Prove the Leibnitz rule of derivatives

[Karol]

Homework Statement

Homework Equations

Newton's binomial's: $(a+b)^n=C^0_n a^n+C^1_n a^{n-1}b+...+C^n_n b^n$

The Attempt at a Solution

I use induction and i try to prove for n+1, whilst the formula for n is given:
$$\frac{d^{n+1}(uv)}{dx^{n+1}}=\frac{d}{dx}\frac{d^{n}(uv)}{dx^n}=$$
The first member of the derivative of the product of each member in $~\displaystyle \frac{d^{n}(uv)}{dx^n}~$+ the whole derivative of the last member in $~\displaystyle\frac{d^{n}(uv)}{dx^n}~$ give the desired $~\displaystyle \frac{d^{n+1}(uv)}{dx^{n+1}}~$.
In $~\displaystyle \frac{d^{n+1}(uv)}{dx^{n+1}}~$ there are n+2 members and my statement also gives the correct number of members: it adds one member to the n+1 members of $~\displaystyle \frac{d^{n}(uv)}{dx^n}$.
But if i consider the rest k-1 members, then deriving makes one more member for each, so $~\displaystyle \frac{d}{dx}\frac{d^{n}(uv)}{dx^n}~$ creates too many members!
Apart from that, i am not sure i can use the induction formula since i don't think the book taught it till this chapter.

 

[Orodruin]

If you cannot use induction, think of the possible ways of selecting which of the derivatives act on each function for a given term.

If you want to use induction I would suggest writing everything in series notation and play with the sum limits.

 

[Karol]

selecting which of the derivatives act on each function for s given term

I don't understand. for example the general term:
$$\frac{d}{dx}\left[ \left( \begin{array}{m} n\\k \end{array} \right) \frac{d^{n-k}u}{dx^{n-k}}\frac{d^kv}{dx^k} \right]$$
$$=\left( \begin{array}{m} n\\k \end{array} \right) \left[ \frac{d^{n+1-k}u}{dx^{n+1-k}}\frac{d^kv}{dx^k}+\frac{d^{n-k}u}{dx^{n-k}}\frac{d^{k+1}v}{dx^{k+1}} \right]$$
Each member, when beeing derived, makes 2 members. i don't understand what you mean.

 

[Orodruin]

I don't understand. for example the general term:
$$\frac{d}{dx}\left[ \left( \begin{array}{m} n\\k \end{array} \right) \frac{d^{n-k}u}{dx^{n-k}}\frac{d^kv}{dx^k} \right]$$
$$=\left( \begin{array}{m} n\\k \end{array} \right) \left[ \frac{d^{n+1-k}u}{dx^{n+1-k}}\frac{d^kv}{dx^k}+\frac{d^{n-k}u}{dx^{n-k}}\frac{d^{k+1}v}{dx^{k+1}} \right]$$
Each member, when beeing derived, makes 2 members. i don't understand what you mean.

No, I am talking about each derivative in $(d/dx)^n(uv)$. Each derivative can act on either $u$ or $v$. How many possibilities are there to get $k$ derivatives acting on $u$ and $n-k$ on $v$.

The other option is doing it by induction. Note that adjacent terms will lead to similar derivatives. You need to collect the derivatives that are the same into one term with one coefficient. For example $d(u'v)/dx = u''v + u'v'$ and $d(uv')/dx = u'v' + uv''$. While both contributions contain two terms, if you sum them you would have three terms as you would collect the $u'v'$ into one term.

 

[Karol]

I still don't think i understand. for example n=5, k=0,1,2... then:
5x0, 4x1, 3x2, 2x3, 1x4, 0x5
There are n+1 possibilities.

 

[Orodruin]

I still don't think i understand. for example n=5, k=0,1,2... then:
5x0, 4x1, 3x2, 2x3, 1x4, 0x5
There are n+1 possibilities.

No, this is not what I am talking about. I am talking about the combinatoric number of possibilities of creating each of those terms. For example, there is only one possibility to generate $u'''''v$, namely that all derivatives act on $u$. This gives you the factor 1 before that term.

Edit: Again, if you have problems I suggest writing the induction step in series notation and play with the summation limits.

 

[Ray Vickson]

I don't understand. for example the general term:
$$\frac{d}{dx}\left[ \left( \begin{array}{m} n\\k \end{array} \right) \frac{d^{n-k}u}{dx^{n-k}}\frac{d^kv}{dx^k} \right]$$
$$=\left( \begin{array}{m} n\\k \end{array} \right) \left[ \frac{d^{n+1-k}u}{dx^{n+1-k}}\frac{d^kv}{dx^k}+\frac{d^{n-k}u}{dx^{n-k}}\frac{d^{k+1}v}{dx^{k+1}} \right]$$
Each member, when beeing derived, makes 2 members. i don't understand what you mean.

To make writing easier, let $D = d/dx$. Look at the coefficient of $D^j v$. You get $C(n,j) D^{n+1-j}u$ from $C(n,j) D (D^{n-j}u D^j v)$ and you get $C(n,j-1) D^{n+1-j} u$ from $C(n,j-1) D(D^{n-j+1} D^{j-1} v)$. What is $C(n,j) + C(n,j-1)$? (Here, $C(a,b)$ means ${a \choose b}$.)

 

[Karol]

$$\frac{d^{n+1}(uv)}{dx^{n+1}}=\left[ \sum_{k=0}^n \left( \begin{array}{m} n\\k \end{array} \right) u^{n-k}v^k \right]'=\sum_{k=0}^n \left( \begin{array}{m} n\\k \end{array} \right) (u^{n+1-k}v^k+u^{n-k}v^{k+1})$$
$$=u^{n+1}v+\sum_{k=0}^n \left[ \left( \begin{array}{m} n\\k \end{array} \right) + \left( \begin{array}{m} n\\k+1 \end{array} \right) \right] u^{n-k}v^{k+1}+uv^{n+1}$$

 

[Orodruin]

Your sum is going one too far. It encompasses one of the terms that you have taken out (the corresponding partner term is zero).

 

[Karol]

$$\frac{d^{n+1}(uv)}{dx^{n+1}}=\left[ \sum_{k=0}^n \left( \begin{array}{m} n\\k \end{array} \right) u^{n-k}v^k \right]'=\sum_{k=0}^n \left( \begin{array}{m} n\\k \end{array} \right) (u^{n+1-k}v^k+u^{n-k}v^{k+1})$$
$$=u^{n+1}v+\sum_{k=0}^{n-1} \left[ \left( \begin{array}{m} n\\k \end{array} \right) + \left( \begin{array}{m} n\\k+1 \end{array} \right) \right] u^{n-k}v^{k+1}+uv^{n+1}$$
But how to do it with the first method:

No, this is not what I am talking about. I am talking about the combinatoric number of possibilities of creating each of those terms. For example, there is only one possibility to generate $~u′′′′′v~$, namely that all derivatives act on $~u$. This gives you the factor 1 before that term.

But the factor before all other terms is also 1 since each can appear only once: $~u''''v'~$, $~u'''v''~$ etc.

 

[Orodruin]

$$\frac{d^{n+1}(uv)}{dx^{n+1}}=\left[ \sum_{k=0}^n \left( \begin{array}{m} n\\k \end{array} \right) u^{n-k}v^k \right]'=\sum_{k=0}^n \left( \begin{array}{m} n\\k \end{array} \right) (u^{n+1-k}v^k+u^{n-k}v^{k+1})$$
$$=u^{n+1}v+\sum_{k=0}^{n-1} \left[ \left( \begin{array}{m} n\\k \end{array} \right) + \left( \begin{array}{m} n\\k+1 \end{array} \right) \right] u^{n-k}v^{k+1}+uv^{n+1}$$
But how to do it with the first method:

But the factor before all other terms is also 1 since each can appear only once: $~u''''v'~$, $~u'''v''~$ etc.

No it is not. It is a combinatorical problem of finding how many different ways you can "construct" each term. Take the n = 2 case as example.

 

[Karol]

$$(uv)''=u''v+2u'v'+uv'',~~(uv)'''=u'''v+3u''v'+3u'v''+uv'''$$
You mean i can construct the term $~u'v'~$ in 2 different ways?
And $~u''v'~$ and $~u'v''~$ each of them in 3 different ways?
I don't know which ways. $~u'v',~u''v',~u'v''~$ are all unique, i can "construct" each only by one way: for example $~u'v''~$ i differentiate u once and i differentiate v 2 times, there is no other way.
How can i construct $~u'v'~$ in 2 different ways?

 

[Orodruin]

How can i construct $~u'v'~$ in 2 different ways?

You can let the first d/dx act on u and the second on v or vice versa - 2 choices.

 

[Ray Vickson]

$$\frac{d^{n+1}(uv)}{dx^{n+1}}=\left[ \sum_{k=0}^n \left( \begin{array}{m} n\\k \end{array} \right) u^{n-k}v^k \right]'=\sum_{k=0}^n \left( \begin{array}{m} n\\k \end{array} \right) (u^{n+1-k}v^k+u^{n-k}v^{k+1})$$
$$=u^{n+1}v+\sum_{k=0}^{n-1} \left[ \left( \begin{array}{m} n\\k \end{array} \right) + \left( \begin{array}{m} n\\k+1 \end{array} \right) \right] u^{n-k}v^{k+1}+uv^{n+1}$$
But how to do it with the first method:

But the factor before all other terms is also 1 since each can appear only once: $~u''''v'~$, $~u'''v''~$ etc.

What s preventing you from simplifying ${n \choose k} + {n \choose k+1}$? It is an entirely standard result, known for hundreds of years.

BTW: you need not use an "array" to typset ${a \choose b}$; the built-in command "{ a \choose b}" will do it.

 

[Karol]

In n=3, i denote the first d/dx as 1 and the second d/dx as 2, and even so there are more possibilities than 3:
$$u'v'': u^1v^{11},~u^1v^{12},~u^2v^{12},~u^2v^{11},~u^2v^{22}$$

 

[Orodruin]

In n=3, i denote the first d/dx as 1 and the second d/dx as 2, and even so there are more possibilities than 3:
$$u'v'': u^1v^{11},~u^1v^{12},~u^2v^{12},~u^2v^{11},~u^2v^{22}$$

No there are not. The choices are $u^1 v^{23}$, $u^2 v^{13}$, and $u^3 v^{12}$. It is a 3 choose 1 situation, you have 3 choices in which derivative acts on $u$, the other derivatives act on $v$. The same derivative cannot act on several functions or multiple times on the same function.

 

[Karol]

I don't know combinatorics but i know now how to calculate the coefficients. but i don't understand the logic.
When i take d/dx of n=2 to get d³/dx³ i don't select which derivative acts, i just derive.
And even so this gives only the coefficients, in the first method of proof of $~d^n(uv)/dx^n~$.

 

[Orodruin]

I don't know combinatorics but i know now how to calculate the coefficients. but i don't understand the logic.
When i take d/dx of n=2 to get d³/dx³ i don't select which derivative acts, i just derive.
And even so this gives only the coefficients, in the first method of proof of $~d^n(uv)/dx^n~$.

I don't understand your concern, if you can derive the coefficients you have your proof. For each derivative, it can act on either $u$ or $v$. In the end, you will therefore end up with a total of $2^n$ terms if you do not collect the terms that have the same derivatives. However, since the derivatives commute, it does not matter in which order they act and you can collect all term with the same number of derivatives on $u$. For $u^{(k)}$, there will be ${n \choose k}$ such terms and therefore the coefficient in front of $u^{(k)} v^{(n-k)}$ must be ${n \choose k}$.

Note that this is also completely compatible with the fact that $\sum_{k=0}^n {n\choose k} = 2^n$.

 

[Orodruin]

When i take d/dx of n=2 to get d³/dx³ i don't select which derivative acts, i just derive.
And even so this gives only the coefficients, in the first method of proof of $~d^n(uv)/dx^n~$.

Also, it is not about selecting where the derivatives act, it is about selecting which terms you look at and how they appear. The argumentation for the $n = 2$ case is as follows: You want to compute
$$
\newcommand{\dd}[2]{\frac{d#1}{d#2}}
\newcommand{\red}[1]{\color{red}{#1}}
\newcommand{\blue}[1]{\color{blue}{#1}}
\dd{^2uv}{x^2} = \red{\dd{}{x}}\blue{\dd{}{x}} uv = \red{\dd{}{x}}(u\blue{'} v + u v\blue{'})
= u\blue{'}\red{'} v + u\blue{'} v\red{'} + u\red{'}v\blue{'} + uv\blue{'}\red{'},
$$
where we have tracked how each term appeared from the differentiations by colouring the primes. (Also note how there are $2^2 = 4$ terms before collecting them together.)
Note how there are two options resulting in the $u'v'$ term, one where the red derivative acts on $u$ and one where the blue derivative acts on $u$. Hence you have ${2 \choose 1} = 2$ possibilities of constructing this term and it is therefore the coefficient in front. For the term $u'' v$, both derivatives must act on $u$ and you have ${2\choose 2} = 1$ possibilities of constructing it.

For arbitrary $n$, imagine the same thing, but with more colours.

 

[Karol]

Thank you very much Orodruin for your thoroughness and patience.
Now i see Ray's remark on how to combine $~\displaystyle {n \choose k}+{n \choose k+1}$, how is it called and where do i find it
(i am not hundreds of years old...)

 

[Orodruin]

Thank you very much Orodruin for your thoroughness and patience.
Now i see Ray's remark on how to combine $~\displaystyle {n \choose k}+{n \choose k+1}$, how is it called and where do i find it
(i am not hundreds of years old...)

Those are the binomial coefficients. The easiest way of simplifying is to write them out in terms of factorials and start simplifying.

 

[Karol]

$${n \choose k}+{n \choose k+1}=\frac{n!}{k!(n-k)!}+\frac{n!}{(k+1)!(n-k-1)!}=\frac{n![(k+1)+(n-k)]}{(k+1)!(n-k)!}=\frac{n!(n+1)}{(k+1)!(n-k)!}$$

 

[I like Serena]

Thank you very much Orodruin for your thoroughness and patience.
Now i see Ray's remark on how to combine $~\displaystyle {n \choose k}+{n \choose k+1}$, how is it called and where do i find it
(i am not hundreds of years old...)

It's called Pascal's Triangle. And I'm afraid being a mere century of age is not going to do the job.
We'll need about 4 centuries. :)
Anyway, we can get a hint from that triangle on what that sum should be.
The numbers in the triangle correspond to $\binom n k$, where $n$ is the row, and $k$ is the column.

 

[SammyS]

$${n \choose k}+{n \choose k+1}=\frac{n!}{k!(n-k)!}+\frac{n!}{(k+1)!(n-k-1)!}=\frac{n![(k+1)+(n-k)]}{(k+1)!(n-k)!}=\frac{n!(n+1)}{(k+1)!(n-k)!}$$

Of course, $\ n!(n+1) = (n+1)!\$ and $\ n-k \$ can be written as $\ (n+1)-(k+1)\$.

 

[Karol]

$$\frac{n!(n+1)}{(k+1)!(n-k)!}=\frac{(n+1)!}{(k+1)![(n+1)-(k+1)]!}=\frac{(n+1)!}{(k+1)!n!}$$

 

[Orodruin]

The last step is not correct.

 

[Karol]

$$\frac{n!(n+1)}{(k+1)!(n-k)!}=\frac{(n+1)!}{(k+1)![(n+1)-(k+1)]!}=\frac{(n+1)!}{(k+1)!k!}$$

 

[Orodruin]

No, that neither. It is arithmetically wrong. What happens if you let $n+1 = N$ and $k+1 = K$ in the step before the last?

 

[Karol]

$$\frac{n!(n+1)}{(k+1)!(n-k)!}=\frac{(n+1)!}{(k+1)![(n+1)-(k+1)]!}=\frac{(n+1)!}{(k+1)!(k+1)!(n-1)}=\frac{(n+1)!}{[(k+1)!]^2(n-1)}$$
$$=\frac{(n+1)n}{(k+1)!(n-1)}$$

 

[Orodruin]

No, now you are just guessing. Why would $[(n+1) - (k+1)]! = (k+1)!(n-1)$? Why don't you do what I suggested in #28?

 

[Karol]

Because i don't know what to do with that:
$$\frac{(n+1)!}{(k+1)![(n+1)-(k+1)]!}=\frac{(n+1)!}{(k+1)!(N-K)!}$$
So?
I am not guessing. n=5, k=3:
$$[(n+1) - (k+1)]! =(1\cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 - 1\cdot 2 \cdot 3 \cdot 4 )=(1\cdot 2 \cdot 3 \cdot 4 )(5 \cdot 6 -1 )=(k+1)![n(n+1)-1]$$

 

[Orodruin]

Because i don't know what to do with that:
$$\frac{(n+1)!}{(k+1)![(n+1)-(k+1)]!}=\frac{(n+1)!}{(k+1)!(N-K)!}$$

You have one more $n+1$ and one more $k+1$. Why did you not make the substitution there as well?

 

[fresh_42]

So?
I am not guessing. n=5, k=3:

$[(n+1)−(k+1)]!=(1⋅2⋅3⋅4⋅5⋅6−1⋅2⋅3⋅4)=(1⋅2⋅3⋅4)(5⋅6−1)=(k+1)![n(n+1)−1]$​

This is completely wrong, simple as that.
$(5-3)!=2!=2\neq 5!-3!=120-6=114$ and $(n-m)!\neq n!-m!$.

Repetition never makes something true, which has been wrong in the first place.

 

[Orodruin]

I am not guessing. n=5, k=3:
$$[(n+1) - (k+1)]! =(1\cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 - 1\cdot 2 \cdot 3 \cdot 4 )=(1\cdot 2 \cdot 3 \cdot 4 )(5 \cdot 6 -1 )=(k+1)![n(n+1)-1]$$

No, this is wrong. The factorial is outside of the parentheses! With $n = 5$ and $k = 3$ you have
$$
[(n+1)-(k+1)]! = [6-4]! = 2! = 2
$$
and
$$
(k+1)! (n-1) = 4! \cdot 4 = 96.
$$
Clearly not the same thing.

Edit: In addition, even if it did hold you cannot just take two arbitrary values for unknowns, show that it is true for that particular choice, and then expect that it holds for an arbitrary choice of the unknowns.

 

[SammyS]

$$\frac{n!(n+1)}{(k+1)!(n-k)!}=\frac{(n+1)!}{(k+1)![(n+1)-(k+1)]!}=\frac{(n+1)!}{(k+1)!k!}$$

$\displaystyle \frac{n!(n+1)}{(k+1)!(n-k)!}=\frac{(n+1)!}{(k+1)![(n+1)-(k+1)]!}\ \ \ne\ \ \frac{(n+1)!}{(k+1)!k!}$

Now, take $\displaystyle \ \frac{(n+1)!}{(k+1)![(n+1)-(k+1)]!}\$ and make the substitution @Orodruin suggested:

...
What happens if you let $n+1 = N$ and $k+1 = K$ ?

 

[Karol]

$$\frac{(n+1)!}{(k+1)![(n+1)-(k+1)]!}=\frac{N!}{K!(N-K)!}=\frac{n!}{k!(n-k)!}=\frac{n!}{(k!)^2}$$

 

[Orodruin]

First step ok. The next two are clearly false. What is your argumentation? This is what I mean by "just guessing" because you can easily try a few values for $n$ and $k$ and see that it is non-sensical.

 

[Karol]

$$\frac{(n+1)!}{(k+1)![(n+1)-(k+1)]!}=\frac{N!}{K!(N-K)!}={n+1 \choose k+1}$$
$$\frac{d^{n+1}(uv)}{dx^{n+1}}=u^{n+1}v+\sum_{k=0}^{n-1} \left( \begin{array}{m} n+1\\k+1 \end{array} \right) u^{n-k}v^{k+1}+uv^{n+1}$$
But i have to prove:
$$\sum_{k=0}^n \left( \begin{array}{m} n\\k \end{array} \right) u^{n-k}v^k\ \ \rightarrow\ \ \sum_{k=0}^{n+1} \left( \begin{array}{m} n+1\\k+1 \end{array} \right) u^{n+1-k}v^{k+1}$$
If i substitute k=0 the first term in the sum gives $~{n+1 \choose 1}~$ and doesn't equal 1, as needed

 

[Orodruin]

$$\frac{(n+1)!}{(k+1)![(n+1)-(k+1)]!}=\frac{N!}{K!(N-K)!}={n+1 \choose k+1}$$
$$\frac{d^{n+1}(uv)}{dx^{n+1}}=u^{n+1}v+\sum_{k=0}^{n-1} \left( \begin{array}{m} n+1\\k+1 \end{array} \right) u^{n-k}v^{k+1}+uv^{n+1}$$
But i have to prove:
$$\sum_{k=0}^n \left( \begin{array}{m} n\\k \end{array} \right) u^{n-k}v^k\ \ \rightarrow\ \ \sum_{k=0}^{n+1} \left( \begin{array}{m} n+1\\k+1 \end{array} \right) u^{n+1-k}v^{k+1}$$
If i substitute k=0 the first term in the sum gives $~{n+1 \choose 1}~$ and doesn't equal 1, as needed

No, again you did not think it through thoroughly. Also, what you wrote down in the end is not what you have to prove (because it is false!). There should be a $k$ and not a $k+1$ in the binomial coefficient as well as in the derivative of $v$. You just want the implication of the theorem being true for $n$ meaning that it is true for $n+1$ - $k$ is a summation variable and does not play a role - nothing says that you should change the way $k$ appears.

 

[Karol]

I have to prove:
$$\sum_{k=0}^n \left( \begin{array}{m} n\\k \end{array} \right) u^{n-k}v^k\ \ \rightarrow\ \ \sum_{k=0}^{n+1} \left( \begin{array}{m} n+1\\k \end{array} \right) u^{n+1-k}v^{k}$$
And i have:
$$u^{n+1}v+\sum_{k=0}^{n-1} \left( \begin{array}{m} n+1\\k+1 \end{array} \right) u^{n-k}v^{k+1}+uv^{n+1}=$$
$$=u^{n+1}v+\sum_{k=1}^{n-1} \left( \begin{array}{m} n+1\\k \end{array} \right) u^{n-k}v^{k}+uv^{n+1}=\sum_{k=0}^{n+1} \left( \begin{array}{m} n+1\\k \end{array} \right) u^{n-k}v^{k}$$

 

[Orodruin]

You forgot to make the shift in the term $u^{n-k}$ when you changed the summation variable.

You really should be able to find these errors yourself and not just assume that things are not the way they are presented in the problem. If you spent 10 minutes looking for this error, you might have found it.

 

[SammyS]

@Karol
You had this in post #10:

$$\frac{d^{n+1}(uv)}{dx^{n+1}}=\left[ \sum_{k=0}^n \left( \begin{array}{m} n\\k \end{array} \right) u^{n-k}v^k \right]'=\sum_{k=0}^n \left( \begin{array}{m} n\\k \end{array} \right) (u^{n+1-k}v^k+u^{n-k}v^{k+1})$$
$$=u^{n+1}v+\sum_{k=0}^{n-1} \left[ \left( \begin{array}{m} n\\k \end{array} \right) + \left( \begin{array}{m} n\\k+1 \end{array} \right) \right] u^{n-k}v^{k+1}+uv^{n+1}$$

You got this (I suppose) by splitting the sum into two sums and taking out the k = 0 term from the first, and the k = n term from the second, then you shifted the index in the first sum so that you could combine them into one sum.

Instead of shifting the index so the sum starts at k = 0, shift it so that it starts at k = 1.

You get:

$\displaystyle \sum_{k=1}^{n} \left[ { n \choose {k-1} } + {n \choose {k}} \right] u^{n+1-k}v^{k}$​

I think you'll find that $\displaystyle \ { n \choose {k-1} } + {n \choose {k}} \$ is easier to simplify than the previous combination, and the index will easier to work with in your final result..

 

[Karol]

Note that i changed also the (n-1) above the ∑
$$=u^{n+1}v+\sum_{k=1}^{n} \left( \begin{array}{m} n+1\\k \end{array} \right) u^{n+1-k}v^{k}+uv^{n+1}=\sum_{k=0}^{n+1} \left( \begin{array}{m} n+1\\k \end{array} \right) u^{n+1-k}v^{k}$$

 

[epenguin]

I suggest that the simplest proof of Leibniz' theorem is by algebrisation. The student will in any case meet before too long in differential equations the very strong analogy if not more between differential operators and polynomials.

We start with
\begin{equation} D^{1}\left( uv\right) =uD^{1}v+vD^{1}u \end{equation}
where D¹ denotes the differential operator d/dx (so that also dⁱ/dxⁱ is $D^{i})$. These differential operators follow the same 'indices rule' as ordinary quantities raised to powers in arithmetic and algebra i.e.
\begin{equation} D^{n}D^{m}=D^{n+m} \end{equation}
All the calculus involved finishes there, and from there on it is algebraic.

I found it easier to reason with a slightly more complicated but explicit notation and rewrite eq (1) as
\begin{equation}D^{1}D^{0}(uv)=D^{0}uD^{1}v+D^{1}uD^0{v} \end{equation}
where $D^{0}$ represents differentiating zero times, so that $D^{0}u =u$ .

In the same way I represent the binomial usually written $(a+b)$ as
\begin{equation}(b^{0}a^{1}+b^{1}a^{0}) \end{equation}
Then if we accept as having been proved the usual binomial expansion formula (I don't know if the OP is there yet) which I will write in the form
\begin{equation}\left ( a^{1}b^{0}+a^{0}b^{1}\right) ^{n}=\sum ^{n}_{i=0}\binom {n} {i}a^{i}b^{n-i} \end{equation}
then Leibniz' theorem immediately follows because the formal algebra of the expansion of $(b^{0}a^{1}+b^{1}a^{0})^{n}$ and of $(D^{0}uD^{1}v+D^{1}uD^0{v})^{n}$ are exactly the same because they follow exactly the same index and other rules, so in equation (5) you can just replace $a^{i}b^{n-i}$ with $D^{i}uD^{n-i}v$ and the result is true.

I think at this argument is essential at some point in order to make clear that that the theorem depends on nothing but Equations (1) and (2) (maybe that is saying too much, but at any rate only on something minimal and familiar).

 

[Karol]

Why do i need to expand $~\left ( a^{1}b^{0}+a^{0}b^{1}\right) ^{n}~$ and not just (a+b)ⁿ?
Why do i need $~(D^{0}uD^{1}v+D^{1}uD^0{v})^{n}~$ and not just $~(D^{1}u+D^{1}v)^{n}~$?
$$(D^{1}u+D^{1}v)^{n}=\sum ^{n}_{i=0}\binom {n} {i}D^nn D^{n-i}v$$

 

[SammyS]

Why do i need to expand $~\left ( a^{1}b^{0}+a^{0}b^{1}\right) ^{n}~$ and not just (a+b)ⁿ?
Why do i need $~(D^{0}uD^{1}v+D^{1}uD^0{v})^{n}~$ and not just $~(D^{1}u+D^{1}v)^{n}~$?

... because
$\displaystyle D^{n+1}(uv)=D^n(D(uv))=... \$

 

[epenguin]

Why do i need to expand $~\left ( a^{1}b^{0}+a^{0}b^{1}\right) ^{n}~$ and not just (a+b)ⁿ?
Why do i need $~(D^{0}uD^{1}v+D^{1}uD^0{v})^{n}~$ and not just $~(D^{1}u+D^{1}v)^{n}~$?
$$(D^{1}u+D^{1}v)^{n}=\sum ^{n}_{i=0}\binom {n} {i}D^iu D^{n-i}v$$

That is just my idea actually which I haven't actually seen anywhere, not something you have to do. You can see, I hope, it's not wrong, at worst unnecessary.
The reason I thought of it is that I think it makes it easier - you can handle the two expressions blindly in exactly the same way., I thought that If you were treating
(uDv + vDu)ⁿ in the same way as (a + b)ⁿ a student might start writing things like (uⁿDⁿv + ...). Or at least wondering why he can't. Or to say it another way it makes the analogy between the two things more complete, and you have to think less when applying it. (Maybe you could do with that because actually the LHS of your last equation is not what we are looking for.).

I even thought one might make the analogy (equivalence? isomorphism?...? mathematicians say) more complete by making the ordinary binomial expansion in terms of operations rather than numbers, whatever they are, defining an operation E
E = (a + b)× where × is the multiplication symbol.

However this is open to mathematicians here to discuss what they think is the best way of formulating the argument, including didactically, and maybe do a better set out proof with the same idea - In fact that was I what hoping for.

 

[Karol]

Thank you all very much, Orodruin, Ray, Serena, Sammy, fresh 42 and epenguin