Calculus Taylor Series?
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The function f(x)=Sigma of((f^n)(c)/n!)*((x-c)^n) is what you start with.  You can replace the Sigma(sum) with an integral.  Derivatives are the basically the opposite of Integrals.  If you derive the integral, you end up back at the start: ((f^n)(c)/n!)*((x-c)^n).  It might be easier to use a graph to see what happens. \n\nBut basically, intervals are merely used to look at a part of whatever is on the graph.  It doesn't matter what you do to that part, you are still looking at that part.  That's why the intervals don't change.\n\nDoes that help you or confuse you?
176564
http://www.math.com/\nhttp://mathworld.wolfram.com/\nhttp://www.calculus-help.com/\nhttp://math.about.com/od/calculus/
375277
Mathworld.com.  Or, if you type the subject your interested in followed by "Forum" in google, you'll find some good help that way.  Fore xample, "calculus forum".  Also, you can ask me whatever you want.
787465
Write out the exponentials and use the series for e^x. Other than that, do your own homework.
970213
Use the Chain Rule repeatedly:\n\nOriginal function: y = (tan(sin x))^2\n\ndy/dx = 2 tan(sin x) (d/dx) tan(sin x)\n\n= 2 tan(sin x) (sec(sin x))^2 (d/dx) sin x\n\n= 2 tan(sin x) (sec(sin x))^2 cos x, QED
365360
Use the ratio test. We want the absolute value of the ratio as k approaches infinity of the (k+1)th term to the kth term to be less than 1. So take the expression x^n/n^(1/5) and let n = k + 1. Stick the result in the numerator. Then let n = k and stick that result in the denominator. Now take the limit as k goes to infinity. You should just end up with x (I don't want to write the whole process out here cuz it's hard to write equations but you should get something that looks like x[k/(k+1)]^(1/5) which approaches x as k -> infinity). So if |x| < 1, you're good to go. If x is plus or minus 1, we're still not sure if the series converges or not. The x = 1 case gives us a p-series with p = 1/5, which diverges. The x = -1 case gives us an alternating series in which the magnitude of each term is always decreasing and approaches zero as n goes to infinity. Thus, it passes the alternating series test for convergence and the series converges for x = -1.\n\nSo the interval of convergence is [-1,1) and the radius of convergence is 1.
438371
same thing as (8x^2+2x-2)^(3/2), right?\nso add one to the power\nmultiply by the inverse of that power\nand divide by the derivative of what's in the parenthesis on the left\ngot it?
748672
using Maclaurin series you find the series for sin and cos and then use long division for the Maclaurin series of tan
750604
As Stuart has said above, you need to use the product rule to work this out.  The product rule states:\n\nthe derivative of f(x)g(x) is f(x)g'(x) + f'(x)g(x)\n\nSo, when you differentiate sinx wrtx you get cos and when you differentiate 2x wrtx you get 2.  Plugging these into the formula above you get:\n\n(f(x)g(x))'=2x(cosx)+2(sinx)\n                =2xcosx + 2sinx\n\nHope that this is helpful!
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http://www.taylor.com
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Do you want how to derive them or what the derivatives are?\n\nF(x)=a^x  ---> f'(x) = a^x (ln a)\n\nf(x) = sin x ---> f'(x) = cos x\n         cos x      --->           -sin x\n         tan x       ---->           sec^2 (x)\n         sec x      ---->           sec x tan x\n         csc x      ---->           - csc x cot x\n         cot x       ---->            - csc^2 (x)
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let y = x^x\nw = ln y = xlnx\nthen y = e^w\ndy/dx = e^w * dw/dx\ndw/dx = x(1/x) + lnx = 1+lnx\ndy/dx = (1+lnx)*x^x
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Check out Stewart's Calculus(refer to the first link). If you are looking for something  more rigorous refer to Apostol's Calculus(refer to second link).
909632
well the series is only convergent if the partial sums approach zero as n approaches infinity\n\nn^2(e^(-n^3) = n^2/(e^(n^3)\n\nnow take  lim n->inf n^2/(e^(n^3)\n\ne is being rasied to a number that gets very large very quickly as opposed to n squared so this should approach zero\n\nto prove that we can just use L'Hopistal's rule once and see that this then becomes lim n->inf 2/(3ne^(n^3)) this is obviously zero\ntherefore since the partial somes become zero eventually.. the total series converges ... you continue to add zero to the total sum as n goes to infinity\n\nrusty on my calc two hopefuly that helps
554917
try Liethold's book...thats our txtbook in school for calculus.
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use derivative rule for exponential as\n\nd/dx(A exp(bx))=(A*b)*exp(bx)\n\nyou got hint. No more Spoon Feeding
919856
It's easier to sum x(15/27)^x (over x) for a general q instead of 15/27, and substitute q = 15/27 only at the end. \nI also change x to n (looks better).\nTwo methods: \n\n1. Method using calculus: \nFor each n, view nq^(n-1) as the derivative, w.r.t. q, of q^n. Note that the (infinite) sum of derivatives = derivative of sum (summation over n, differentiation wrt q). So, instead of summing the derivatives, we'll differentiate the sum (because the sum of q^n is easy). So, sum q^n over n from 0 to infty to get 1/(1-q) -1 (or, which is the same, q/(1-q)), differentiate this wrt q to get 1/(1-q)^2, and multiply by q (because we started from nq^(n-1) instead of nq^n).\n\nAns: q/(1-q)^2.\nFinally, sub q = 15/27 in this.\n\n2. If you want to avoid calculus, view nq^n as the sum of n equal terms (so now you have a double summation), change the order of the double summation, and you'll only have easy sums to do (the geometric series).