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