# Proof (x^n)'=nx^n-1 by complete induction

1. Apr 12, 2012

### Hernaner28

1. The problem statement, all variables and given/known data
Prove that $$({x^n})' = n{x^{n - 1}}$$ by COMPLETE induction.

2. Relevant equations
Nothing.

3. The attempt at a solution
I did the base step for n=1 and then I suppctosed that it was true for n-1 but then I thought about it and I wonder why the professor told us to prove this by induction when in fact induction is useful for natural numbers, and that is true for real numbers.
Anyway, I think he asked for it because we know that:
\begin{align} & x\text{ is differentiable}\Rightarrow xx={{x}^{2}}\text{ is dif}\text{.}\Rightarrow {{x}^{2}}x={{x}^{3}}\Rightarrow ...{{x}^{n}}\text{ is differentiable if }n\in \mathbb{N} \\ & c\in \mathbb{R}\Rightarrow c\cdot {{x}^{n}}\text{ is dif}\text{.}\Rightarrow P(x)={{a}_{0}}{{x}^{n}}+{{a}_{1}}{{x}^{n-1}}+...+{{a}_{n}}\text{ is differentiable} \\ \end{align}

So therefore we could finally prove at the same time that any polynomial is differentiable. But what to do on n-1? Just an idea please!

Thank you!

Last edited: Apr 12, 2012
2. Apr 12, 2012

### Staff: Mentor

"true for real numbers"? Is that a typo?
Assume that the statement is true for n = k. I.e., that (xk)' = kxk-1.

Now, use this assumption to show that the statement is true for n = k + 1. I.e., that (xk+1)' = (k + 1)xk. The product rule will come in handy.

3. Apr 12, 2012

### HallsofIvy

Staff Emeritus
I'm not sure what you mean by "what to do on n-1?" "n- 1" never arises in induction. Nor do I see why you feel the need to prove a polynomial is differentiable when the problem says nothing about polynomials.

It is easy to prove for n= 1. If n= 1, we have f(x)= x and its derivative is $1= 1(x^0)= n(x^{n-1})$.

Now, for fixed k, assume that, for $n\le k$, the derivative of $x^n$ is $nx^{n-1}$. Write $x^{k+1}$ as $x(x^k)$ and use the product rule.
Your professor suggested induction because the problem is to show that the derivative of $x^n$ is $nx^{n-1}$ for n an integer, not a general real number.

(I honestly don't see any reason to use the "complete" induction rather than simple induction.)

4. Apr 12, 2012

### Hernaner28

I mean that the derivative function shown of x^n is true for any real number, not only naturals but now I realize why he wants to prove it only for naturals. I didn't understand why you want me to do that strange product kIe, I have to prove it using complete induction. So I want to prove it for n, I know it's true for n=1, so I assume that is true for n-1 (as far as I know this is the process of complete induction).

This is true for all n real:$$({{x}^{n}}{)}'=n{{x}^{n-1}}$$
But the professor want to prove that only for naturals so that we could finally prove that a polynomial is differentiable (because in class we had previously proven that the product of differentiable functions is differentiable). Why do you mention integers? A polynomial doesn't have negative power terms. Apart from that isn't induction only for naturals and not integers?
I want to do it by complete induction as he asked for.

Thank you both!

5. Apr 12, 2012

### Staff: Mentor

kIe? What is that?

It's not clear to me that you understand how an induction proof goes.
1) Establish that the statement you want to prove is true for some base case, such as k = 1.
2) Assume that the statement is true for n = k.
3) Using the assumption in #2, show (prove) that the statement is true for n = k + 1.
Yes, we're talking about positive integers.

6. Apr 12, 2012

### Hernaner28

hahahaha!! Great misunderstaning: when you wrote n=k .I.e I thought you were telling me to do that product k.I.e... I didn't understand what i.e. mean until now that I remembered (I don't speak english so...).

I know that your method works but you're using the simple induction. Apart from that simple induction we have 2 types of complete induction. The traditional complete induction says that in the second step we may assume not only that the statement holds for n = m but also that it is true for all n less than or equal to m like HallsofIvy explained.

Another proof by complete induction uses the hypothesis that the statement holds for all smaller n more thoroughly which I definately do not understand.

http://en.wikipedia.org/wiki/Mathematical_induction

Thanks!

Last edited: Apr 12, 2012
7. Apr 12, 2012

### Hernaner28

Done:

\begin{align} & {{x}^{n-1}}={{x}^{-1}}({{x}^{n}}) \\ & ({{x}^{n-1}}{)}'={{\left( \frac{1}{x}({{x}^{n}}) \right)}^{\prime }}={{\left( \frac{1}{x} \right)}^{\prime }}{{x}^{n}}+\frac{1}{x}({{x}^{n}}{)}' \\ & (n-1){{x}^{n-2}}=\underbrace{\frac{-{{x}^{n}}}{{{x}^{2}}}}_{-{{x}^{x-2}}}+\frac{1}{x}({{x}^{n}}{)}' \\ & (n-1){{x}^{n-2}}+{{x}^{n-2}}=\frac{1}{x}({{x}^{n}}{)}' \\ & n{{x}^{n-2}}=\frac{1}{x}({{x}^{n}}{)}' \\ & n{{x}^{n-1}}=({{x}^{n}}{)}' \\ \end{align}

Anyway I still don't understand that "more thoroughly" complete induction proof which mentions the wikipedia. I'll stick with the ordinary one I guess...

Last edited: Apr 12, 2012
8. Apr 13, 2012

### Staff: Mentor

I.e. isn't English - it's the Latin abbreviation for id est, which means "that is."

9. Apr 13, 2012

### Joffan

Wow... that's a tough approach, and slightly suspect as it anticipates the derivative of negative powers. And not really COMPLETE induction either; that might start something like:

$n ≥ 2 \implies$ we can choose k,m both ≥ 1 that make $k+m=n$ (which also gives $k<n$ and $m<n$)

$$x^n = x^kx^m \\ (x^n)' = (x^k)'x^m + x^k(x^m)'$$

etc.