Scrolling Game Development Kit Forum
General => Off-Topic => Topic started by: Zorb Burger on 2006-01-20, 07:42:37 PM
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i be readin book o on calculus an it not make two muc sense. thin yall can hel me teh unerstan better??
thanks! ;)
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Maybe if we could actually understand what you're saying better...
(Not trying to be rude, but you're posts are either spam, unneeded insults, or just illegible.
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y don ya jus sit down lil boy, an quit poutin. the onyl troll her is u in ma topic. try to be stayin on topic, ya fool!! :D
edti: so wat yer sayin is dat ya could read ma post? yer jus a dern hypocrite! ;D
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i nee help wit somtin calld derivitives or howmevr ya spell it...
tink yawl cin help??
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I don't exactly believe that you're enrolled in a Calculus class/course. [Not meant to be offensive!]
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ho u be nown ya freakn foo!1 if i coud i wood be putin ya on ma hate lis!!!
i be good in math, im lik a god at math. prolly better din al dese low ifes on teh broad!!(I don min if yawl includ me in dat catagore of low lifes!1 DAYUM!!)
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god at math?
need help with derivatives?
now you're being a little contradictory. ;)
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god at mos subjects in math, dummy. why else i be axin fer help. common sense in freakin nerds dese days... maks a hill billy wit som sense, seem smarter den yawl. since yawl an't got any common sense.
an boy is dis board filld wit air headd yankees, f'ing low lives... :)
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Simply put, a derivative of a function is the rate of change (slope) of a function at any given point. For example, the function f(x) = x has a constant slope of 1 when graphing it (y = x). The notation of a derivative uses the Prime symbol ('). So if f(x) = x then f'(x) = 1.
It's rather simple for constant-slope functions, since the derivative is simply equal to the constant slope. For a variable-slope function, such as x^2, it's slightly more difficult. However, there are tricks to learn to make calculating the derivative of a function much easier than actually using lim(f(x)) as x-> infinity (or something like that. I forget exactly what it is). For example, using the function f(x) = x^2, the slope of the graph of f(x) at any point x is 2x, that is f'(x) = 2x.
In general: if f(x) = ax^b then f'(x) = abx^(b-1)
So for another example: f(x) = 2x^3 then f'(x) = 6x^2 and f''(x) = 12x (second derivative; That is, the derivative of the derivative function).
And of course, if f(x) = C, then f'(x) = 0
If, however, you get other functions, such as f(x) = sin(x) or f(x) = ln(x), then there are other rules you need to apply.
Specifically, here are a couple of general rules:
if f(x) = sin(x) then f'(x) = cos(x)
if f(x) = cos(x) then f'(x) = -sin(x)
if f(x) = ln(x) then f'(x) = 1/x
And there are product rules, quotient rules, and some other kind of rules but I forget the names. In words:
The derivative of a product is the derivative of the first times the second, plus the first times the derivative of the second.
Example: if f(x) = 2x * x^2 then f'(x) != 2 * 2x as you might initially think. Instead, it must be seperated into g(x) = 2x and h(x) = x^2 so:
f(x) = g(x) * h(x). So, applying the rule, we get
f'(x) = g'(x) * h(x) + g(x) * h'(x)
= 2 * x^2 + 2x * 2x
= 6x^2
The derivative of a quotient is the derivative of the top times the bottom, minus the top times the derivative of the bottom, all over the bottom squared.
Example:
f(x) = 2x/x^2
f'(x) = (2(x^2) - 2x(2x)) / x^4
= -2x^2 / x^4
= -2 / x^2
And the other one, which I don't remember, is: The derivative of a function (like sin(x^2)) is the natural derivative of the function (like cos(x^2)) times the derivative of the inside (2x). So if f(x) = sin(x^2) then f'(x) = cos(x^2) * 2x. Or if f(x) = ln(x^2) then f'(x) = 2/x (skipping simplification steps)
If you need more help, this website may be able to help. It can solve integrals, which are the opposite of derivatives (so if you put in cos(x^2) * 2x it will give you sin(x^2)) and it used to be able to solve derivativesm but I tried it this morning, and it didn't seem to be working. In any case, it can help: QuickMath (http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=calculus&s2=integrate&s3=basic)
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dern man yous got a lo of time on yer hands... ;)
but ya help me out an i be respectin dat... so i be takin back dem remarks... :)