Proof by induction of polynomial differentiability

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SUMMARY

The discussion focuses on proving the derivative of the polynomial function (ax^n) using mathematical induction. The base case is established as (ax)' = a, and the proof progresses by assuming the statement holds for n = k, leading to the conclusion that (ax^(k+1))' = (n+1)ax^n. The importance of using the definition of the derivative, particularly limits, is emphasized, alongside the application of the product rule and the constant multiple rule.

PREREQUISITES
  • Understanding of polynomial differentiation
  • Familiarity with mathematical induction
  • Knowledge of the definition of the derivative, including limits
  • Proficiency in applying the product rule and constant multiple rule
NEXT STEPS
  • Study the concept of mathematical induction in depth
  • Review the definition of the derivative and its application in calculus
  • Learn how to apply the product rule in differentiation
  • Practice proving derivatives of polynomial functions using induction
USEFUL FOR

Students in calculus or analysis courses, particularly those struggling with differentiation proofs and mathematical induction techniques.

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Homework Statement


Prove that (ax^n)' = nax^n-1 using induction.

I am very weak with induction proof, and I haven't had much trouble proving the basis step, but I can't seem to finish it...

Homework Equations





The Attempt at a Solution


1. Prove (ax)' = a

(a(x+h) - a(x))/h = (ax + ah - ax)/h = (ah)/h = a

2. Prove (ax^n)' = nax^n-1
? I can't seem to get my algebra right...
 
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In proving your base case using the definition of the derivative, you need to do this as a limit. Are you required to use the definition of the derivative in this problem? If not it's much simpler to show that d/dx(ax) = a d/d(x) = a, using the constant multiple rule and the power rule.

The next step is to assume that d/dx(axn) = naxn+1. Then use that to prove that d/dx(axn+1) = (n+1)axn+2. If you're not required to use the definition of the derivative, you can do this using the product rule, keeping in mind that axn+1 = axn * x.
 
This is for an analysis class, so yes, I would think that I would have to use the definition of derivative in this one...
 
Just because it's an analysis class doesn't necessarily mean that you have to use the definition of the derivative. That's an assumption you are making that may or may not be justified.

If if turns out that you do have to use the definition, it shouldn't be that hard for this proof. If you use the definition, however, you need to include limits. What you showed for your base case is very sloppy, not using limits at all. It should look something like this.
\frac{d(ax)}{dx}~=~\lim_{h \rightarrow 0}\frac{a(x + h) -ax}{h}~=~\lim_{h \rightarrow 0}\frac{ax +ah - ax}{h}~=~\lim_{h \rightarrow 0}\frac{ah}{h}~=~a
 
Let P(n) be the statement that for the natural number n, (ax^n)' = nax^{n-1}. Since you've already proven that P(1) holds, assume that P(k) holds and complete the proof by showing that this implies that P(k+1) holds. So you should now start with,

(ax^{k+1})' = (ax^kx)'

Can you see the next step that you should take?
 
Haha, fantastic! Thanks to both of you. That makes things much easier to understand for me.
 

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