How would I go about generalizing the product rule to the product of $n$ functions $\psi_1(x), \ \psi_2(x), ..., \ \psi_n(x)$? That is, I'm hoping to obtain an expression for
$$ \frac{d}{dx} \prod_{j = 1}^n \psi_j(x) $$
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$\begingroup$ Experimentation with $3$ and $4$ terms will give you the answer. A quick but not entirely justified alternative way is to let $y$ be the product, take the logarithm, and differentiate implicitly. $\endgroup$– André NicolasCommented Nov 2, 2013 at 21:02
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$\begingroup$ @AndréNicolas, your logarithm technique is one of those things that looks pretty until you take the lid off and find it's full of worms. $\endgroup$– dfeuerCommented Nov 2, 2013 at 21:36
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$\begingroup$ @deufer : Of course, Andre's technique gives the answer, then you can prove it's correct using induction. $\endgroup$– Stefan SmithCommented Nov 3, 2013 at 0:19
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2 Answers
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You have
$$\frac{d}{dx} \prod_{j=1}^n \psi_j(x) = \sum_{k=1}^n \left(\psi_k'(x)\prod_{\substack{j=1\\j \neq k}}^n \psi_j(x)\right).$$
If none of the $\psi_j$ has zeros, you can also write it in the form
$$\frac{d}{dx} \prod_{j=1}^n \psi_j(x) = \left(\sum_{k=1}^n \frac{\psi_k'(x)}{\psi_k(x)}\right)\prod_{j=1}^n \psi_j(x).$$
answered Nov 2, 2013 at 20:59
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$(fgh)'=f'gh+f(gh)'=f'gh+fgh'+fg'h$. Already we can see how this will go. Can you prove it by induction?
