Combining Binomial Theorem, Differentiation and Induction !! Differentiation of the product of two functions $n$ times - A Formula

This is outside of syllabus but many of you should extend yourself beyond the syllabus a bit. Is there a formula for differentiating the product of two functions $$y=uv$$  many times over (say, $n$ times)

To get an idea of what the answer to this might be, actually perform the differentiation several times over to see what you get. So find the first derivative, the second derivative, the third derivative, and the fourth derivative. Is there a pattern? Here are the answers.

First derivative,
${d \over dx } (uv) = u'v+uv'$

Second derivative,
$\begin{array}{rl}{d^2 \over dx^2 } (uv) & = u''v+u'v'+u'v'+uv''\\ &= u''v+2u'v'+uv''\end{array}$

Third derivative,
$\begin{array}{rl}{d^3 \over dx^3 } (uv)
& = u'''v+u''v'+2u''v'+2u'v''+u'v''+uv'''\\
& = u'''v+3u''v'+3u'v''+uv'''\end{array}$

Fourth derivative,
$\begin{array}{rl}{d^4 \over dx^4 } (uv)
& = u''''v+u'''v'+3u'''v'+3u''v''+3u''v''+3u'v'''+u'v'''+v''''\\
& = u''''v+4u'''v'+6u''v''+4u'v'''+v''''\end{array}$

Can you see that the coefficients (numbers in front) follow the pattern of Pascal's Triangle below? 

1

1   1

1  2  1

1  3   3  1

1  4  6   4   1

Here is the general case. It uses the binomial theorem and sigma notation, so hopefully you have seen these before.

THEOREM:
If $f(x) = uv$ where $u$ and $v$ are functions of $x$ then the
nth derivative of this product is given by
$$f^{(n)}(x)={d^n \over dx^n } (uv)=\sum^n_{r=0} {n\choose r} u^{(r)}v^{(n-r)}$$
where we use the notation for the $r^{th}$ derivative

$\displaystyle u^{(r)}={ d^r\over dx^r} u$

(When it's a derivative there are brackets around the 'index')

Using this notation differentiating one more time looks like this
$$\displaystyle{ d\over dx}\left( u^{(r)}\right) = { d\over dx} { d^r\over dx^r} u={ d^{r+1}\over dx^{r+1}}=u^{(r+1)}$$
PROOF: 
By induction on $n$.
For $n=0$, $f^{(0)}(x)=f(x)=uv$ true. and

$ {0\choose 0} u^{(0)}v^{(0)}=uv$ and so it is true for $n=0$.

[Remark, $n=0$ corresponds to not differentiating at all, so it should be the original product $uv$.]

For $n=1$,

$ {1\choose 0} u^{(0)}v^{(1)}+{1\choose 1} u^{(1)}v^{(0)}=uv'+u'v$ and so it is true for $n=1$.


[Remark, $n=1$ corresponds to differentiating once, the result is nothing more than the usual product rule for two functions.]

Assume the statement is true for $n=k$, prove it is true for $n=k+1$.


$\begin{array}{rcl}f^{(k+1)}(x)&=&{ d\over dx} f^{(k)}(x)\\
{ \;}&=&\displaystyle{ d\over dx} \sum^k_{r=0} {k\choose r} u^{(r)}v^{(k-r)}\\
&=&\displaystyle
\sum^k_{r=0} {k\choose r} u^{(r+1)}v^{(k-r)}+
\sum^k_{r=0} {k\choose r} u^{(r)}v^{(k+1-r)}\\
&=&\displaystyle
\sum^{k+1}_{s=1} {k\choose s-1} u^{(s)}v^{(k+1-s)}+

\sum^k_{r=0} {k\choose r} u^{(r)}v^{(k+1-r)}\\\end{array}$

where we changed variables to $s=r+1$ in the first summation.

$\begin{array}{lcl}
f^{(k+1)}(x) &= &\displaystyle{k\choose k} u^{(k+1)}v^{(0)}+{k\choose 0} u^{(0)}v^{(k+1)}\\&&\displaystyle +\sum^{k}_{s=1} {k\choose s-1} u^{(s)}v^{(k+1-s)}\\&&\displaystyle +
\sum^k_{r=1} {k\choose r} u^{(r)}v^{(k+1-r)}\\&= &\displaystyle u^{(k+1)} + v^{(k+1)}\\&&\displaystyle +\sum^{k}_{r=1} \left({k\choose r-1} +{k\choose r}\right) u^{(r)}v^{(k+1-r)} \\
&= &\displaystyle \sum^{k+1}_{r=0} \left({k\choose r-1} +{k\choose r}\right) u^{(r)}v^{(k+1-r)}
\\&= &\displaystyle \sum^{k+1}_{r=0} \left({k\choose r-1} +{k\choose r}\right) u^{(r)}v^{(k+1-r)}
\\&= &\displaystyle \sum^{k+1}_{r=0} {k+1\choose r}u^{(r)}v^{(k+1-r)}\end{array}$
and so it is proven true for $n=k+1$. Therefore by induction the statement is true for all integers $n\ge 0$.

COROLLARY?

Can you apply this formula (or otherwise) and find an expression for the nth derivative of the function $\displaystyle y={u\over v}$?

Find a formula for the $n$-th derivative of $f(x) = \displaystyle {e^x\over 1 - x}$.

Question
Find a formula for the $n$-th derivative of  $f(x) = \displaystyle {e^x\over 1 - x}$. That is find a formula for $f^{(n)}(x)$. Also write out the derivative for $n = 1, 2, 3$.

A formula that is used here for the $n$-th derivative of a function that is a product of two other functions, $f(x)=u(x).v(x)$ was proven in ANOTHER POST (CLICK ME). The result is that

$${d^n \over dx^n } (uv)=\sum^n_{r=0} {n\choose r} u^{(r)}v^{(n-r)}$$
This formula is used to write down the $n$-th derivative of 
$f(x) = \displaystyle{e^x\over 1 - x}$. 

Theorem (harder) If $f(x) = uv$ where $u$ and $v$ are functions of $x$ then the nth derivative of this product is given by $$f^{(n)}(x)={d^n \over dx^n } (uv)=\sum^n_{r=0} {n\choose r} u^{(r)}v^{(n-r)}$$

This is probably outside of most high school syllabuses but many of you should extend yourself beyond the syllabus a bit. So here goes!To get an idea of what the result says, differentiate the product $y=uv$ several times. So find the first derivative, the second derivative, the third derivative, and see what pattern you get there..fill in the details here..

First derivative,
${d \over dx } (uv) = u'v+uv'$

Second derivative,
${d^2 \over dx^2 } (uv)=$

Third derivative,
${d^3 \over dx^3 } (uv)=$

Fourth derivative,
${d^4 \over dx^4 } (uv)=$

Here is the general case when you differentiate $n$ times.

THEOREM (harder):
If $f(x) = uv$ where $u$ and $v$ are functions of $x$ then the
nth derivative of this product is given by
$$f^{(n)}(x)={d^n \over dx^n } (uv)=\sum^n_{r=0} {n\choose r} u^{(r)}v^{(n-r)}$$
where we use the notation for the $r^{th}$ derivative
$$u^{(r)}={ d^r\over dx^r} u$$
(When it's a derivative there are brackets around the 'index')
And notation wise we will then have
$$\displaystyle{ d\over dx}\left( u^{(r)}\right) = { d\over dx} { d^r\over dx^r} u={ d^{r+1}\over dx^{r+1}}=u^{(r+1)}$$

PROOF: 

By induction on $n$.
For $n=0$, $f^{0}(x)=f(x)=uv$ true. and

$ {0\choose 0} u^{(0)}v^{(0)}=uv$ and so it is true for $n=0$.

[Remark, $n=0$ corresponds to not differentiating at all, so it should be the original product $uv$.]

For $n=1$,

$ {1\choose 0} u^{(0)}v^{(1)}+{1\choose 1} u^{(1)}v^{(0)}=uv'+u'v$ and so it is true for $n=1$.


[Remark, $n=1$ corresponds to differentiating once, the result is nothing more than the usual product rule for two functions.]

Assume the statement is true for $n=k$, prove it is true for $n=k+1$.


$\begin{array}{rcl}f^{(k+1)}(x)&=&{ d\over dx} f^{(k)}(x)\\
{ \;}&=&\displaystyle{ d\over dx} \sum^k_{r=0} {k\choose r} u^{(r)}v^{(k-r)}\\
&=&\displaystyle
\sum^k_{r=0} {k\choose r} u^{(r+1)}v^{(k-r)}+
\sum^k_{r=0} {k\choose r} u^{(r)}v^{(k+1-r)}\\
&=&\displaystyle
\sum^{k+1}_{s=1} {k\choose s-1} u^{(s)}v^{(k+1-s)}+

\sum^k_{r=0} {k\choose r} u^{(r)}v^{(k+1-r)}\\\end{array}$

where we changed variables to $s=r+1$ in the first summation.

$\begin{array}{lcl}
f^{(k+1)}(x) &= &\displaystyle{k\choose k} u^{(k+1)}v^{(0)}+{k\choose 0} u^{(0)}v^{(k+1)}\\&&\displaystyle +\sum^{k}_{s=1} {k\choose s-1} u^{(s)}v^{(k+1-s)}\\&&\displaystyle +
\sum^k_{r=1} {k\choose r} u^{(r)}v^{(k+1-r)}\\&= &\displaystyle u^{(k+1)} + v^{(k+1)}\\&&\displaystyle +\sum^{k}_{r=1} \left({k\choose r-1} +{k\choose r}\right) u^{(r)}v^{(k+1-r)} \\
&= &\displaystyle \sum^{k+1}_{r=0} \left({k\choose r-1} +{k\choose r}\right) u^{(r)}v^{(k+1-r)}
\\&= &\displaystyle \sum^{k+1}_{r=0} \left({k\choose r-1} +{k\choose r}\right) u^{(r)}v^{(k+1-r)}
\\&= &\displaystyle \sum^{k+1}_{r=0} {k+1\choose r}u^{(r)}v^{(k+1-r)}\end{array}$
and so it is proven true for $n=k+1$. Therefore by induction the statement is true for all integers $n\ge 0$.

EXAMPLE OF USE IS HERE <-- CLICKME