The 58th derivative of the function (1+x^3)^(30) evaluated at x=0 is definitively zero. This conclusion is reached through the application of the binomial series expansion and the Maclaurin series. The relevant terms in the expansion yield only powers of x that are multiples of 3, and since 58 is not a multiple of 3, the derivative at that order results in zero. The calculations confirm that no non-zero terms contribute to the 58th derivative.
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freshman2013 said:Homework Statement
find the 58th derivative of (1+x^3)^(30) at zero (The Answer is zero)
Homework Equations
The binomial series expansion
the maclaurin series (f^(n)(0)/n!)x^n
The Attempt at a Solution
So I expanded the given equation into the binomial series and at at the 19th term, I got 30(30-1)...(30-19+1)/57! * x^(57). But I don't get how that gets me any closer to the answer of zero.