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Let $f_k:(a,b)\to \mathbb{R}$ be differentiable and non-zero $(f_k(x)\neq0$ for all $a<x<b),\ k=1,\cdots,n$

Find $\frac{d}{dx}f_1(x)f_2(x)\cdots f_n(x)$.

I know that the problem should be solved by using induction and $(f_1f_2)' = f_1'f_2 + f_1f_2'$. But how am I supposed to start?

Help appreciated!

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marked as duplicate by Henry T. Horton, Pedro Tamaroff, rschwieb, Arkamis, Alexander Gruber Mar 28 '13 at 19:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

This is the sort of problem logarithmic differentiation is intended for. – anon Mar 28 '13 at 19:02
Why did you double post? – Pedro Tamaroff Mar 28 '13 at 19:03

4 Answers 4

up vote 5 down vote accepted

$$(f_1 f_2 f_3)'=(f_1 f_2)' f_3 + f_1 f_2 f_3'=f_1' f_2 f_3 + f_1 f_2'f_3 + f_1 f_2 f_3'$$ Do you see the pattern now? Once you see it, applying induction will be easier.

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treat $(f_1f_2f_3)'$ like $[(f_1f_2)f_3]'$ so from this this is obvious now: $[(f_1f_2)f_3]'=(f_1f_2)'f_3+(f_1f_2)f_3'$

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use the product rule. induction step: $$ \frac{d(f_1\cdots f_{n-1}f_n)}{dx}=\frac{d(f_1\cdots f_{n-1})}{dx}f_n+f_1\cdots f_{n-1}\frac{df_n}{dx} $$

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$[f_1(x)f_2(x)f_3(x)f_4(x)...]'$ = $f_1(x)'[f_2(x)f_3(x)f_4(x)...]+f_1(x)[f_2(x)f_3(x)f_4(x)...]'$ = $f_1(x)'[f_2(x)f_3(x)f_4(x)...]+f_1(x)f_2(x)'[f_3(x)f_4(x)...]+f_1(x)f_2(x)[f_3(x)f_4(x)...]'$... $$[\prod_k{f_k}]'=\sum_{k=1}^nf_k(x)'\prod_{j=1, j\ne{k}}^{n}f_j(x)$$

For each function is used determine its derivation $f_i(x)'$. Then perform $n$ products where is gradually replaced on place i=1..n the f_i(x) his derivation $f_i(x)'$. These products are then summed together.

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