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Prove that for any polynomial $P(x)= a_nx^n + \cdots +a_1x+a_0,P$ is differentiable, and $P'(x) = na_nx^{n-1}+\cdots+2a_2x+a_1.$

I am trying to figure out a way to prove this with out having to use induction. I feel like I should be able to utilize the fact that an additive combination of derivatives is differentiable at a point. Maybe stating that as a lemma and then proceeding with a proof but I am not exactly positive what that would look like or how to start.

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  • $\begingroup$ To prove that an additive combination of derivatives is differentiable at a point, you would need to use induction. The same with an integer power of $x$ $\endgroup$ – Omnomnomnom Nov 9 '15 at 21:26
  • $\begingroup$ If I am aloud to assume that power functions are differentiable, and I already have a written proof for the additive combinations bit, then what should I do? (I am trying to avoid having to develop the induction proof for this question) $\endgroup$ – B ry Nov 9 '15 at 21:29
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Hint:

If you want prove this without indiction you can do two steps:

1) prove that the derivative of $x^n$ is $D(x^n)=nx^{n-1}$ . This can be done using the binomial expansion formula in the definition of derivative. ( note that this formula can be proved with combinatorial arguments, without induction).

2) proof that the derivative operator $D$ is a linear operator, i.e. $ aD(f(x)) = D(af(x))$ and $D(f(x)+g(x))=D(f(x))+D(g(x))$, so that: $$ D \left(\sum_{i=1}^n f_i(x)\right)=\sum_{i=1}^nD(f_i(x)) $$

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