[SOLVED] how do you find derivative equations?

[viet_jon]

I'm not sure if this is the right sub forum.

1. The problem statement, all variables and given/known data


f(x) = x^3 +6X^2 + 9x + 4


2. Relevant equations



3. The attempt at a solution

in my book, the first derrivative equation is 3x^2 + 12x + 9
and second equation as 6x + 12



how do you get to these equations?

I know how to calculate the difference........ I'm guessing somehow that's how you get it.


I'm not even sure if my question makes sense. I'm working ahead of class, so there's alot of gaps I need to fill in.

[rohanprabhu]

what you are actually looking at is 'differential calculus' and it isn't as simple as getting the difference or some such method. You should get a textbook on 'Differential Calculus'.

[Rainbow Child]

Do you know the derivative of g(x)=a\,x^n ?

[AngusYoung93]

The rule you should use is the power rule
f(x) = x^a
f'(x) = ax^(a-1)
The derivative of a constant is 0.

So the derivative of x^3 is 3x^2.
Also, the derivative of the sum is the sum of the derivative. So, you can take the derivative of each part of the equation separately.
So x^3 + 6x^2 becomes 3x^2 + 12x.

[viet_jon]

tnkx for the quick reply....


k....this is weird then......differential calculus? I'm taking a grade 12 class.

with the function in the first post.....I'm suppose to find coordinates and nature of the critical points. And in the example, it says differentiate ...... f'(x) = 3x^2 + 12x +9 = 0
without explaining how to I get that equation.

rainbow child...I have no clue what that means. That looks way ahead of me.


angus....give me a few moments with that.

[AngusYoung93]

A critical point is where the first or second derivative is 0 or undefined. In the first derivative, the critical points will tell you where the original equation changes from increasing to decreasing. While the second derivative will tell you where the equation changes concavity.

[viet_jon]

The rule you should use is the power rule
f(x) = x^a
f'(x) = ax^(a-1)
The derivative of a constant is 0.

So the derivative of x^3 is 3x^2.
Also, the derivative of the sum is the sum of the derivative. So, you can take the derivative of each part of the equation separately.
So x^3 + 6x^2 becomes 3x^2 + 12x.

thnkx....


i got it now.....

[Gib Z]

If you haven't heard of differential calculus before but are attempted this, i would advise you to stop straight away. Learn from the foundations, get a good textbook on Differential Calculus.

[viet_jon]

I've heard of it, but from this forum only.

Differential Calculus is probably way ahead of me though. I don't understand why my grade 12 workbook required me to do it....and even worse, not explaining how it's derived.

I browsed through my brother's first year Uni calculus book, I have a better understanding of it now. The stuff looks really interesting btw..... it would probably be more interesting if I understood it all.

[rohanprabhu]

I've heard of it, but from this forum only.

Differential Calculus is probably way ahead of me though. I don't understand why my grade 12 workbook required me to do it....and even worse, not explaining how it's derived.

I browsed through my brother's first year Uni calculus book, I have a better understanding of it now. The stuff looks really interesting btw..... it would probably be more interesting if I understood it all.

i just don't understand how differential calculus is 'way ahead' of you if you are in 12th grade. I had differential calculus from my 11th grades beginning. Which education board does your school follow?