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$$f: \Bbb R \to \Bbb R; x \mapsto f(x)=x(x-1)(x-2)\cdots(x-10)$$ Evaluate $f'(0)$!

I've tried to set the factors apart, but I only know that $(fg)'=f'g+fg'$. I don't know how I should apply that rule for any $n$ amount of factors. I also thought of actually doing the multiplication, but I don't know what shortcut I should use, and multiplicating one after the other takes extremely long.

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    $\begingroup$ look at the edited text to see how to make your equations looks nice $\endgroup$
    – Surb
    Commented Sep 12, 2016 at 19:47
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    $\begingroup$ Do you want $f^{\prime}(0)$ factorial, or are you just very excited about $f^{\prime}(0)$? $\endgroup$ Commented Sep 12, 2016 at 19:47
  • $\begingroup$ @Surb thanks, I'll take a look at it! I think Jack gave me the perfect answer below, thanks! $\endgroup$
    – bp99
    Commented Sep 12, 2016 at 19:52
  • $\begingroup$ Another method: logarithmic differentiation. $\endgroup$
    – GEdgar
    Commented Sep 12, 2016 at 19:53
  • $\begingroup$ 10!! looks like a super awesome number $\endgroup$
    – cronos2
    Commented Sep 12, 2016 at 20:01

6 Answers 6

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Apply the definition: $$ f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}= \lim_{x\to0}\,(x-1)(x-2)\dots(x-10)=10! $$ More generally, if $f(x)=x(x-1)\dots(x-n)$, the same argument shows $$ f'(0)=(-1)^n\cdot n! $$

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    $\begingroup$ Interestingly, the definition of the derivative is useful at times! $\endgroup$
    – Wojowu
    Commented Sep 12, 2016 at 20:07
  • $\begingroup$ @Wojowu Who might have suspected it? ;-) $\endgroup$
    – egreg
    Commented Sep 12, 2016 at 20:12
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Let $g(x) = x$ and $h(x) = (x-1)\cdots (x-10)$. Then $f'(x) = g(x) h'(x) + g'(x)h(x)$. Since $g(0)=0$ and $g'(0)=1$, we have $$f'(0) = h(0) = (-1)(-2)\cdots (-10) = 10!$$

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  • $\begingroup$ Your answer was the most intuitive for me, thank you! It's the simplest IMO and I've only begun calculus :) $\endgroup$
    – bp99
    Commented Sep 12, 2016 at 20:19
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Hint: A polynomial is always an entire function, and in a neighbourhood of the origin: $$ x(x-1)\cdot\ldots\cdot(x-10) = 10!\,x+O(x^2) $$ hence $$ \frac{d}{dx}\left. x(x-1)\cdot\ldots\cdot(x-10)\right|_{x=0} = \color{red}{10!}. $$

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Welcome to Math SE. To format your question you can use LaTeX.

Now coming to your question, the derivative of a product is just the sum of the $n$ products where only one of the members is differentiated. In your case

$((x)... (x-10))' = [(x-1)...(x-10)]+[x(x-2)...(x-10)]+...+[x(x-1)...(x-9)]$

Note that all but the first expression will evaluate to $0$ at $x=0$ since you're multiplying by 0 ($x$) so $f'(0)=(-1)(-2)...(-10)=10!$

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  • $\begingroup$ Thanks for your answer! However, I don't fully understand this part, unfortunately: $((x)... (x-10))' = [(x-1)...(x-10)]+[x(x-2)...(x-10)]+...+[x(x-1)...(x-9)]$ How does this work exactly? $\endgroup$
    – bp99
    Commented Sep 12, 2016 at 20:06
  • $\begingroup$ @bertalanp99 Try it by hand with just three terms. Differentiate $(x-1)(x-2)(x-3)$ using the product rule (twice). The form should pop out. Your question just adds more terms in the product. $\endgroup$
    – John
    Commented Sep 12, 2016 at 20:14
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There is an $n$-term version of the product rule that is definitely worth knowing about. You'll see the pattern from the $n = 3$ case. If $f(x) = f_1(x) f_2(x) f_3(x)$, then \begin{equation} f'(x) = f_1'(x) f_2(x) f_3(x) + f_1(x) f_2'(x) f_3(x) + f_1(x) f_2(x) f_3'(x). \end{equation} (Do you see what the formula would be for a product of four functions?)

In your case, $f'(x)$ is a sum of $11$ terms, and all but one of those terms vanish when you plug in $x = 0$.

Richard Feynman made a big deal about the usefulness of this $n$-term product rule in The Feynman Tips on Physics.

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More generally, if $f(x) =\prod_{k=1}^n (x-a_k)^{b_k} $, then $\ln f(x) =\sum_{k=1}^n b_k \ln(x-a_k) $.

Differentiating, $\dfrac{f'(x)}{f(x)} =\sum_{k=1}^n \dfrac{b_k}{x-a_k} $, so $f'(x) =f(x)\sum_{k=1}^n \dfrac{b_k}{x-a_k} $.

Setting $x=0$,

$\begin{array}\\ f'(0) &=f(0)\sum_{k=1}^n \dfrac{b_k}{-a_k}\\ &=\prod_{j=1}^n (-a_j)^{b_j}\sum_{k=1}^n \dfrac{b_k}{-a_k}\\ &=\sum_{k=1}^n b_k(-a_k)^{b_k-1}\prod_{j=1, j \ne k}^n (-a_j)^{b_j}\\ \end{array} $

If $a_k = k-1$ and $b_k = 1$ as in your case,

$\begin{array}\\ f'(0) &=\sum_{k=1}^n \prod_{j=1, j \ne k}^n (-j+1)\\ &=(-1)^{n-1}\sum_{k=1}^n \prod_{j=1, j \ne k}^n (j-1)\\ &=(-1)^{n-1}\left(\prod_{j=2}^n (j-1)+\sum_{k=2}^n \prod_{j=1, j \ne k}^n (j-1)\right)\\ &=(-1)^{n-1}(n-1)! \qquad\text{since all terms with }k\ge 2 \text{ have j=1 so are zero}\\ \end{array} $

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  • $\begingroup$ Why would anyone downvote this answer must be a deep mistery. It is correct, it gives the correct answer in the particular case asked by the OP and also, for whoever is interested, it gives a nice though slightly abstract development to deal with these things. And anyone aspiring to cope with mathematics at any level above $\;2\cdot2=4\;$ cannot get scared of a little abstraction. +1 $\endgroup$
    – DonAntonio
    Commented Sep 12, 2016 at 20:28
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    $\begingroup$ @DonAntonio I'm not the downvoter. The answer glosses over two facts: the first is that taking the logarithm is not really possible (but it's a formal method, so nothing really bad if we know what we're doing). However, the computation is done assuming $a_k\ne0$; it can easily be adjusted, though, with doing a limit or some simpler trick. $\endgroup$
    – egreg
    Commented Sep 12, 2016 at 20:52
  • $\begingroup$ @egreg You're right, indeed. +1 $\endgroup$
    – DonAntonio
    Commented Sep 12, 2016 at 20:55
  • $\begingroup$ In my answer, I separated out the case $a_k = 0$ in the first term; this caused all the other terms to vanish at $x = 0$. $\endgroup$ Commented Sep 12, 2016 at 21:42

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