Proof by induction of polynomial differentiability

1. Nov 7, 2009

ssayan3

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

2. Relevant equations

3. 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....

2. Nov 7, 2009

Staff: Mentor

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.

3. Nov 7, 2009

ssayan3

This is for an analysis class, so yes, I would think that I would have to use the definition of derivative in this one....

4. Nov 7, 2009

Staff: Mentor

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.
$$\frac{d(ax)}{dx}~=~\lim_{h \rightarrow 0}\frac{a(x + h) -ax}{h}~=~\lim_{h \rightarrow 0}\frac{ax +ah - ax}{h}~=~\lim_{h \rightarrow 0}\frac{ah}{h}~=~a$$

5. Nov 7, 2009

jgens

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?

6. Nov 7, 2009

ssayan3

Haha, fantastic! Thanks to both of you. That makes things much easier to understand for me.