3.5 Higher Order Derivatives

Larry Musolino

3.5 Higher Order Derivatives

Higher Order Derivatives

The derivative function f'(x) provides a rate of change for a function f(x). In some cases, we may also be interested in the rate of change for the derivative function f'(x) itself.

This leads us to the notation of higher order derivatives. For example if we take the derivative of f'(x), assuming the first derivative exists, we arrive at the second derivative f”(x).

We can differentiate the second derivative f”(x) and arrive at the third derivative f”'(x) and so on.

In the next chapter we will see that the second derivative provides an indication for the concavity of a graph.

Second Derivative

Let $y^{'} = f^{'} (x) .$
The second derivative of $f$ is the derivative of $y^{'} = f^{'} (x) .$
Using prime notation, this is $f^{″} (x)$ or $y^{″} .$ You can read this aloud as “y double prime.”
Using Leibniz notation, the second derivative is written $\frac{d^{2} y}{d x^{2}}$ or $\frac{d^{2} f}{d x^{2}} .$ This is read aloud as “the second derivative of f”.

Example 1

Find $f^{″} (x)$ for $f (x) = 3 x^{7}$

First, we need to find the first derivative:
$f^{'} (x) = 21 x^{6}$

Then we take the derivative of that function to determine the second derivative:
$f^{″} (x) = \frac{d}{d x} (f^{'} (x)) = \frac{d}{d x} (21 x^{6}) = 126 x^{5}$

If $f (x)$ represents the position of a particle at time $x,$ then $v (x) = f^{'} (x)$ will represent the velocity (rate of change of the position) of the particle and $a (x) = v^{'} (x) = f^{″} (x)$ will represent the acceleration (the rate of change of the velocity) of the particle.

You are probably familiar with acceleration from driving or riding in a car. The speedometer tells you your velocity (speed). When you leave from a stop and press down on the accelerator, you are accelerating – increasing your speed.

Example 2

The height (feet) of a particle at time $t$ seconds is $f (t) = t 3 - 4 t 2 + 8 t .$ Find the height, velocity and acceleration of the particle when t = 0, 1, and 2 seconds.

$f (t) = t 3 - 4 t 2 + 8 t s o f (0) = 0 f e e t, f (1) = 5 f e e t, a n d f (2) = 8 f e e t .$

The velocity is $v (t) = f ‘ (t) = 3 t 2 - 8 t + 8$ so $v (0) = 8 f t / s, v (1) = 3 f t / s, a n d v (2) = 4 f t / s .$ At each of these times the velocity is positive and the particle is moving upward, increasing in height.

The acceleration is $a (t) = f ” (t) = 6 t - 8$ so $a (0) = - 8 f t / s 2, a (1) = - 2 f t / s 2 a n d a (2) = 4 f t / s 2 .$

At time 0 and 1, the acceleration is negative, so the particle’s velocity would be decreasing at those points – the particle was slowing down. At time 2, the velocity is positive, so the particle was increasing in speed.

Example 3

Find $f^{″} (x)$ for $f (x) = 4 x^{5} + 3 x^{2} + 2 x$

First, we need to find the first derivative:
$f^{'} (x) = 20 x^{4} + 6 x + 2$

Then we take the derivative of that function to determine the second derivative:
$f^{″} (x) = 80 x^{3} + 6$

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