Proof of Various
In this section we’re going to prove many of the various
derivative facts, formulas and/or properties that we encountered in the early
part of the Derivatives chapter. Not all of them will be proved here and some
will only be proved for special cases, but at least you’ll see that some of
them aren’t just pulled out of the air.
Definition of Derivative
Sum/Difference of Two Functions :
This is easy enough to prove using the definition of the
derivative. We’ll start with the sum
of two functions. First plug the sum
into the definition of the derivative and rewrite the numerator a little.
Now, break up the fraction into two pieces and recall that
the limit of a sum is the sum of the limits.
Using this fact we see that we end up with the definition of the
derivative for each of the two functions.
The proof of the difference of two functions in nearly
identical so we’ll give it here without any explanation.
Proof of Constant
Times a Function :
This is property is very easy to prove using the
definition provided you recall that we can factor a constant out of a
limit. Here’s the work for this
Proof of the
Derivative of a Constant :
This is very easy to prove using the definition of the
derivative so define and the use the definition of the
Power Rule :
There are actually three proofs that we can give here and
we’re going to go through all three here so you can see all of them. However, having said that, for the first two
we will need to restrict n to be a
positive integer. At the time that the
Power Rule was introduced only enough information has been given to allow the
proof for only integers. So, the first two proofs are really to be read at that
The third proof will work for any real number n.
However, it does assume that you’ve read most of the Derivatives chapter
and so should only be read after you’ve gone through the whole chapter.
In this case as noted above we need to assume that n is a positive integer. We’ll use the definition of the derivative
and the Binomial Theorem in this theorem.
The Binomial Theorem tells us that,
are called the binomial coefficients and is the factorial.
So, let’s go through the details of this proof. First, plug into the definition of the derivative and
use the Binomial Theorem to expand out the first term.
Now, notice that we can cancel an and then each term in the numerator will
have an h in them that can be
factored out and then canceled against the h in the denominator. At
this point we can evaluate the limit.
For this proof we’ll again need to restrict n to be a positive integer. In this case if we define we know from the alternate limit form of the
definition of the derivative that the derivative is given by,
Now we have the following formula,
You can verify this if you’d like by simply multiplying
the two factors together. Also, notice
that there are a total of n terms
in the second factor (this will be important in a bit).
If we plug this into the formula for the derivative we see
that we can cancel the and then compute the limit.
After combining the exponents in each term we can see that
we get the same term. So, then
recalling that there are n terms in
second factor we can see that we get what we claimed it would be.
To completely finish this off we simply replace the a with an x to get,
In this proof we no longer need to restrict n to be a positive integer. It can now be any real number. However, this proof also assumes that
you’ve read all the way through the Derivative chapter. In particular it needs both Implicit Differentiation and Logarithmic
Differentiation. If you’ve not
read, and understand, these sections then this proof will not make any sense
So, to get set up for logarithmic differentiation let’s
first define then take the log of both sides, simplify
the right side using logarithm properties and then differentiate using
Finally, all we need to do is solve for and then substitute in for y.
Before moving onto the next proof, let’s notice that in all
three proofs we did require that the exponent, n, be a number (integer in the first two, any real number in the
third). In the first proof we couldn’t have
used the Binomial Theorem if the exponent wasn’t a positive integer. In the second proof we couldn’t have factored
if the exponent hadn’t been a positive
integer. Finally, in the third proof we
would have gotten a much different derivative if n had not been a constant.
This is important because people will often misuse the power
rule and use it even when the exponent is not a number and/or the base is not a
Product Rule :
As with the Power Rule above, the Product Rule can be proved
either by using the definition of the derivative or it can be proved using
Logarithmic Differentiation. We’ll show
both proofs here.
This proof can be a little tricky when you first see it so
let’s be a little careful here. We’ll
first use the definition of the derivative on the product.
On the surface this appears to do nothing for us. We’ll first need to manipulate things a
little to get the proof going. What
we’ll do is subtract out and add in to the numerator. Note that we’re really just adding in a
zero here since these two terms will cancel.
This will give us,
Notice that we added the two terms into the middle of the
numerator. As written we can break up
the limit into two pieces. From the
first piece we can factor a out and we can factor a out of the second piece. Doing this gives,
At this point we can use limit properties to write,
The individual limits in here are,
The two limits on the left are nothing more than the
definition the derivative for and respectively. The upper limit on the right seems a little
tricky, but remember that the limit of a constant is just the constant. In this case since the limit is only
concerned with allowing h to go to
zero. The key here is to recognize
that changing h will not change x and so as far as this limit is
concerned is a constant. Note that the function is probably not a
constant, however as far as the limit is concerned the function can be
treated as a constant. We get the
lower limit on the right we get simply by plugging into the function
Plugging all these into the last step gives us,
This is a much quicker proof but does presuppose that
you’ve read and understood the Implicit
Differentiation and Logarithmic Differentiation
sections. If you haven’t then this
proof will not make a lot of sense to you.
First write call the product y and take the log of both sides and use a property of logarithms
on the right side.
Next, we take the derivative of both sides and solve for .
Finally, all we need to do is plug in for y and then multiply this through the
parenthesis and we get the Product Rule.
Quotient Rule :
Again, we can do this using the definition of the derivative
or with Logarithmic Definition.
First plug the quotient into the definition of the
derivative and rewrite the quotient a little.
To make our life a little easier we moved the h in the denominator of the first step
out to the front as a . We also wrote the numerator as a single
rational expression. This step is
required to make this proof work.
Now, for the next step will need to subtract out and add in
to the numerator.
The next step is to rewrite things a little,
Note that all we did was interchange the two
denominators. Since we are multiplying
the fractions we can do this.
Next, the larger fraction can be broken up as follows.
In the first fraction we will factor a out and in the second we will factor a out.
We can now use the basic properties of limits to write
The individual limits are,
The first two limits in each row are nothing more than the
definition the derivative for and respectively. The middle limit in the top row we get
simply by plugging in . The final limit in each row may seem a
little tricky. Recall that the limit
of a constant is just the constant.
Well since the limit is only concerned with allowing h to go to zero as far as its
concerned and are constants since changing h will not change x. Note that the function
is probably not a constant, however as far as the limit is concerned the
function can be treated as a constant.
Plugging in the limits and doing some rearranging gives,
There’s the quotient rule.
Now let’s do the proof using Logarithmic
Differentiation. We’ll first call the
quotient y, take the log of both
sides and use a property of logs on the right side.
Next, we take the derivative of both sides and solve for .
Next, plug in y
and do some simplification to get the quotient rule.
We’ll start off the proof by defining and noticing that in terms of this
definition what we’re being asked to prove is,
Let’s take a look at the derivative of (again, remember we’ve defined and so u
really is a function of x) which we
know exists because we are assuming that is differentiable. By definition we have,
Note as well that,
and notice that
and so is continuous at
Now if we
assume that we can rewrite the definition of to get,
Now, notice that (1)
is in fact valid even if we let and so is valid for any value of h.
Next, since we
also know that is differentiable we can do something
similar. However, we’re going to use a
different set of letters/variables here for reasons that will be apparent in
a bit. So, define,
we can go
through a similar argument that we did above so show that is continuous at and that,
Do not get
excited about the different letters here all we did was use k instead of h and let . Nothing fancy here, but the change of
letters will be useful down the road.
Okay, to this
point it doesn’t look like we’ve really done anything that gets us even close
to proving the chain rule. The work
above will turn out to be very important in our proof however so let’s get
going on the proof.
What we need to
do here is use the definition of the derivative and evaluate the following
Note that even
though the notation is more than a little messy if we use instead of u we need to remind ourselves here that u really is a function of x.
Let’s now use (1)
to rewrite the and yes the notation is going to be
unpleasant but we’re going to have to deal with it. By using (1),
the numerator in the limit above becomes,
If we then define and we can use (2)
to further write this as,
Notice that we were able to cancel a to simplify things up a little. Also, note that the was intentionally left that way to keep the
mess to a minimum here, just remember that here as that will be important here in a
bit. Let’s now go back and remember
that all this was the numerator of our limit, (3). Plugging this into (3)
Notice that the
h’s canceled out. Next, recall that and so,
But, if ,
as we’ve defined k anyway, then by
the definition of w and the fact
that we know is continuous at we also know that,
that . Using all of these facts our limit becomes,
This is exactly
what we needed to prove and so we’re done.