Proof by Induction of the Power Rule of Differentiation

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Homework Help Overview

The discussion revolves around using mathematical induction to prove a differentiation rule involving products of functions. The original poster presents a formula for the derivative of a product of functions and expresses uncertainty about how to proceed with the inductive step, particularly in handling derivatives.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the inductive step and the application of the product rule for differentiation. There is a focus on whether certain assumptions can be made regarding the product of functions and the validity of using the product rule in this context.

Discussion Status

Some participants have provided guidance on using the product rule and the associative property of multiplication to simplify the problem. There is an acknowledgment of the original poster's confusion regarding the application of these concepts. Multiple interpretations of the problem's title and the differentiation rule are being explored.

Contextual Notes

There is a mention of the original poster's uncertainty about assumptions in the proof process and a reference to checking with a professor for clarification. The distinction between the "power rule" and the "product rule" is also noted as a point of confusion.

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?
 

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