Welcome to Calculus, I'm Professor Ghrist. We're about to begin lecture
11 on differentiation rules. From your previous exposure to calculus, you probably learned how to
compute derivatives quickly without the definition by following
a certain set of rules or laws. We're going to cover those
rules in this lesson. Such a topic has the potential
to be boring, but pay attention because we're going
to cover why the rules hold, and in doing so, we'll get a better
grasp of the notion of a derivative. Recall from our previous
lesson that we had two different ways of defining the derivative. One is in terms of a rate of change,
it's a limit as the input approaches a, the change in the output over
the change in the input. Our second definition,
a little bit stronger definition, is in terms of first order variation. One changes the input to f
by a small amount, h and looks at how the change in
the output depends upon h. If you like, it's the coefficient
of the first order term in the Taylor Series of f at a. From these definitions flows
the interpretation of the derivative as the rate of change of the output
with respect to change in the input. And this is the interpretation that
you will want to have memorized. A few remarks are in order. First of all, the derivative,
as you know, most certainly depends upon the input value
a that you are examining. Secondly, a derivative
concerns rates of change. We're not actually measuring
how much the output changes. We are looking at the rate of change,
as the change in the input is getting closer and
closer to zero. Lastly, the derivative at a is
telling you a rate of change. It's telling you how fluctuations,
or changes, h, in the input are perhaps amplified, if the derivative is positive and
larger than 1. Or damped, if the derivative is,
say, less than 1. Or reversed, in the case where
the derivative is negative. That is a good interpretation for
the derivative, one that will serve you
well in this course. Now there are several rules associated
with computing derivatives. I'm going to present these rules
compactly using a simplified differential notation that you're familiar with from
your previous exposure to calculus. The first rule is that of linearity. This is consists of two parts. First what you might call a summation
rule, that the derivative of the function u plus v is the derivative
of u plus the derivative of v. The second part of linearity
says that if we multiply u by a constant c and take its derivative, what we get is that constant,
c times the derivative of u. The second rule, the product rule, states that the derivative of the product
of two functions, u and v, is u times the derivative of v plus
v times the derivative of u. Lastly, and perhaps most importantly,
the chain rule states that the derivative of a composition of two functions is
the product of the individual derivatives. We'll have some more to say
about that in a moment. For now,
let's focus on the rules of linearity. The summation rule can be
visualized rather simply, keeping track of how u changes and
how v changes. And keeping track of how u plus v
changes is really not that difficult. Likewise, when you
multiply u by a constant, what happens to its rate of change? It is similarly scaled. That picture is at least reasonable. There's another way to think
about these rules, as well. This is a way that I often think about it. If you take the sum of the two
functions and then differentiate, it's the same as differentiating
the pieces and then adding them together. Or likewise, you can multiply by
a constant and then differentiate, or you can differentiate and
then multiply by a constant, whichever path you travel,
you will get to the same place. These pictures illuminate but do not
really justify the differentiation rules. To do a better justification,
let's use our definition of the derivative as a first order of variation and
employ the language big-O. First of all,
u + v evaluated at x + h means what? Why don't we take u, evaluate it at x + h, and add to it v evaluate it at x + h? Now by our definition
of the derivative of u, we know that the term on
the left is u plus the derivative of u times h plus something
in big-O of h squared. Likewise with v. Now, let's rearrange this
sum by ascending order of h. The zeroth order terms are u + v. What are the first order terms,
those that have an h? Well, on the left, we have du/dx, and we add to it the term from the right,
dv/dx. Everything else is something
in big-O h squared plus something in big O of h squared. That is in big-O of h squared, naturally. So we see that the first order coefficient is the sum of the derivatives,
as expected. Likewise, when you multiply
u by a constant, c, and evaluate that at x + h,
it is simply c times u + du/dx times h + something
big-O of h squared. The zeroth order term is c times u. The first order term is
c times du/dx times h. Everything else is a constant
times big-O of h squared. But remember in big-O,
constants don't count, and so we see that the first order term,
the derivative is c times du/dx. Likewise for the product rule,
it's not hard to visualize the first order variation of u
times v in terms of du and dv. But to write this out using our
language of big-O is very simple. If we take (u x v) and
evaluate it at (x + h), that is u at (x + h)v at (x + h). We can substitute in the expansions for
u and v and perform this multiplication. If we multiply these terms as if they
were polynomials, what do we get? The zeroth order term is simply u times v. All of those terms that are of first order in h have coefficients u
times dv/dx + v times du/dx. Everything else in that multiplication,
as you can check, is going to be in big-O of h squared. So the derivative can be read off
from the first order of the term, and we get the product rule. Lastly comes the chain rule. This is perhaps the most important
differentiation rule, and it deserves a little bit of exposition. The chain rule concerns the composition
of two functions, u composed with v. That means you do v first and
feed the output of v into the input of u. Now, what do the rates of change, what does the derivative look like,
in this case? Well, dv is telling you at
what rate the output of v changes with the respect to
change in the input, du tells you at what rate the out of u changes
with respect to change in the input. What is the derivative of the composition? Well, it tells you. When you change the input
to v at a certain rate, at what rate does the output, the final
output of u composed with v, change? And the answer, as one can intuit is
that these rates of change multiply. The justification of the chain rule
is going to follow the same ideas but with a bit more manipulation involved. If we consider u composed with v and evaluate the input at x + h, then, clearly, our first step should
be to expand out v using what we know about the derivative
of v with respect to x. Everything else in the variation
has big-O of h squared. Now, let's change perspective and
think about this from the point of view of u, which takes as its input v. Now we're evaluating u, not at v,
but at v plus some perturbation, some variation term which
we know has the structure of dv/dx times h plus something
in big-O of h squared. Forget about that structure for
the moment. We're going to expand this out. The zeroth order term is u(v). Next comes the derivative of u with
respect to v times this perturbation term plus something in big-O of that
perturbation term squared. And now, here comes the final steps. If we look at that big-O on the right, and view it as a function of h,
then squaring everything in sight, we see that everything has
powers of h that are at least 2. And so we can replace that
by a big-O of h squared. Now for the terms in the middle,
when we distribute the multiplication of du/dv,
with all of the terms in the parentheses, well the latter term,
du/dv times something in big-O of u squared yields again
something in big-O of h squared. And so
combining those big-O's of h squared, we are left with a first
order term in h that has coefficient du/dv x dv/dx. And that is the proper
form the chain rule. Now one has to be careful. The differential notation is
especially misleading at this point. You must evaluate your derivatives
at the correct inputs. Let's look at a simple example. Let's say that f(x) is
the exponential function e to the x. What is the derivative of
f composed with itself? Well, according to the chain rule, it should be df/dx x df/dx. Well, that's not exactly what it is, because we have to be
careful about the inputs. If we were to say evaluate
this derivative at x = 0, then the second term must
be evaluated at x = 0. The first term in this
product is evaluated not at x = 0, but at x = e to the 0, or 1. That would give us a value of e x 1,
which is, of course, e. Now this looks awfully confusing. What's really going on? Well, you know how to
differentiate functions, and you know how to use the chain rule. Let's think about what
f composed with f is. It's really e to the e to the x. And if I were to ask you how do you
differentiate that, you would say well, first, I differentiate the e to the x and
evaluate that at e to the x. That gives me e to the e to the x,
but because it's the chain rule, I have to multiply by the derivative of
that exponent, in this case, e to the x. Now if you evaluate
both of these at x = 0, you get the same computation as
above with the final value of e. Be careful about where you
evaluate your derivatives. There are a number of other
differentiation rules that you may have seen in your previous exposure to
calculus, the reciprocal rule, the quotient rule, the inverse rule. If you remember seeing these,
then take a brief look. If you've not seen them before, then you might wanna work
through the following examples, such as showing that the derivative
of secant is secant times tangent. Or that the derivative of tangent
is really secant squared. Or that the derivative of log of x,
that is the inverse of the exponential function,
is, in fact, 1/x. I'll leave it to you to take a more
careful look at some of these rules and to practice in the homework sets. And so we see the basic rules for
differentiation. But wait, there's more. If you look at the bonus material,
you'll see how some of the ghosts of these rules haunt other
realms of mathematics. In our next lesson, we'll cover one
of the canonical applications of derivatives to linearization.