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

[Jayden1]

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

 

[Plato2]

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 $$2x+h+3.$$

 

[Jayden1]

Can you please show me the steps you took in simplifying to I know what I did wrong.

 

[Jameson]

Can you please show me the steps you took in simplifying to I know what I did wrong.

\(\displaystyle \frac{(x+h)^2+3(x+h)+6-(x^2+3x+6)}{h}\)

\(\displaystyle \frac{x^2+2xh+h^2+3x+3h+6-x^2-3x-6}{h}\)

\(\displaystyle \frac{2xh+h^2+3h}{h}\)

 

[Jayden1]

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.

 

[Sudharaka]

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.

 

[Jayden1]

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 -.-

 

[Jameson]

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".

 

[Jayden1]

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?

 

[Plato2]

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?

 

[Jameson]

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. :)

 

[Jayden1]

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...

 

[Jameson]

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 \(\displaystyle f(x)=ax^2+b\) 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!