# Proof by induction help.

1. Sep 23, 2008

### Ed Aboud

1. The problem statement, all variables and given/known data

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

2. Relevant equations

Product rule
$$\frac{d}{dx} (uv) = u \frac{dv}{dx} + v \frac{du}{dx}$$

3. 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.

2. Sep 23, 2008

### Dick

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.