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The Second Derivative
When we take the derivative of a function f (x), we get a derived function f ′(x), called the deriva- tive or first derivative. If we now take the derivative of this function f ′(x), we get another derived function f ′′(x), which is called the second derivative of f. In differential notation this is written d^2 f dx^2. If we think of^
d dx as an operator, we can think of^
d^2 dx^2 as representing the operator being applied twice. The second derivative of f (x) tells us the rate of change of the derivative f ′(x) of f (x).
More specifically, the second derivative describes the curvature of the function f. If the function curves upward, it is said to be concave up. If the function curves downward, then it is said to be concave down. The behavior of the function corresponding to the second derivative can be summarized as follows
- The second derivative is positive (f ′′(x) > 0): When the second derivative is positive, the function f (x) is concave up.
- The second derivative is negative (f ′′(x) < 0): When the second derivative is negative, the function f (x) is concave down.
- The second derivative is zero (f ′′(x) = 0): When the second derivative is zero, it corresponds to a possible inflection point. If the second derivative changes sign around the zero (from positive to negative, or negative to positive), then the point is an inflection point. This corresponds to a point where the function f (x) changes concavity. If the second derivative does not change sign (ie. it goes from positive to zero to positive), then it is not an inflection point (x = 0 with f (x) = x^4 is an example of this).
Let us consider the following functions, and look at how their derivatives correspond to their graphs.
0
1
2
3
4
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2
x^
x
0
1
2
3
4
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2
x^
x
0
1
2
3
4
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2
2x
x
0
2
4
6
-1 -0.5 0 0.5 1 1.5 2 2.5 3
x/(x-1)
x
Example 1 Find f ′(x) and f ′′(x) if f (x) = x^3. Compare these derivatives to the graph above.
Solution By repeated applications of the power rule, we find that f ′(x) = 3x^2 , and f ′′(x) = 6x. For all x, the first derivative f ′(x) > 0, so the function f (x) is always increasing. Considering the second derivative, we see that for x < 0 we have f ′′(x) < 0, so f (x) is concave down. For x > 0 we have f ′′(x) > 0, so f (x) is concave up. At x = 0, f ′′(x) = 0, and since the second derivative changes signs around 0, this is an inflection point, as can be seen above.
Example 2 Find f ′(x) and f ′′(x) if f (x) = x^2. Compare these derivatives to the graph above. Solution By repeated applications of the power rule, we find that f ′(x) = 2x, and f ′′(x) = 2. For all x, the second derivative f ′′(x) > 0, so the function f (x) is always concave up. Considering the first derivative, we see that for x < 0 we have f ′(x) < 0, so f (x) is decreasing. For x > 0 we have f ′(x) > 0, so f (x) is increasing. At x = 0, f ′(x) = 0, which corresponds to a critical point, where f (x) is not changing, and is in fact, the minimum of f (x).
Example 3 Find f ′(x) and f ′′(x) if f (x) = 2x. Compare these derivatives to the graph above. Solution By repeated applications of the power rule, we find that f ′(x) = 2, and f ′′(x) = 0. For all x, the first derivative f ′(x) > 0, so the function f (x) is always increasing. Also, for all x, the second derivative is 0. This corresponds to a graph that does not have any concavity, such as the line above.
Example 4 Find f ′(x) and f ′′(x) if f (x) = (^) x−x 1. Compare these derivatives to the graph above. Solution In this situation we cannot just use the product rule to calculate the derivatives. Instead, we must use the quotient rule. Let u(x) = x and v(x) = x − 1. We find that u′(x) = 1 = v′(x), as these are both linear functions. We find the derivative as
f ′(x) =
v(x)u′(x) − u(x)v′(x) v(x)^2
x − 1 − x (x − 1)^2
(x − 1)^2
In order to calculate the second derivative, we once again need to use the quotient rule. Let u(x) = −1 and v(x) = (x − 1)^2 = x^2 − 2 x + 1. We find that u′(x) = 0 and v′(x) = 2x − 2. Using the quotient rule we find
f ′′(x) =
v(x)u′(x) − u(x)v′(x) v(x)^2
(x − 1)^2 · 0 − (−1)(2x − 2) ((x − 1)^2 )^2
2(x − 1) (x − 1)^4
(x − 1)^3
In summary, f ′(x) = (^) (x−−^1 1) 2 and f ′′(x) = (^) (x−^2 1) 3. Now that we’ve completed the arduous task
of calculating the above derivatives, we can continue to compare them to the graph above. The derivative is negative for all x 6 = 1, and is not defined for x = 1. This indicates that for all x 6 = 1 the function f (x) is decreasing. However, when we look at the graph of f , we see that values of f for x > 1 are greater than values of f for x < 1. This strange behavior occurs because the derivative is not defined for x = 1, where the function value essentially increases by an infinite amount. The behavior of f ′′(x) is not so complicated, and it shows us that for x < 1 the function is concave down, as f ′′(x) < 0. For x > 1, f ′′(x) > 0, so the function is concave up. At x = 0 the function and its derivatives are not defined, so it makes little sense to talk about the influence of the second derivative. Nevertheless, this point is something of an inflection point (although not tech- nically), as the concavity of the function changes here. This example illustrates that functions and their derivatives may have very unexpected behavior at points where their denominator’s go to zero.
In the previous examples we found the derivatives and compared their behavior to the graphs of the function that we already knew. Based on this insight, we should be able to sketch a function based on knowledge of its derivatives. The general procedure is as follows
0
5
10
-2 -1 0 1 2 3 4
2x^3 - 7x^2 + 5
x
Our final application of the second derivative is less mathematical, and more physical. If we have a function of position, say y(t), the first derivative corresponds to velocity, and the second derivative corresponds to acceleration. Thus, we can rewrite Newton’s force equation as
F = ma = m
d^2 y dt^2
If we know the force and the mass of the object, then the above is a differential equation which we can solve in order to find the acceleration, velocity, and finally position. Recall that a falling object has position described by
y(t) = 9. 8 t^2
velocity described by
v(t) = y′(t) = 19. 6 t
and finally acceleration described by
a(t) = y′′(t) = 19. 6
where the above quantities are measured in meters and seconds (velocity is meters per second, where acceleration is meters per second squared).
