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.