Explore the Derivation of f'(x) from f(x) = x^2 + 3x + 6

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Discussion Overview

The discussion revolves around the derivation of the derivative of the function f(x) = x^2 + 3x + 6. Participants explore the steps involved in calculating the derivative using the limit definition, addressing simplifications, and clarifying misunderstandings related to the process. The conversation includes aspects of mathematical reasoning and conceptual clarification.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents the function f(x) = x^2 + 3x + 6 and attempts to derive its derivative, initially simplifying the expression incorrectly.
  • Another participant corrects the simplification, suggesting it should be 2x + 3 instead of 2x + h^2 + 3h.
  • Several participants request clarification on the steps taken in the simplification process, indicating confusion about handling the variable h.
  • There is a discussion about the necessity of canceling h to avoid division by zero, with some participants expressing uncertainty about the correct approach.
  • A participant questions the expected value of the derivative at x = 0, suggesting it should be 0, while others argue that it can be different based on the function.
  • Another participant explains that the gradient of a quadratic function at its vertex is 0, but acknowledges that the gradient at x = 0 for the given function is 3.
  • Participants discuss the implications of the derivative's value at specific points and whether it aligns with their expectations based on the function's characteristics.

Areas of Agreement / Disagreement

There is no consensus on the simplification steps, as participants express differing views on the correct form of the derivative. Additionally, there is disagreement regarding the expected gradient of the tangent line at x = 0, with some participants asserting it should be 0 while others maintain it can be 3.

Contextual Notes

Participants express uncertainty about the simplification process and the handling of the variable h, indicating potential limitations in their understanding of the order of operations and factoring techniques.

Who May Find This Useful

Students learning about derivatives, individuals seeking clarification on the limit definition of a derivative, and those interested in exploring common pitfalls in calculus derivations.

Jayden1
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So I started with this function: f(x) = x^2 + 3x + 6

I then got this from it:

m = ((((x + h)^2) + (3*(x + h)) + 6) - (x^2 + 3x + 6)) / ((x + h) - x)

Which I eventually simplified to 2x + h^2 + 3h

And then: limit h -> 0: 2x + h^2 + 3h = 2x

So f ' (x) = 2x

Is that right? By doing this, I ended up getting a tangent line gradient of 19 (for x value of 8).

The reason I keep thinking this is wrong is because I thought f(x) = x^2 also ended up with f ' (x) = 2x
 
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Jayden said:
So I started with this function: f(x) = x^2 + 3x + 6
I then got this from it:
m = ((((x + h)^2) + (3*(x + h)) + 6) - (x^2 + 3x + 6)) / ((x + h) - x)
Which I eventually simplified to 2x + h^2 + 3h WRONG
It should have simplified to 2x+h+3.
 
Can you please show me the steps you took in simplifying to I know what I did wrong.
 
Jayden said:
Can you please show me the steps you took in simplifying to I know what I did wrong.

[math]\frac{(x+h)^2+3(x+h)+6-(x^2+3x+6)}{h}[/math]

[math]\frac{x^2+2xh+h^2+3x+3h+6-x^2-3x-6}{h}[/math]

[math]\frac{2xh+h^2+3h}{h}[/math]
 
What h are you supposed to get rid of at this point?

We need to get rid of h because you can't divide by 0. It's just unclear which one of the numerator h's should be divided out...

Or are you supposed to divide every part by h. So you get 2x + h + 3?

is h^2 / h = h?

EDIT: I get it now. Just blame my terrible fractions skills.
 
Jayden said:
Can you please show me the steps you took in simplifying to I know what I did wrong.

Hi Jayden, :)

\[m=\frac{((x + h)^2 + 3(x + h) + 6) - (x^2 + 3x + 6)}{(x + h) - x}\]

\[\Rightarrow m=\frac{x^2+2hx+h^2+3x+3h+6-x^2-3x-6}{h}\]

\[\Rightarrow m=\frac{2hx+h^2+3h}{h}\]

\[\Rightarrow m=\frac{h(2x+3+h)}{h}\]

\[\therefore m=2x+3+h\]

By the way Jayden, as I http://www.mathhelpboards.com/threads/1180-Help-with-limits-tangent-line?p=5743&viewfull=1#post5743, it would be better if you can refresh your knowledge about the order of operations. :)

Kind Regards,
Sudharaka.
 
Sudharaka said:
Hi Jayden, :)

\[m=\frac{((x + h)^2 + 3(x + h) + 6) - (x^2 + 3x + 6)}{(x + h) - x}\]

\[\Rightarrow m=\frac{x^2+2hx+h^2+3x+3h+6-x^2-3x-6}{h}\]

\[\Rightarrow m=\frac{2hx+h^2+3h}{h}\]

\[\Rightarrow m=\frac{h(2x+3+h)}{h}\]

\[\therefore m=2x+3+h\]

By the way Jayden, as I http://www.mathhelpboards.com/threads/1180-Help-with-limits-tangent-line?p=5743&viewfull=1#post5743, it would be better if you can refresh your knowledge about the order of operations. :)

Kind Regards,
Sudharaka.

Saved my life with that. It's not that I don't know the order. It's that I forget things like what you just did with factoring out h. I always forget these things and it makes me get them wrong -.-
 
Jayden said:
Saved my life with that. It's not that I don't know the order. It's that I forget things like what you just did with factoring out h. I always forget these things and it makes me get them wrong -.-

Computing derivatives like this always depends on factoring out an "h" at the end and canceling it with an "h" in the denominator, otherwise you divide by 0 and it won't work. So with these problems always be looking for a way to factor out and cancel an "h".
 
1 question. If you end up with 2x + h + 3.

Then limit -> 0 to get 2x + 3...

f ' (0) = 2*0 + 3

f ' (0) = 3.

Shouldn't the tangent line to x = 0 have a gradient of 0 not 3?
 
  • #10
Jayden said:
f ' (0) = 3.
Shouldn't the tangent line to x = 0 have a gradient of 0 not 3?
Why in the world would you say that?
Why should the tangent at x=0 be any particular value?
 
  • #11
Jayden said:
1 question. If you end up with 2x + h + 3.

Then limit -> 0 to get 2x + 3...

f ' (0) = 2*0 + 3

f ' (0) = 3.

Shouldn't the tangent line to x = 0 have a gradient of 0 not 3?

It doesn't always have to be 0. You'll find tons of examples where it's not 0 so don't worry. In this case it should be 3 at x=0 so everything is good. :)
 
  • #12
lets say we plot the point a on (0, f(0)) (the function being f(x) = x^2 + 3x + 6)

now the derivative of f(x) = x^2 + 3x + 6 is apparently f ' (x) = 2x + 3

f ' (a) = 2*a + 3
f ' (a) = 2*0 + 3
f ' (a) = 3

Is there a different method of getting the derivative with functions like x^2 and f(x) = x^2 + 3x + 6 ?

EDIT: Awesome. Thanks Jameson. I just asumed that the turning point of a quadratic would always have a tangent line with a gradient of 0...
 
  • #13
On a quadratic graph, the gradient will be 0 at the vertex which is also where the graph changes directions. If you have a quadratic in the form of [math]f(x)=ax^2+b[/math] then the gradient at x=0 is 0 but your equation is different. I think the original question is answered so keep the questions coming if you want and make a new thread!
 

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