One Rule To Multiply Them: The Product Rule

At this point, if we are presented with functions that look like this: we are not to be sent away without a derivative (should we choose to seek one).  If on the other hand, we are presented with a function more like this , we may be more troubled in finding an answer.  Given only our rule pertaining to sums, we may be tempted (I was anyway) to think of multiplication as an extension of addition.

If for instance, instead of we had a constant value of 3, the equation would become .  Because we know how multiplication works, we could turn that into with our rule for sums, we could then derive that as . Again, because we know how multiplication works, that looks a lot like .

Alas, we don't have a constant in place of .  Thinking it through though, instead of the 3, if we just plug back in, it almost seems like .  When I got here, it felt like I was on to something. But, what if we'd done it the other way and replaced with a constant?  That would seem to imply that .  So, which is it?  Both!  In fact, when it means .

None of the proceeding is help towards the proof.  It's really just rambling to show the logic I used when I was thinking through the proof.  I also cheated and used some theories of limits I haven't proven yet (and don't intend to any time soon).

Moving on to the proof.  We'll start with our equation

Given what we know about derivatives, we know we need to get it looking like this:

If we do that to the left side, we know we have to do it to the right side, so becomes

This is where trickery part 1 comes into play.  We're going to add and subtract a value.  This doesn't change the equation, because it zeroes itself out, but it does allow for the algebra we will need to proceed.  The value we're adding is
Now, we'll slap that thing into the middle of our existing equation (bolded for your pleasure) and we get

Holy monster equation batman!  Actually it's big enough that we're going to split it down the middle like so:



Now it's time for some simplification to make it easier to read. Time for factoring!  If you look hard enough you'll see that on the left side, we have a common factor of and on the right side, there is a common factor of


So, pulling those out, we now have an equation that looks like this:



Getting closer to what we need.  Here's where the limit trickery comes into play.

Because of the aforementioned limit theorems, our previous step is the equivalent of this:



One more piece of limit mystery and we get this:



Those outside limits tale care of themselves leaving us with:


Adding the left side of our equation back in for context we have:



Which of course proves that