The second derivative

If we differentiate a function and then differentiate the derivative we get what is called the second derivative.  If the original function is called f(x), we find f´(x) and then differentiate  f´(x) . The outcome is written  f´´(x) .

Example 1

Find the second derivative of the following functions:

a)    f(x) = x³ + x² + x + 1

       f´(x) = 3x² + 2x + 1

       f´´(x) = 6x + 2

b)    f(x) = x⁴ + x³ + x^–1 + 1

       f´(x) = 4x³ + 3x² – x^–2

       f´´(x) = 12x² + 6x + 2x^–3

c)    f(x) = sin x

       f´(x) = cos x

       f´´(x) = –sin x

d)    f(x) = cos x

       f´(x) = –sin x

       f´´(x) = –cos x

e)    f(x) = e^–x

       f´(x) = –e^–x

       f´´(x) = e^–x

f)    f(x) = ln x

      f´(x) = 1/x = x^–1

       f´´(x) = –x^–2

If we define the function f(t) as the position of an object at time t we get the following definitions in physics.

    f(t) is the position of an object after time t.

    f´(t) gives us the velocity of the object at time t.

    f´´(t) gives the acceleration of the object at time t.

Velocity is a measure of the change in position per unit  time. Acceleration is a measure of the change in velocity per unit time.

If  f(t) is measured in metres and t in seconds then  f´(t) is measured in ^m/s and f´´(t) in ^m/s².

Example 2

When an arrow is shot vertically into the air we can find a function h(t) that gives the height of the arrow above the ground at time t. This function depends on various factors such as the strength of the shooter and the type of bow and arrow.
The following function gives the height, h, of the tip  of the arrow, in metres,  t seconds after it has been shot from the bow.

h(t) = – 4.9t² + 24.5t + 2.4.

a)   Find the time it takes for the arrow to hit the ground.

Imagine a coordinate system with the height h on the vertical (y) axis and the time t on the horizontal (x) axis.  Imagine that the shooter stands at the origin of the system and the tip of the arrow leaves the bow at a height of  2.4 m. The arrow goes up and then comes down again,  reaching the ground when the height h is 0. To solve the problem we need to find when the function  h(t) = 0.  We can do this by using a calculator or the quadratic formula.

      The value t ≈ 0 is not quite accurate as we start measuring the time when the arrow is  at a height of 2.4 m .
      The other value of t tells us that the arrow strikes the ground after 5 seconds.

b)   Now we will calculate the height that the arrow reaches.

The arrow slows down as it rises and reaches it's maximum height when the velocity is zero. h´(t) gives the velocity at time t.  So we need to find h´(t)  and solve the equation  h´(t)  = 0.

          h´(t) = –9.8t + 24.5 = 0

           9.8t = 24.5

                t = ^24.5/_9.8 = 2.5 s

        We conclude that the arrow reaches it's maximum height after 2.5 seconds. Now we can use the formula for h(t) to find the maximum height.
         h(2.5) = – 4.9·2.5² + 24.5·2.5 + 2.4 = 33.025 m

       The arrow reaches a height of 33 m.

c)   Now we find the acceleration by differentiating again.

          h´(t) = –9.8t + 24.5

          h´´(t) = –9.8 m/s²

This is a very important result, a value that gives the acceleration due to the gravitational field of the earth. The negative value is because the arrow is slowing down.

 

Let's examine what information the second derivative gives us about the graph of a function. The first derivative tells us about the slope of the graph. If the derivative is positive the slope is increasing, if negative the slope is decreasing.
The second derivative tells us how the slope changes. If the second derivative is positive it means the slope of the graph is increasing if negative the slope is decreasing.
When the sign of the second derivative changes from positive to negative it means the gradient stops increasing and starts decreasing. In other words the graph changes from being convex , (bending up) to being concave (bending down). The reverse happens when the second derivative changes from negative to positive. A  point where the bend of the curve changes is called a point of inflexion. When a continuous function changes between being positive and negative it must go through zero, so in a point of inflextion the second derivative is zero. If you try to draw a tangent in a point of inflexion you will find that it cuts the graph.

Summarising :

When the second derivative is positive the gradient of the graph is increasing and the curve bends upwards, is convex. .
When the second derivative is negative the gradient of the graph is decreasing and the curve bends downwards, is concave.

The derivative of the function f(x) = e^x is f´(x) = e^x and the second derivative f´´(x) = e^x. The second derivative is always positive and therefore the graph is always convex. See the diagram below.

The function g(x) = ln x is only defined for x greater than zero. The derivative is g´(x) = 1/x = x^–1 and the second derivative  g´´(x) = –x^–2 = –1/x². The second derivative is always negative so the graph bends downwards or is concave as the diagram shows.

f(x) = ex is convex . f´´(x) is positive   ( +).

g(x) = ln x is concave. g´´(x) is negative   (–).

Example 3

Find the first and second derivatives of the functions f(x) = x² + 4x + 3 and g(x) = –x² + 4x – 3.

    f(x) = x² + 4x + 3                       g(x) = –x² + 4x – 3

    f´(x) = 2x + 4                             g´(x) = –2x + 4

    f´´(x) = 2                                    g´´(x) = –2

f´´(x) is positive so the graph of f(x) is convex. g´´(x) is negative so the graph of g(x) is concave.

Use the derivative to  find the vertex of each function. The vertex occurs when the gradient of a tangent is zero so:

   f´(x) = 2x + 4 = 0                        g´(x) = –2x + 4 = 0

      2x = –4                                     –2x = –4

        x = –2                                         x = 2

  f(–2) = (–2)² + 4(–2) + 3 = –1      g(2) = –2² + 4·2 – 3 = 1

The graph of  f(x) is convex and therefore has a minimum point  (–2, –1).

The graph of g(x) is concave and therefore has a maximum point  (2, 1). The graphs are shown below.

Example 4

Find the first and second derivatives of the function f(x) = x³ – 3x² + 4, the maximum and minimum points and the point of inflexion (if they exist ). Then find the equation of the tangent in the point of inflexion.

   f(x) = x³ – 3x² + 4

   f´(x) = 3x² – 6x

   f´(x) = 3x² – 6x = 3x(x – 2) = 0

        x = 0 or 2

Now we look at how the signs of the derivative change around the zeros:

  f´(–1) = 3(–1)² – 6(–1) = 3 + 6 = 9    (+)

   f´(1) = 3·1² – 6·1 = –3                       (–)

   f´(3) = 3·3² – 6·3 = 9                         (+)

 We do the same for the second derivative:

 f´´(x) = 6x – 6 = 0

       6x = 6

         x = 1

  f´´(0) = 6·0 – 6 = –6                          (–)

   f´´(2) = 6·2 – 6 = 6                            (+)

Putting all this information together we can make the following table:

The graph is concave when x is less than 1, shown by a curve bending downwards. The graph is convex when x is greater than 1 shown by a curve that bends upwards.  We see that there is a maximum when  x = 0 , a minimum when x = 2 and a point of inflexion when  x = 1.

Now we can calculate the coordinates of these points.

   f(0) = 0³ – 3·0² + 4 = 4    Maximum point  = (0, 4)

   f(2) = 2³ – 3·2² + 4 = 0    Minimum point  = (2, 0)

   f(1) = 1³ – 3·1² + 4 = 2    Point of inflexion = (1, 2)

Now to find the equation of the tangent in the point of inflexion.

The general equation of a tangent to the function f(x) in the point ( a, b ) is  y = f´(a)(x – a) + b. In this case (a, b ) is the point of inflexion, (1, 2).
We have already found that  f´(1) = –3 so we simply need to put this information into the equation for a tangent.

   y = –3(x – 1) + 2

      = –3x + 3 + 2

      = – 3x + 5

The equation of the tangent in the pont of inflexion is y = –3x + 5.

We can check the working by drawing the graph of  f(x) and of the tangent on the graphical calculator. Notice how the tangent cuts the graph in the point of inflextion.