The Second Derivative.
Lecture 19
As we have seen, the derivative, f 0(x), of a function can provide us
with very important information about the function, f (x), itself. In
fact, for many problems the derivative can play a more important role
than the original function. That suggests that maybe the derivative
of the derivative is worth some careful study. And indeed it is. Sure,
enough, that is our next challenge.
So let y = f (x) be some function. Then its derivative, variously
denoted by
d
d
dy
,
f (x) and
f,
y 0, f 0, f 0(x),
dx dx
dx
is itself a function. The derivative of this derivative, called the second derivative of f (x), is denoted by
d2y
d2
d2
y , f , f (x),
,
f (x), and
f.
dx2
dx2
dx2
And, of course, this second derivative measures the rate of change of
the rate of change. More about that later, but first, you should note
that computing this second derivative requires no new techniques —
to compute the second derivative one just computes the derivative of
the derivative:
00
00
00
Example 1. If y = 4x3, then
d2y
d
dy
2
2
= 12x , so
=
(12x
) = 24x.
dx
dx2 dx
Similarly, if f (x) = x3 − 4x + 1/x + 5, then
d
f (x) = f 0(x) = 3x2 − 4 − x−2,
dx
so
d2
d 0
−3
f
(x)
=
f
(x)
=
6x
+
2x
.
dx2
dx
As we mentioned, the derivative f 0(x) of a function f (x) measures the rate of change of f with x, so the second derivative f 00(x)
measures the rate of change of the rate of change of f (x). A legitimate question at this point is, “What does that mean in practice?”
Perhaps the example most familiar to most of us is simply that of
distance, velocity, and acceleration.
Example 2. An airplane starts down the runway en route to
taking off. It’s distance in feet from the beginning of the runway as
a function of time, measured in seconds, is
s(t) = 2.5t2.
So, for example, after 45 seconds the plane will have travelled about
a mile. The velocity of the plane during this takeoff run is the rate
of change of its distance from the start of the runway, that is, its
velocity is simply the derivative of s(t). Thus,
d
s(t) = 5t ft/sec.
dt
For example, after 45 seconds the velocity will be 225 ft/sec or about
150 mph. That is, after 45 seconds the rate of change of the distance
of the plane from the beginning of the runway is about 150 mph.
But how fast is the plane speeding up as it travels down the runway?
That is, the rate of change of the velocity, or the acceleration, which
is just the derivative of v(t), the second derivative of s(t)!
v(t) =
d
d2
a(t) = v(t) = 2 s(t) = 5 ft/sec2.
dt
dt
feet per second per second. This means that every second during the
takeoff run, the plane’s velocity increases about 5 ft/sec or about 3.5
miles per hour.
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Remark. Distance, such as s(t) in this example, is something
we see. But it is static; you see two runway lights ahead and sense
that one is, say, twice as far away as the other. The rate of change
of distance, that is, the velocity, is also something you see, but it is
dynamic. As the plane zips down the runway you see the velocity in
how fast your distance is changing, in how the runway lights move
across the window. The acceleration, however, is something you
probably don’t see, but, boy, you feel it. There must be some force to
change the velocity and that is what you feel in your back. (Of course,
when the plane lands, it slows down, so the velocity is decreasing and
the acceleration is negative. This you feel too — you just hope it
isn’t so great that you feel it as your nose smashes against the seat
in front you.)
One of the most effective ways to understand the relationships
among functions and their derivatives and second derivatives is through
an analysis of what they mean graphically. So let’s first go back and
review the graphical significance of the derivative. Here it is in a
nutshell:
f (x)
increasing
decreasing
horizontal tangent
We remind you of this graphically:
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f 0(x)
f 0(x) > 0
f 0(x) < 0
f 0(x) = 0
f 0(x)
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f (x)
1
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1
1
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1
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But there is something else we might note from these examples.
What happens to the graph y = f (x) when the derivative f 0(x) is
increasing? Or decreasing? And sure enough, When f 0(x) is increasing, the graph of f (x) is turning upward and when f 0(x) is decreasing
the graph of f (x) is turning downward. Let’s be a bit more precise.
Let y = f (x) be differentiable on some interval. Then at
some point x = a on that interval we say that f (x) is
•
concave upward at x = a if at points x near a the
graph of f (x) lies above the line tangent at x = a;
•
concave downward at x = a if at points x near
a the graph of f (x) lies below the line tangent at x = a.
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concave upwards
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f 0(x) increasing
f 00(x) > 0
concave downwards
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f 0(x) decreasing
f 00(x) < 0
Let’s summarize what we have so far in the table:
f (x)
f 0(x)
increasing
f 0(x) > 0
decreasing
f 0(x) < 0
horizontal tangent
f 0(x) = 0
concave upward
f 0(x) increasing
concave downward f 0(x) decreasing
no concavity
horizontal tangent
f 00(x)
—
—
—
f 00(x) > 0
f 00(x) < 0
f 00(x) = 0
If the function f is continuous and changes direction from increasing
to decreasing or vice-versa at x = a, then f has an extreme value
at x = a. If f changes concavity, from up to down or vice-versa, at
x = a, then we say that f has an inflection point at x = a.
So
If y = f (x) is continuous and has an inflection
point at x = a, then f 0(x) has a critical point
at x = a.
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You should be alert for a couple of things. The first, and pretty
obvious one, is that end points don’t get in the picture for inflection
points — these simply don’t happen at end points. The other, more
subtle, is that inflection points occur at critical points of f 00(x), but
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they need not occur there.
Example 3. The derivative of f (x) = x4, namely, f 0(x) = 4x3,
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has
a critical point at x = 0, but y = f (x) is always concave upwards
and so has no inflection points.
f 0(x) = 4x3
f (x) = x4
However. if we can compute the second derivative, then it tips us
off to all inflection points. In that case:
The function f (x) has an inflection point at
x = a if and only if f 00(x) changes sign at x = a.
So let’s conclude this with a few examples.
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Example 4. Consider f (x) = 3x5 − 5x4 + 4x. Then
f 0(x) = 15x4 − 20x3 + 4,
and
f 00(x) = 60x3 − 60x2 = 60x2(x − 1).
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So the only critical points for the derivative are at x = 0 and x = 1.
Thus, to find the inflection points of f (x) we need only check the
signs of the second derivative f 00(x) at these points. And we find
 00
(x < 0
 f (x) < 0
f 00(x) < 0 ,
When
0 < x < 1 we have
 00
f (x) > 0
x>1
so there is an inflection point for f (x) at x = 11 but not at x = 0.
And sure enough, here are their graphs: Untitled-1
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y = f (x)
1
y = f 0(x)
y = f 00(x)
Example 5. This time consider f (x) = xe−x. Now we have
f 0(x) = e−x − xe−x = (1 − x)e−x,
and
f 00(x) = −e−x − (e−x − xe−x) = (x − 2)e−x.
So the only critical point for the derivative f 0(x) occurs when f 00(x) =
0 and that is only at x = 2. And since f 00(x) changes sign from
negative to positive at x = 2, that is an inflection point for y = f (x).
Note, also that the concavity changes there from down to up. And
again the graphs make this clear — try it!!
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