How do you find derivative equations?

[viet_jon]

[SOLVED] how do you find derivative equations?

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

Homework Statement

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

Homework Equations

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 a lot 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?