Proving by Induction: Product Rule for Derivatives with Multiple Functions

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SUMMARY

The forum discussion centers on proving the product rule for derivatives of multiple functions using mathematical induction. The statement to prove is that the derivative of the product of k functions, represented as \(\frac{d}{dx} \prod_{i=1}^k f_i (x)\), equals \((\sum_{i=1}^k \frac{ \frac{d}{dx} f_i (x)}{f_i (x)} ) \prod_{i=1}^k f_i (x)\). The discussion emphasizes the importance of establishing the base case for \(n=1\) and correctly applying the product rule, \(\frac{d}{dx} (uv) = u \frac{dv}{dx} + v \frac{du}{dx}\), to prove the induction step.

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Homework Statement



Prove by induction on k that for all integers
[tex]\frac{d}{dx} \prod_{i=1}^k f_i (x) = (\sum_{i=1}^k \frac{ \frac{d}{dx} f_i (x)}{f_i (x)} ) \prod_{i=1}^k f_i (x)[/tex]

Homework Equations



Product rule
[tex]\frac{d}{dx} (uv) = u \frac{dv}{dx} + v \frac{du}{dx}[/tex]

The Attempt at a Solution



I am honestly not sure how to start this. It is the first one like this that I have tried.
 
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This problem is harder to write than it is to solve. Let's write "'" for d/dx and call P(k) the product of the f_i and S(k) the sum of the f'_i/f_i. Then what you have above is (P(k))'=S(k)P(k). Assume that's true. Now you want to prove (P(k+1))'=S(k+1)P(k+1), correct? P(k+1)=P(k)*f_(k+1) and S(k+1)=S(k)+f'_(k+1)/f_(k+1), also ok? Apply the product rule to P(k)*f_(k+1) and see if you can match the two sides up. And don't forget to prove the n=1 case to start the induction.
 

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