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Let $f(x) = x^{2n} - x^{2n-1} + \cdots + x^4 - x^3 + x^2 - ax + b$

We wish to compute f'(1). The way I did this was to differentiate term by term, and then pair the terms of the form $-(2m + 1)x^{2m} + (2m) x^{2m-1}$ to get $-1$ summed $n-1$ times to get a $1-n$ term. Considering the first and last terms, we see that $2n$ and $-a$ contribute a term. So we get f'(1) = n+1-a

My book just states that f'(1) = n-1+2-a. They might have a quicker way than my reasoning, and I am not sure where the $+2$ comes from. Does anyone see how the book did it?

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    Or maybe $f(x)=(-ax +b)-(1-x)+[1-x+x^2-x^3+\cdots +x^{2n}]$. The sum in square brackets is $\frac{1+x^{2n+1}}{1+x}$, easy to differentiate.2011-10-06

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They are just pairing neighboring terms the opposite way from you, namely $(2k)1^{2k-1}-(2k-1)1^{2k-2}=2k-(2k-1)=1$ There are $n-1$ full pairs, and $2-a$ is the derivative of $x^2-ax$.

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If $f(x) = x^{2n} - x^{2n-1} + \cdots + x^4 - x^3 + x^2 - ax + b$, $f(x) = x^{2n} - x^{2n-1} + \cdots + x^4 - x^3 + x^2 - x + 1 + x - 1 - ax + b$, so $f(x) = \frac{x^{2n+1}+1}{x+1} + x - 1 - ax + b = \frac{x^{2n+1}+1}{x+1} + x(1-a) - 1 + b$.

Differentiating, f'(x) = \frac{(2n+1)x^{2n}(x+1)-(x^{2n+1}+1)}{(x+1)^2} + 1-a = \frac{2nx^{2n+1}+(2n+1)x^{2n}-1}{(x+1)^2} + 1-a.

Setting $x=1$, f'(1) = \frac{2n+(2n+1)-1}{4}+1-a =n+1-a.

At no extra charge, $f(1) = \frac{1+1}{1+1} + (1-a) - 1 + b = 1 - a + b$ (which you can see from cancellation, but it's nice to have a check).

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    Very good, but, in my opinion, more complicated than the others because of the need to invoke the quotient rule.2011-10-06