Finding the Derivative: Find (h^(10))(2), or the 10th derivative when x=2. h(x)=1=2(x-2)+6(x-2)^2+...+(n+1)!(x-2)^n+...

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Finding the Derivative: Find (h^(10))(2), or the 10th derivative when x=2. h(x)=1=2(x-2)+6(x-2)^2+...+(n+1)!(x-2)^n+...

Mathematics
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the only part we need consider is the that with\[n\ge10\]because all other terms vanish. That would be\[11!(x-2)^{10}+12!(x-2)^{11}+\dots+(n+1)!(x-2)^n\]after the tenth derivative this will be\[11!\cdot10!(x-2)^0+12!\cdot11!(x-2)^1+\dots\]\[\dots+(n+1)!\cdot\frac{n!}{(n-10)!}(x-2)^{n-10}\]all the terms after the zero exponent are junk and vanish when x=2, so we are left with\[f^{10}(2)=11!\cdot10!\]
but what about the zero exponent? if 2 is plugged into the second to last step, or11!⋅10!(x−2)^0, it will become 11!10!(0)^0. Does the 0^0 just cancel out?
that is a very good question...

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wolfram considers it defined http://www.wolframalpha.com/input/?i=d%5E10%2Fdx%5E10%5B%28x-2%29%5E10%5D+at+x%3D2 but I have no real insight on it :/

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