + 3x^2h + 3xh^2 + h^3 - 3x^2 - 6xh - 3h^2 - x^2 + 3x^2)/h
There's still at least one error/typo there .

Should be :

(x^3 + 3x^2h + 3xh^2 + h^3 - 3x^2 - 6xh - 3h^2 - x^3 + 3x^2)/h

Right ?
 
  • #8
Oh! Typo. So f(x+h) - f(x) should be 3x^2 + 3xh + h^2 - 6x - 3h
And the limit would be
3x^2 - 6x
?
 
  • #9
pbonnie said:
Oh! Typo. So f(x+h) - f(x) should be 3x^2 + 3xh + h^2 - 6x - 3h
And the limit would be
3x^2 - 6x
?

That's really (f(x+h)-f(x))/h. But yes, that's the right limit.
 
  • #10
Great, thank you both :)
 

Related to Derivative using first principle definition

1. What is the first principle definition of a derivative?

The first principle definition of a derivative is the mathematical formula used to find the instantaneous rate of change of a function at a specific point. It is also known as the limit definition of a derivative.

2. How is the first principle definition of a derivative calculated?

The first principle definition of a derivative is calculated by finding the limit of the difference quotient as the change in x approaches 0. This is represented by the formula f'(x) = lim (h → 0) (f(x+h) - f(x))/h.

3. What is the significance of the first principle definition in calculus?

The first principle definition is significant in calculus because it is the foundation for finding the derivative of any function. It allows us to calculate the instantaneous rate of change of a function at any point, which is essential in many real-world applications.

4. Can the first principle definition be used to find the derivative of any function?

Yes, the first principle definition can be used to find the derivative of any function. However, it can be a time-consuming process and is often reserved for finding the derivatives of more complex functions that cannot be easily solved using other methods.

5. What are the advantages of using the first principle definition to find derivatives?

Using the first principle definition allows us to find the derivative of any function, even those that cannot be solved using other methods. It also provides a deeper understanding of the concept of a derivative and can be used to prove other derivative rules and formulas.

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