Can You Help Verify This Calculus Derivative Solution?

  • Thread starter Jake Minneman
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In summary, the conversation is about a 14-year-old trying to teach themselves calculus and asking for help with a derivative. A summary of the conversation includes the person's thought process, a correction of a typo, a suggestion to use LaTeX formatting, and a final comment about the helpfulness of the conversation.
  • #1
Jake Minneman
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Hello, I am fourteen and "attempting" to teach myself calculus I was wondering if anyone with prior calculus knowledge would be willing to solve this derivative in order to check my answer.d
---- (x^3-x^2)+(3x)
dx
(My thought process)

d-----------------d
=--- (x^3-x^2)+---(3x)
dx---------------dx

=3x^(3-1)-2x^(3-1)+3x

=3x^2-2x+3

=3x^2-2x+3Sorry if I screwed that up.
 
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  • #2
I wouldn't worry too much about embarrassing yourself. Everyone here had to learn this stuff at some time and probably made the same kinds of mistakes as everybody else.

You can use www.wolframalpha.com to find derivatives. Just enter "derivative of (x^3-x^2)+(3x)".
 
  • #3
Thanks for the help
 
  • #4
You have a couple of typos in your work, but your method and final answer are correct.
 
  • #5
Jake Minneman said:
Thanks for the help
Yes I got it
 
  • #6
Jake Minneman said:
Hello, I am fourteen and "attempting" to teach myself calculus I was wondering if anyone with prior calculus knowledge would be willing to solve this derivative in order to check my answer.


d
---- (x^3-x^2)+(3x)
dx
(My thought process)

d-----------------d
=--- (x^3-x^2)+---(3x)
dx---------------dx

=3x^(3-1)-2x^(3-1)+3x
In the line above, it should be
=3x^(3-1)-2x^(2-1)+3[STRIKE]x[/STRIKE]
Jake Minneman said:
=3x^2-2x+3

=3x^2-2x+3


Sorry if I screwed that up.
 
  • #7
Mark44 said:
In the line above, it should be
=3x^(3-1)-2x^(2-1)+3[STRIKE]x[/STRIKE]

Thank you the typing of the question has been giving me trouble.
 
  • #8
You might want to look into how to format things using LaTeX. You can click any of the lines below to see what I did.
[tex]\frac{d}{dx}(x^3 - x^2 + 3x)[/tex]
[tex]=\frac{d}{dx}x^3 - \frac{d}{dx}x^2 + \frac{d}{dx}3x[/tex]
[tex]=3x^{3 -1} - 2x^{2 -1} + 3x^{1 - 1}[/tex]
[tex]=3x^2 - 2x + 3[/tex]
 
  • #9
Mark44 said:
You might want to look into how to format things using LaTeX. You can click any of the lines below to see what I did.
[tex]\frac{d}{dx}(x^3 - x^2 + 3x)[/tex]
[tex]=\frac{d}{dx}x^3 - \frac{d}{dx}x^2 + \frac{d}{dx}3x[/tex]
[tex]=3x^{3 -1} - 2x^{2 -1} + 3x^{1 - 1}[/tex]
[tex]=3x^2 - 2x + 3[/tex]

This helped tremendously
 
  • #10
Glad to hear it!

There are a few LaTeX tags that I use a lot.
Integral - [ tex] \int f(x) dx [ /tex]
Definite integral - [ tex] \int_{a}^{b} f(x) dx [ /tex]
Fraction or rational expression - [ tex] \frac{x + 2}{x^2 - 4}[ /tex]
Limit - [ tex] \lim_{x \to 0} f(x) [ /tex]

If you remove the extra spaces I added in the tex tags, the expressions above render like this:

[tex] \int f(x) dx [/tex]
[tex] \int_{a}^{b} f(x) dx [/tex]
[tex] \frac{x + 2}{x^2 - 4}[/tex]
[tex] \lim_{x \to 0} f(x) [/tex]
 
  • #11
thats sweet
 

FAQ: Can You Help Verify This Calculus Derivative Solution?

1. What is a derivative in algebra?

A derivative in algebra is a mathematical tool used to find the rate of change of a function at a specific point. It measures how much the output of a function changes when the input changes by a small amount.

2. How do I solve an algebraic derivative?

To solve an algebraic derivative, you first need to identify the function and its independent variable. Then, you can use the rules of differentiation, such as the power rule, product rule, quotient rule, and chain rule, to find the derivative. Finally, evaluate the derivative at the given point to get the slope of the function at that point.

3. Why is finding derivatives important?

Finding derivatives is important because it allows us to analyze the behavior of functions and solve real-world problems involving rates of change. It is also a fundamental concept in calculus and is used in many fields such as physics, engineering, and economics.

4. Is there a shortcut to finding derivatives?

Yes, there are many shortcuts or rules that can be used to find derivatives quickly and efficiently. These rules, such as the power rule and chain rule, can help simplify the process of finding derivatives and make it easier to solve more complex problems.

5. Can derivatives be negative?

Yes, derivatives can be negative. The sign of a derivative depends on the behavior of the function at a specific point. A negative derivative indicates that the function is decreasing at that point, while a positive derivative indicates that the function is increasing at that point.

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