One Rule to Divide Them All: The Quotient Rule

Now that we can find the derivatives of equations containing sums or products, it's time we move on to equations involving quotients. For instance, given function which represents how are we to find the derivative?  The Quotient Rule is designed for just such a purpose.  So, how do we find the rule?  Well, you could Bing it, or we can think through it using what we know about derivatives.

For starters, we need to remember that the derivative is a rate of change.  We can recall that



Looking at our original equation, , we can think about it as the quotionent of two functions and .  This means that .  Applying the definition of a derivative, we get

 That's pretty messy, so we're just going to think about the top part for now: .  Algebra being what it is, we can work some trickery up here then add it back into the full equation later.

The first thing to do is to combine the two fractions utilizing common denominators giving us:

Simple is usually better, but to make any progress, we need to add something to this equation.  Keeping in mind we're only looking at the top of the equation, we need to find a way to get the patterns and to show up in some meaning full way.  I know it's a vague goal, but it helps to explain why our next step is to add  and subtract (so that it zeroes out) the value .  Given this new addition and some rearranging, our equation now looks like this:



Simplifying, we get:
Now we factor out a from the left side and from the right side and we get:

It's now time to plug this back into our full equation which gives us:

Simplifying, we get:

Using some limit trickery, we know this equal to

Further limit work lets us pull out a part of the denominator on both sides because the limit of a product is the product of the limits:

We're left with two limits we can solve, and two that are very close to the difference equation we're looking for.  We'll solve the simple limits first to clean it up and we have:

We can now pull out part of the numerator on both sides using the same limit theorem. On the left we pull out  while on the right we pull out .  These likewise are easy to solve and making our full equation:

We can now clearly see the difference quotients in there and can reduce to or even more simply:


We have now added the Quotient Rule to our bag of tricks.  We know that when it means

Let's apply this to our original function:

This means that and .  While and . Using our new found rule, we know that

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