Proof by Induction of the Power Rule of Differentiation

tinylights
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Homework Statement


Okay, the concept here is to use induction to prove that for n, (f1 x f2 x ... x fn-1 x fn)' = (f'1 x f2 x ... x fn) + (f1 x f'2 x ... x fn) + ... + (f1 x f2 x ... x f'n).

2. Homework Equations / 3. The Attempt at a Solution

I solved the initial step, which was quite easy. I started to set up the inductive step, by stating that:

(f1 x f2 ... x fn x fn+1)' = (f'1 x f2 x ... x fn+1) + (f1 x f'2 x ... x fn+1) + ... + (f1 x f2 x ... x f'n+1).

And I do understand how induction works - I know I am supposed to plug in what I have for the "n" equation into part of my "n+1" equation and find equality. I just don't know HOW I'm supposed to do that for some reason. I think it's the derivatives throwing me off because in the last few problems that I did, it was fine. Please help me simplify?
 
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tinylights said:

Homework Statement


Okay, the concept here is to use induction to prove that for n, (f1 x f2 x ... x fn-1 x fn)' = (f'1 x f2 x ... x fn) + (f1 x f'2 x ... x fn) + ... + (f1 x f2 x ... x f'n).

2. Homework Equations / 3. The Attempt at a Solution

I solved the initial step, which was quite easy. I started to set up the inductive step, by stating that:

(f1 x f2 ... x fn x fn+1)' = (f'1 x f2 x ... x fn+1) + (f1 x f'2 x ... x fn+1) + ... + (f1 x f2 x ... x f'n+1).

And I do understand how induction works - I know I am supposed to plug in what I have for the "n" equation into part of my "n+1" equation and find equality. I just don't know HOW I'm supposed to do that for some reason. I think it's the derivatives throwing me off because in the last few problems that I did, it was fine. Please help me simplify?

(f1 x f2 ... x fn x fn+1)=(f1 x f2 ... x fn) x fn+1. Use the usual two factor product rule on that.
 
Okay, that makes it simpler. I didn't know if I was "allowed" to assume that, or if I had to go from the ground up. I'll double-check with my professor but for now I'll work it out that way. Thanks!
 
tinylights said:
Okay, that makes it simpler. I didn't know if I was "allowed" to assume that, or if I had to go from the ground up. I'll double-check with my professor but for now I'll work it out that way. Thanks!

I didn't "assume" anything. (fg)'=f'g+fg' is the base case for your induction. It's the usual product rule. Sure, you can assume that. And (f1 x f2 ... x fn x fn+1)=(f1 x f2 ... x fn) x fn+1 is just using the associative rule for multiplication. I think you are pretty safe in using that!
 
You titled this "power rule" but talk about the "product rule". Which are you trying to prove?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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