Proof by Induction of the Power Rule of Differentiation

In summary, the conversation was about using induction to prove the equation (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). The individual was having trouble with the derivatives but was helped by another individual who suggested using the product rule. The individual then asked if they were allowed to assume certain rules, to which the other individual replied that it was safe to use the associative rule for multiplication. Finally, the conversation ended with a question about which rule was being proved.
  • #1
tinylights
18
0

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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  • #2
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.
 
  • #3
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!
 
  • #4
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!
 
  • #5
You titled this "power rule" but talk about the "product rule". Which are you trying to prove?
 

FAQ: Proof by Induction of the Power Rule of Differentiation

1. How does proof by induction work?

Proof by induction is a mathematical technique used to prove that a statement holds for all natural numbers. The proof is divided into two steps: the base case, where the statement is shown to hold for the first natural number, and the induction step, where it is shown that if the statement holds for a particular natural number, it also holds for the next natural number.

2. What is the power rule of differentiation?

The power rule of differentiation is a formula used to find the derivative of a power function. It states that if a function, f(x), can be written as xn, where n is a constant, then its derivative is nxn-1. This rule is a fundamental concept in calculus and is used to find the slope of a curve at a specific point.

3. How is the power rule of differentiation proven using induction?

The proof by induction of the power rule of differentiation is done by using the definition of the derivative and the binomial theorem. The base case is shown by directly computing the derivative of x1, which gives the result of 1. The induction step involves showing that if the power rule holds for xk, then it also holds for xk+1. By using the binomial theorem and the assumption that the power rule holds for xk, the result of (k+1)xk is obtained, which proves the induction step.

4. Why is proof by induction important in mathematics?

Proof by induction is important in mathematics because it allows us to prove statements that hold for an infinite number of natural numbers. It provides a rigorous and systematic way of proving these statements, which can be applied to various mathematical concepts and theories. It is also used in computer science and physics to prove the correctness of algorithms and mathematical models.

5. Are there any limitations to proof by induction?

Yes, there are some limitations to proof by induction. It can only be used to prove statements that hold for all natural numbers, and it is not suitable for proving statements that involve real numbers or other types of numbers. Additionally, it requires a clear understanding of the base case and the induction step, which can be challenging for some proofs. Therefore, it is essential to use other proof techniques in conjunction with proof by induction to prove more complex statements.

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