Professor Dave here, let’s learn about higher
derivatives.
We’ve learned how to take the derivative
of a variety of functions, and we know that
the derivative of a function is itself a function.
But if the derivative is a function, what
is to stop us from taking the derivative of
the derivative? Well, nothing, in fact. We
can do this, and we call it the second derivative,
with respect to the original function. So
if we have some function f, and we take the
derivative, we get f prime. And if we try
to get f prime, prime, or f double prime,
which is the derivative of the derivative
of f, we must call this the second derivative
of f. In Leibniz notation, if we take the
derivative of some function y equals f of
x, that would be dy over dx, so if we take
the derivative of that, by placing d over
dx out here, that can be represented as d
squared y over dx squared. So this notation
also indicates a second derivative, just like
f double prime.
For simple polynomials, finding the second
derivative is not hard to do at all. Take
three x squared plus four x minus five. Taking
the derivative of this by using the power
rule, we get six x plus four as the first
derivative, and then we just take the derivative
of this to get six as the second derivative.
All we did was take the derivative twice.
Of course this could get a little trickier.
Take x sine x. To get the first derivative
we need the product rule, so remembering how
that works from a previous tutorial, let’s
put the functions where they go and take these
two derivatives. We should get x cosine x
plus sine x. Then to get the second derivative,
we just take the derivative of this new function.
For the first term, again we need the product
rule, so we set it up just as we are used
to, and evaluate the derivatives. That should
give us negative x sine x plus cosine x, and
then we can’t forget this other term here,
which simply becomes cosine x. So that gives
us negative x sine x plus two cosine x as
our answer.
So why is this useful? Well let’s recall
that a derivative represents the rate of change
of a function. This means that a second derivative
represents the rate of change of the rate
of change of the function. That may sound
confusing, but there is an extremely good
use for this in kinematics, the study of motion,
which we covered in the classical physics
course. We can make a graph of the position
of an object versus time. This could be a
falling object, a speeding car, pretty much
anything. As we’ve mentioned before, the
derivative of this position function will
be the rate of change in this function, or
the change in position with respect to time,
and that’s velocity. So velocity is the
derivative of position. But the derivative
of this velocity function will be the change
in velocity with respect to time, and that’s
acceleration. Position is meters, velocity
is meters per second, and acceleration is
meters per second, per second, or meters per
second squared. That means that acceleration
is the second derivative of position, and
that A of t equals v prime of t equals s double
prime of t. This is one way of understanding
the acceleration due to gravity on earth as
it relates to Galileo’s law for the position
of an object in freefall. He derived the equation
4.9 t squared by observation, and if we take
the derivative of this, we get 9.8 t, and
if we take the derivative of that, we get
9.8, and this is why acceleration due to gravity
on earth is a constant equal to 9.8 meters
per second squared. This understanding makes
it extremely simple to answer questions in
physics that would be tedious without calculus.
For example, say we have this function, s
of t, that represents the position of an object,
t cubed minus five t squared plus eight t
minus one. What will be the acceleration of
the object after five seconds? Without calculus,
this would be very cumbersome to solve. But
it’s easy for us now, because we just find
the second derivative, and plug in the number
five. The first derivative, v of t, will be
three t squared minus ten t plus eight. Then
the second derivative, A of t, must be six
t minus ten. Since this represents the acceleration
at any time t, then to find the acceleration
at precisely five seconds, we just plug in
five. A of five will be six times five, or
thirty, minus ten, which is twenty meters
per second squared.
Now although we won’t go beyond the second
derivative all that often, there is nothing
stopping us from taking a third derivative,
or a fourth, or as many as we like. The third
derivative will simply be the derivative of
the second derivative, which we can write
as f triple prime, or as d cubed y over d
x cubed. For the fourth derivative, it’s
getting to be too many primes, so we just
write f and then the number 4 in parentheses
as a superscript. From there, taking the nth
derivative will look like this, with the n
up here, and we can get it by differentiating
a function n times. So that’s all there
is to higher derivatives. These will return
later when we go back to graphing functions,
but for now let’s check comprehension.
