Proving the power rule by induction

[heimdal]

Homework Statement

If a is a natural number, prove by induction that

y = [g(x)]^a => y' = a[g(x)]^(a-1) * g'(x)

Homework Equations

Let a = 2

y' = (2)[g(x)]^(2-1) g(x)
= 2g(x)g'(x)

Let a = 3

y' = (3)[g(x)]^(3-1) g(x)
= 3g(x)^2 * g'(x)

Let k be any natural number

a(k) = y' = ak[g(x)]^(ak-1) * g'(x)

The Attempt at a Solution

What I did in the above equation was to substitute 2 and 3 (both natural numbers) as a, in order to prove that every natural number k is applicable.

I'm not all too familiar with induction, but am I on the right track? Or am I completely off?

How do I prove the power rule through induction?

 

[dx]

This is how induction works:

1. First, you prove that it is true for a = 1.

2. Then, you assume that it is true for a = k, and then show that this implies that it is true for a = k + 1.

This is enough to show that it is true for all natural numbers.

 

[HallsofIvy]

Homework Statement

If a is a natural number, prove by induction that

y = [g(x)]^a => y' = a[g(x)]^(a-1) * g'(x)

Homework Equations

Let a = 2

y' = (2)[g(x)]^(2-1) g(x)
= 2g(x)g'(x)

Let a = 3

y' = (3)[g(x)]^(3-1) g(x)
= 3g(x)^2 * g'(x)

Why a= 2 and 3? You haven't shown the statement to be true for 1 and that's the critical starting point.

Let k be any natural number

a(k) = y' = ak[g(x)]^(ak-1) * g'(x)

This is what you were supposed to prove. You are simply asserting it.

The Attempt at a Solution

What I did in the above equation was to substitute 2 and 3 (both natural numbers) as a, in order to prove that every natural number k is applicable.

I'm not all too familiar with induction, but am I on the right track? Or am I completely off?

How do I prove the power rule through induction?

What you want to prove is:
If for some k, $(ag(x)^k)'= ag(x)^{k-1}g'(x)$ then $(ag(x)^{k+1})'= ag(x)^k g'(x)$.

That is, if the statement is true for k, it is true for k+1. Then, if you have also proved it true for k= 1, it must true for k+1= 2. Then, since it is true for k= 2, it is true for k+1= 3, etc.

Try writing $(g(x))^{k+1}$ as $g(x)(g(x))^k$ and use the product rule.