# Prove: if $A(x)$ is divisible by $(x-a)^m$, its derivative is divisible by $(x - a)^{m-1}$

I have this question in my textbook:

If the polynomial $A(x)$ is divisible by $(x - a)^m$, then it's derivative is divisible by $(x - a)^{m - 1}$. Prove this.

I have really no clue on how to tackle this question. I probably have to use the formula $D(x^n) = nx^{n-1}$, but I have no clue were to start applying it.

Any tips would be greatly appreciated.

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Hints: First, $(PQ)'=P'Q+PQ'$. Second, the derivative of $(X-a)^m$ is $m(X-a)^{m-1}$.

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The first tip was what I needed. Do you write the answer or do I do? – user21385 Jun 14 '12 at 12:32
Personally, I think it's fine for this sort of question to leave the answer as is. If anyone else comes here wondering the same thing, the hint should be sufficient. – Harald Hanche-Olsen Jun 14 '12 at 12:37

Hint $\rm\,\ f = (x\!-\!a)^m\:\Rightarrow\: f'\,|\:f,\:$ thus we can pull a factor of $\rm\:\color{green}{f'}$ out of the product rule

$$\rm\quad \color{#C00}{f'\,|\:f}\:\Rightarrow\:f'\,|\:(fg)' =\, f'g+fg' =\, \color{green}{f'}\,(g+(\color{#C00}{f/f'})g')\ \$$

Analogous remarks hold true for similar product rules, e.g. the difference product rule

$$\rm \Delta(fg)\, =\, f\, \Delta g + g\, \Delta f + \Delta f\, \Delta g$$

which has a factor of $\rm\:\Delta f\:$ if $\rm\:\Delta f\:|\:f,\:$ and a factor of $\rm\:f\:$ if $\rm\:f\:|\:\Delta f.\:$ This comes in handy when manipulating products, e.g. factorials.

These properties prove fundamental in algorithms for indefinite integration and summation, since they reveal the effect of differentiation and differencing on multiplicity of factors, which allows us to derive simple effective bounds on multiplicity (necessary for effective algorithms).

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