y=(x^3)+(2x^2)
y'=?
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y=(x^3)+(2x^2)
y'=?
EDIT: I hope this is right...This is like...Calculus 200!
y=x^3 + 2x^2
dy/dx=3x^2 + 4x
Last edited by -akichan-; 02-27-2009 at 12:57 AM.
y'=(3x^2)+(4x) ?
I fail at primary education level mathematics so if there's some sort of catch don't hurt me.
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y = not =D
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y'= 3x^2+4x
I COULD have found the answer by using the definition of a derivative (like Real Man), but since the answer came to me almost instantaneously (I am studying Aerospace Engineering; they make us do derivatives all the time), there was no point in doing so. I can't even remember the last time I found a derivative by using the definition (I just use the laws and get it over with)!
Actually, why did I do this? I hate hearing the word "derivative", let alone do them. CALCULUS IS TAKING OVER MY BRAIN!!!
Last edited by wolfgirl90; 02-27-2009 at 09:43 AM.
This is my war face.
This is what happens to trolls who mess with me.
I learned to calculate in different ways, it shouldn't be something that makes me look dumb cos that's what I learned.
Btw Wio, no offense...but...it's hard to read your hand-writing...and you put that on as your profile picture xD The answer is the same, no matter in what ways. If you want me to show you, that's find with me.
Last edited by -akichan-; 02-27-2009 at 12:26 PM.
It has very important theoretical uses in higher levels of day to day mathematics. About the same time you start doing derivatives and integrals in your head without thinking about it, you know so much mathematics you can't possibly keep it all in your head, so you learn the basic definitions and derive what you need to know from that, when you need it.
If you know the definition, and how to use it, you can derive the rules for differentiation of whatever sort of function on the fly when you need it.
The best way of solving the problem is to prove the derivative of an arbitrary power of x, and then using the result to solve the problem. This is physicist math though. Showing it to a mathematician may lead to being drawn and quartered by angry mathematicians (they are just grumpy because they don't have a Nobel prize).
The second step is binomial expansion, and the dots are terms that are multiples of /\x^2 (and thus vanish as /\x ~ 0)
Last edited by Eris; 02-27-2009 at 01:21 PM.
Hey look, Japan made a movie about me!
You're partially right Eris, but you'd still have to prove that the derivative of a sum is equal to the derivative of it's addends, and that you can pull that constant 2 out of the 2x^2. It's kinda picky, but simply going through the definition avoids making sure you've proven all of the differentiation rules you might use.
Xero, all the rules were created to make differentiation quick, not to prove what the actual derivative is. It's just like how a delta epsilon proof can prove what the limit is, even through finding the limit can be as easy as substituting a number in.
Both those things are properties of the basic arithmetics that build the definition of the derivative, such as the distributive nature of multiplication.
Using the definition will lead to errors down the line, due to the sheer amount of terms it amasses fairly quickly. It is horribly ineffective to solve the same problem over and over again. Especially when you need to differentiate powers of 100 or greater. Merely writing down the binomial expansion will likely take several pages, and getting a single term wrong means your result is corrupted.
Hey look, Japan made a movie about me!
*Scoff*
I always hated mathematics.
I know what is needed in my life and that is all I need.
I do applaud those who can do this sort of thing, it boggles my mind too much to bother unfortunately.
Battalion 316
3x^2 + 4x.
Yay for AP Calculus!
I know ONE reason!
For the AP Calc exam. Haha.
They will ask you to find the derivative of something using the definition of the derivative. But even then, you can use the shortcuts, as long as you can spot the definition of a derivative. There's one teeny little reason right there.
I'm talking about after the mediocre calc1/AP calc classes.
But, yeah, I can definitely understand what Eric said. Though I was more refering to a problem such as the one Lavos provided, for which just seems excessive. Basic derivatives (and integrals) can be done in about 5 seconds in your head, with a bit of experience in doing them.
"The color fades along the intervals I follow."
True, I thought you might be referring to that...
So calculus in college doesn't make use of the definition at all? That's nice... I hope I'm not getting too off subject here, but is advanced (like advanced advanced) calc difficult? *is trying to decide on a major atm*
Well, you do use it to prove other things from time to time. It's of theoretical value, but largely not very practical to work with in day to day calculations.
Whether calculus is difficult is highly individual. If you find mathematics simple, it will probably remain that way most of college. It helps a lot to have an inflated sense of self-esteem when studying any subject. I can't stress this enough. It's 10% intelligence and 90% self-esteem.
Hey look, Japan made a movie about me!
First: I don't know what a derivative is.
Second: How am I supposed to even do that equation?
Third: Love your ava there, Lavos!
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A derivative is the rate of change of a function. Say you have a function x^2, and then you increment it ever so slightly, to (x+h)^2 ... how much has it changed?
Well, (x+h)^2 - x^2 is the change between x and x+h... but we want how steep it is, so we're after (the change in (x+h)^2 - x^2) / (the change between x and x+h = h), which is
((x+h)^2 - x^2) / h
If we make h really tiny, we have found the derivative
In the example with x^2, we have that
(x+h)^2 = h^2 + 2*x*h + x^2
So (x+h)^2 - x^2 = 2*x*h + h^2
Putting it back in the derivative, we need to divide it by h, so you get 2*x + h, but h is really tiny, so we can drop the h term and get that the derivative is 2*x.
Hey look, Japan made a movie about me!
Wow Eris.
Wow.
That explanation was epic.
Btw, I don't really understand what you said before about having 90% self-esteem when studying a subject. If you had too much self-esteem, wouldn't you be overconfident of yourself and fail at the subject? (I'm sure you meant the 90% thing in another way though...)
(3x^2)+4x
I'm only in Pre-Calc and I got the answer to that as soon as I saw it.
What is all this discussion even for?
I never play to win. But I do play not to lose.
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