Definition of Higher-Order Derivatives

If $$${f{{\left({x}\right)}}}$$$ is a differentiable function, its derivative $$${f{'}}{\left({x}\right)}$$$ is also a function, so it may have a derivative (either finite or not). This function is called the second derivative of $$${f{{\left({x}\right)}}}$$$, because it is the derivative of a derivative, and is denoted by $$${f{''}}$$$. So, $$${f{''}}={\left({f{'}}\right)}'$$$.

The following notations are used for the second derivative: $$$\frac{{{{d}}^{{2}}{y}}}{{{d}{{x}}^{{2}}}}$$$, $$$\frac{{{{d}}^{{2}}}}{{{d}{{x}}^{{2}}}}{\left({f{{\left({x}\right)}}}\right)}$$$, $$${y}''$$$, $$${f{''}}{\left({x}\right)}$$$, $$${{D}}^{{2}}{y}$$$, $$${{D}}^{{2}}{f{{\left({x}\right)}}}$$$.

Example 1. Find the second derivative of $$${f{{\left({x}\right)}}}=\frac{{1}}{{{x}+{1}}}$$$.

We have that $$${f{'}}{\left({x}\right)}=-\frac{{1}}{{{\left({x}+{1}\right)}}^{{2}}}\cdot{\left({x}+{1}\right)}'=-\frac{{1}}{{{\left({x}+{1}\right)}}^{{2}}}$$$.

Now, $$${f{''}}{\left({x}\right)}={\left({f{'}}{\left({x}\right)}\right)}'={\left(-\frac{{1}}{{{\left({x}+{1}\right)}}^{{2}}}\right)}'=\frac{{2}}{{{\left({x}+{1}\right)}}^{{3}}}$$$.

$$${f{''}}{\left({x}\right)}$$$ can be interpreted as the slope of $$${f{'}}{\left({x}\right)}$$$ at $$${\left({x},{f{'}}{\left({x}\right)}\right)}$$$. In other words, it is the rate of change of the slope of $$${f{{\left({x}\right)}}}$$$.

If $$${s}={s}{\left({t}\right)}$$$ is the position function of an object that moves along a straight line, we know that its first derivative represents the velocity $$${v}{\left({t}\right)}$$$ of the object as a function of time: $$${v}{\left({t}\right)}={s}'{\left({t}\right)}=\frac{{{d}{s}}}{{{d}{t}}}.$$$

The instantaneous rate of change of velocity with respect to time is called the acceleration $$${a}{\left({t}\right)}$$$ of the object. Thus, the acceleration function is the derivative of the velocity function and is therefore the second derivative of the position function: $$${a}{\left({t}\right)}={v}'{\left({t}\right)}=\frac{{{d}{v}}}{{{d}{t}}}={s}''{\left({t}\right)}=\frac{{{{d}}^{{2}}{s}}}{{{d}{{t}}^{{2}}}}$$$.

Example 2. If the position function of the object is $$${s}{\left({t}\right)}={3}{{t}}^{{2}}+{\ln{{\left({t}+{1}\right)}}}$$$, find the acceleration of the object after $$$1$$$ second.

We need to find the second derivative first.

$$${s}'{\left({t}\right)}={\left({3}{{t}}^{{2}}+{\ln{{\left({t}+{1}\right)}}}\right)}'={6}{t}+\frac{{1}}{{{t}+{1}}}$$$.

$$${s}''{\left({t}\right)}={\left({s}'{\left({t}\right)}\right)}'={\left({6}{t}+\frac{{1}}{{{t}+{1}}}\right)}'={6}-\frac{{1}}{{{\left({t}+{1}\right)}}^{{2}}}$$$.

So, the acceleration after $$$1$$$ second is $$${s}''{\left({1}\right)}={6}-\frac{{1}}{{{\left({1}+{1}\right)}}^{{2}}}={5.75}\ \frac{{m}}{{{s}}^{{2}}}$$$.

Similarly, if the function $$${y}={f{{\left({x}\right)}}}$$$ has a finite second derivative, then its derivative (finite or not) is called the third derivative of $$${y}={f{{\left({x}\right)}}}$$$ and is denoted in the following ways: $$$\frac{{{{d}}^{{3}}{y}}}{{{d}{{x}}^{{3}}}}$$$, $$$\frac{{{{d}}^{{3}}}}{{{d}{{x}}^{{3}}}}{\left({f{{\left({x}\right)}}}\right)}$$$, $$${y}'''$$$, $$${f{'''}}{\left({x}\right)}$$$, $$${{D}}^{{3}}{y}$$$, $$${{D}}^{{3}}{f{{\left({x}\right)}}}$$$.

The third derivative is the derivative of the second derivative: $$${f{'''}}={\left({f{''}}\right)}'$$$. So, $$${f{'''}}{\left({x}\right)}$$$ can be interpreted as the slope of $$${f{''}}{\left({x}\right)}$$$ or as the rate of change of $$${f{''}}{\left({x}\right)}$$$.

We can interpret the third derivative physically in the case when the function is the position function $$${s}={s}{\left({t}\right)}$$$ of an object that moves along a straight line. Because $$${s}'''={\left({s}''\right)}'={a}'$$$, the third derivative of the position function is the derivative of the acceleration function and is called jerk: $$${j}=\frac{{{d}{a}}}{{{d}{t}}}=\frac{{{{d}}^{{3}}{s}}}{{{d}{{t}}^{{3}}}}$$$.

Thus, the jerk $$${j}$$$ is the rate of change of acceleration. It is aptly named because a large jerk means a sudden change in acceleration, which causes an abrupt movement in a vehicle.

Similarly, we can define the fourth derivative, the fifth derivative, etc.

In general, if the $$${\left({n}-{1}\right)}-{t}{h}$$$ derivative of the function $$${y}={f{{\left({x}\right)}}}$$$ exists and is finite, its derivative is called the n-th derivative or the derivative of n-th order of the function $$${y}={f{{\left({x}\right)}}}$$$.

So, $$${\color{red}{{{{y}}^{{{\left({n}\right)}}}={\left({{y}}^{{{\left({n}-{1}\right)}}}\right)}'}}}$$$.

The following notations are used for the n-th derivative: $$$\frac{{{{d}}^{{n}}{y}}}{{{d}{{x}}^{{n}}}}$$$, $$$\frac{{{{d}}^{{n}}}}{{{d}{{x}}^{{n}}}}{\left({f{{\left({x}\right)}}}\right)}$$$, $$${{y}}^{{{\left({n}\right)}}}$$$, $$${{f}}^{{{\left({n}\right)}}}{\left({x}\right)}$$$, $$${{D}}^{{{\left({n}\right)}}}{y}$$$, $$${{D}}^{{{\left({n}\right)}}}{f{{\left({x}\right)}}}$$$.

In Lagrange's and Cauchy's notations $$${{y}}^{{{\left({n}\right)}}}$$$, $$${{f}}^{{{\left({n}\right)}}}$$$, $$${{D}}^{{{\left({n}\right)}}}{y}$$$, $$${{D}}^{{{\left({n}\right)}}}{f{}}$$$, if we want to explicitly state with respect to what variable we take the derivative, we write $$${{y}_{{{{x}}^{{n}}}}^{{{\left({n}\right)}}}}$$$, $$${{f}_{{{{x}}^{{n}}}}^{{{\left({n}\right)}}}}$$$, $$${{D}_{{{{x}}^{{n}}}}^{{{\left({n}\right)}}}}{y}$$$, $$${{D}_{{{{x}}^{{n}}}}^{{{\left({n}\right)}}}}{f{}}$$$.

For example, $$${y}''_{{{{x}}^{{2}}}}$$$ is the second derivative of $$${y}$$$ with respect to $$${x}$$$, $$${f{'''}}_{{{{t}}^{{3}}}}$$$ is the third derivative of $$${f{}}$$$ with respect to $$${t}$$$. Note that $$${{x}}^{{2}}$$$ is not $$${x}$$$ squared, it is just the short record for $$${x}{x}$$$, it denotes that the derivative is taken two times. The same is true for $$${{t}}^{{3}}$$$.

So, acceleration can be written as $$${a}={s}''_{{{{t}}^{{2}}}}$$$, and jerk is $$${j}={s}'''_{{{{t}}^{{3}}}}$$$.

Example 3. Find the fourth derivative of $$${y}=\frac{{1}}{{2}}{{x}}^{{4}}-\frac{{1}}{{6}}{{x}}^{{3}}+{2}{{x}}^{{2}}+\frac{{4}}{{3}}{x}-\frac{{1}}{{2}}$$$.

We have that $$${y}'={2}{{x}}^{{3}}-{2}{{x}}^{{2}}+{4}{x}+\frac{{4}}{{3}}$$$.

$$${y}''={\left({y}'\right)}'={\left({2}{{x}}^{{3}}-{2}{{x}}^{{2}}+{4}{x}+\frac{{4}}{{3}}\right)}'={6}{{x}}^{{2}}-{4}{x}+{4}$$$.

$$${y}'''={\left({y}''\right)}'={\left({6}{{x}}^{{2}}-{4}{x}+{4}\right)}'={12}{x}-{4}$$$.

$$${{y}}^{{{\left({4}\right)}}}={\left({y}'''\right)}'={\left({12}{x}-{4}\right)}'={12}$$$.

Note that the derivatives of an order higher than 4 will be zero in this case.