What is an Anti-Derivative?
481212
the area under the curve f(x)!
460400
You would need to know how to integrate (that is, find anti-derivatives). The proceedure forthis type of problem can usually be understood after first semester calculus. The specific integrations may need some second semester calculus.
751800
The 1st derivative is the slope of the original function.  If it is positive, you know the original function was rising (from left to right).  If it's positive but leveling off, the original still rising but not as much.  If it hits the x axis, there is a critical point on the original function (relative max/min or inflection point).
912927
1D\n2D (acceleration can be negative in which case you mihgt call it deceleration)\n3B
980090
Let g(x)=1/(1-2x)^2. An antiderivative (indefinite integral) of this is G(x)=1/2*1/(1-2x), as you may verify. Then, because you can write h(x)=1/(1-x) as the sum of the geometric series sum (k=0..infinity) of x^k if abs(x)<1, it is possible to write \nG(x) = 1/2*h(2x) = 1/2*sum(k=0..infinity) of (2x)^k if abs(2x)<1, that is, if abs(x)<1/2. \n\nBecause g(x)=G'(x), you need to differentiate the power series of G(x), and this can be done term by term inside the radius of convergence, and the radius of convergence of the differentiated power series will be the same as the radius of convergence of the original series (it is a theorem). \n\nThus g(x) = d/dx[1/2*sum(k=0..infinity) of (2x)^k] = 1/2*sum(k=0..infinity) of d/dx(2x)^k = 1/2*sum(k=0..infinity) of\nk*(2x)^(k-1)*2 = sum(k=0..infinity) of k*(2x)^(k-1) =\n= sum(k=1..infinity) of k*(2x)^(k-1) (you can omit the first term because it is equal to zero), if abs(x)<1/2.\n\nFinally, x^2/(1-2x)^2 = x^2*g(x) = x^2*sum(k=1..infinity) of k*(2x)^(k-1) = sum(k=1..infinity) of k*2^(k-1)x^(k+1) if abs(x)<1/2.\nThe radius of convergence is 1/2, because the radius of convergence r is the greatest number such that the series is convergent if abs(x)<r.
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A derivative is a function that provides the limit of the slope of a section of a primary function as the length of that section approaches zero.
224930
anti- tank or anti personell?? I can find out but give me a couple of days
38734
(2/3)x^3 + (5/2)x^2\nyou just take the antiderivitive of each term.  so you backtrack. instead of taking away one from the power, you add one and then divide the whole term by the new power. so the first term would be to the third power. now if you took the derivitive you would multiply by that power of three, so we need to cancel that out by dividing by three. so it's (2/3)x^3  because when you take the derivitive and that three comes down, it will be (3)(2)/(3) and the 3's will cancel out giving you 2x^2 which is what you started with. so the antiderivitive is (2/3)x^3.\n\nnow we do the second term. so we add a power, so it would be squared. now we have to cancel that power out so we'll divide the whole thing by two(or rather, multiply by a half) so the antiderivitive of 5x would be (5/2)x^2. this can be checked by finding the derivitive. multiply the (5/2) times 2 and you are left with 5 and take a power away to get 5x.  integrals are easier than you think, its just the opposite of differentiation. the general formula is:  X^n integrated is (X^n+1)/n+1 so if it's x^3 it would become x^4/4 or (1/4)x^4. again, you can check this by differentiating it. the exponent of 4 is multiplied and cancels out the fraction, and the exponent turns to a 3 and you're left with what you started with...\nhope you understand integrals now :)
549061
1/x^2 can be rexpressed as x^(-2). As you might know, the formula for a derivative of x^n is nx^(n-1). The antiderivative of x^n is (x^(n+1))/n+1. In this case, it would be x^-1/-1, or -x^-1. To check this, we take the derivative, which gives us (-1)(-1)x^-2, or x^-2. The answer checks out. \n\nSo, in summary, the antiderivative of x^-2 is -x^-1, with the remainder C.
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By definition of the derivative\ng'(x) = lim h=>0 [g(x + h) - g(x)] /h =  lim h=>0 [3xh + h^2]/h = \nlim h=>0 [3x + h] = 3x
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Use the second derivative test, not sure exactly what it is still but it should be easy to find on the internet.
1064323
The previous poster got it wrong. You need to take a derivative before you square.\n\nFor arc length, you integrate\nsqrt[1+(y')^2]\nbetween the x-values given.\nIn this case, the derivative is\n(1/2)x^2 -(1/2)x^(-2).\nIf you square this and add one, you will find the\nresult to be a perfect square of\n(1/2)x^2 +(1/2)x^(-2).\nSo you can take the square root and integrate the result from 1 to 3.\nThe anti-derivative is\n(1/6)x^3 - 1/(2x),\nso you evaluate this at 3 and subtract what you get when\nyou evaluate it at 1.
3246
So for the first derivative:\n\n4(x-1)\n-----------\n3x^(2/3)\n\nThe second derivative:\n\n4(x+2)\n-----------\n9x^(5/3)
535169
Matter and Anti-Matter
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If you mean this (e^x)^x, then that is e^(x^2).  There is no closed form expression for the antiderivative of this with respect to x.\n\nHowever if you mean e^(x^x), then I don't know.\n\nUPDATE: Hey discoganya, I'm getting d/dx(x^x) to be \n(x^x)(ln x + 1), not x^x.  Now (d/dx)e^x = e^x.\n\nUPDATE (again): I think that part of the problem with that, discoganya, is that you are assuming that x is a whole number.\n\nThis is how I got my dervative.\n\n(d/dx)(x^x) = (d/dx)(exp(xln x))\n= exp(xln x)(1*ln x + x(1/x))\n= (x^x)(ln x + 1), for x > 0\n\nI have run this through Mathematica, and it confirms it.\n\nI believe the only function such that f'(x) = f(x) is f(x)=  a*exp(x), where a is an arbitrary constant.\n\nBy the way, Mathematica also could not find the integral of e^{x^x), honey, so it also probably does not exist in a closed form.
507403
Behind all the nomenclature, this is just a simple minimization problem. You want to find local minima for A(q). Thus, take the derivative of A(q) to look for critical points (places where the graph is locally flat). These will be places where the tangent line has slope 0; so set the derivative equal to 0 and solve for q. You will get two possible choices. Use the second derivative to see what the behavior of the first derivative is at each point (increasing or decreasing). An increasing first derivative around a critical point (where the derivative is zero) implies that the slope locally changed from negative values to positive values. Visualize this to see that it is a minimum point for the graph of A(q). The q that gives you the positive second derivative and has 0 first derivative is thus q of the Wilson lot size formula.\nFor b, simply replace k with k + bq in A(q) and do a similar analysis as above.\n\nPS. The answerer above has forgotten the negative value of q that also satisfies the first equation. Although a negative q does not correspond to the problem, it is unrigorous to say that there is only one answer to the equation. It should at least be explained as being unphysical.
897617
This is an equation with separated variables. Rewrite as \n\ndu/u^3 = (t^3-t)dt \n\nand integrate (find antideritive) of each side separately.