Higher order derivatives

You already know how to find derivatives of different functions. Also, you know that the derivative of a function is itself a function. So, you can take a derivative of a derivative! It is called the second derivative with respect to the original function, and in this topic, you are going to learn more about it.

Second derivative

Suppose we have some function $f(x)$. If we take the derivative, we'll get $f'(x)$. If we take the derivative again, we'll get $(f'(x))'$ or $f''(x)$ $-$ it is the derivative of the derivative of $f(x)$, and it is called the second derivative of $f(x)$.

We know that if we have a function $y = f(x)$, the first derivative of $y$ is denoted by $\dfrac{dy}{dx}$. If we take the derivative of that, it may be represented as

$\dfrac{d}{dx} \left(\dfrac{dy}{dx}\right) = \dfrac{d^2y}{dx^2}.$Thus, this notation indicates the second derivative. We can also write it as $f''(x)$.

Let's consider an example. Suppose we have a function

$f(x) = 6x^3 + 2x + 8.$Let's find the first derivative:

$f'(x) = 18x^2 + 2.$Now, let's find the derivative of that function, that is, the second derivative:

$f''(x) = 36x.$

What is the physical meaning of the second derivative?

We already know, that the derivative represents the rate of change of a function. Consequently, the second derivative represents the rate of change of the rate of change of our function. This is very often used in kinematics $-$ the study of motion. Imagine a graph of the position of an object versus time. It may be a falling object, a speeding cyclist, or something else. The derivative of this position function will be the rate of change in position with respect to time $-$ that is, velocity.

The derivative of this velocity function is the change of velocity with respect to time, that is, acceleration. Position is expressed in meters, velocity is expressed in meters per second, and acceleration is expressed in meters per second per second, or meters per second squared. Thus, acceleration is the second derivative of position.

Let's consider an example:

$s(t) = t^4 + 5t - 7.$What will be the acceleration of the object after 8 seconds? The concept of the second derivative will help us answer this question. We just need to find the second derivative of that function and plug in $t=8$:

$s'(t) = 4t^3 + 5 \ \Rightarrow \ \ s''(t) = 12t^2.$If $t=8$, then $s''(8) = 768 \ \text{m/s}^2$.

Higher order derivatives

Similarly, we can take a third derivative, a fourth, and so on. The third derivative of $y=f(x)$ is the derivative of the second derivative, which is denoted by $f'''(x)$ or $\dfrac{d^3 y}{dx^3}$. For the fourth derivative, it can be tedious to write four primes, so we usually write $f^{(4)}$. Thus, the $n$-th derivative is denoted by $f^{(n)}$, and we get it by differentiating the function $n$ times.

Conclusion

In this topic, we talked about higher order derivatives. An $n$-th derivative is found by differentiating a function $n$ times. Also, we examined the physical meaning of the second derivative $-$ acceleration – and considered several examples.