Definition of Higher-Order Derivatives

If f(x){f{{\left({x}\right)}}} is a differentiable function, its derivative f(x){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(x){f{{\left({x}\right)}}}, because it is the derivative of a derivative, and is denoted by f{f{''}}. So, f=(f){f{''}}={\left({f{'}}\right)}'.

The following notations are used for the second derivative: d2ydx2\frac{{{{d}}^{{2}}{y}}}{{{d}{{x}}^{{2}}}}, d2dx2(f(x))\frac{{{{d}}^{{2}}}}{{{d}{{x}}^{{2}}}}{\left({f{{\left({x}\right)}}}\right)}, y{y}'', f(x){f{''}}{\left({x}\right)}, D2y{{D}}^{{2}}{y}, D2f(x){{D}}^{{2}}{f{{\left({x}\right)}}}.

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

We have that f(x)=1(x+1)2(x+1)=1(x+1)2{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(x)=(f(x))=(1(x+1)2)=2(x+1)3{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(x){f{''}}{\left({x}\right)} can be interpreted as the slope of f(x){f{'}}{\left({x}\right)} at (x,f(x)){\left({x},{f{'}}{\left({x}\right)}\right)}. In other words, it is the rate of change of the slope of f(x){f{{\left({x}\right)}}}.

If s=s(t){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(t){v}{\left({t}\right)} of the object as a function of time: v(t)=s(t)=dsdt.{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(t){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(t)=v(t)=dvdt=s(t)=d2sdt2{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(t)=3t2+ln(t+1){s}{\left({t}\right)}={3}{{t}}^{{2}}+{\ln{{\left({t}+{1}\right)}}}, find the acceleration of the object after 11 second.

We need to find the second derivative first.

s(t)=(3t2+ln(t+1))=6t+1t+1{s}'{\left({t}\right)}={\left({3}{{t}}^{{2}}+{\ln{{\left({t}+{1}\right)}}}\right)}'={6}{t}+\frac{{1}}{{{t}+{1}}}.

s(t)=(s(t))=(6t+1t+1)=61(t+1)2{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 11 second is s(1)=61(1+1)2=5.75 ms2{s}''{\left({1}\right)}={6}-\frac{{1}}{{{\left({1}+{1}\right)}}^{{2}}}={5.75}\ \frac{{m}}{{{s}}^{{2}}}.

Similarly, if the function y=f(x){y}={f{{\left({x}\right)}}} has a finite second derivative, then its derivative (finite or not) is called the third derivative of y=f(x){y}={f{{\left({x}\right)}}} and is denoted in the following ways: d3ydx3\frac{{{{d}}^{{3}}{y}}}{{{d}{{x}}^{{3}}}}, d3dx3(f(x))\frac{{{{d}}^{{3}}}}{{{d}{{x}}^{{3}}}}{\left({f{{\left({x}\right)}}}\right)}, y{y}''', f(x){f{'''}}{\left({x}\right)}, D3y{{D}}^{{3}}{y}, D3f(x){{D}}^{{3}}{f{{\left({x}\right)}}}.

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

We can interpret the third derivative physically in the case when the function is the position function s=s(t){s}={s}{\left({t}\right)} of an object that moves along a straight line. Because s=(s)=a{s}'''={\left({s}''\right)}'={a}', the third derivative of the position function is the derivative of the acceleration function and is called jerk: j=dadt=d3sdt3{j}=\frac{{{d}{a}}}{{{d}{t}}}=\frac{{{{d}}^{{3}}{s}}}{{{d}{{t}}^{{3}}}}.

Thus, the jerk j{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 (n1)th{\left({n}-{1}\right)}-{t}{h} derivative of the function y=f(x){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(x){y}={f{{\left({x}\right)}}}.

So, y(n)=(y(n1)){\color{red}{{{{y}}^{{{\left({n}\right)}}}={\left({{y}}^{{{\left({n}-{1}\right)}}}\right)}'}}}.

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

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

For example, yx2{y}''_{{{{x}}^{{2}}}} is the second derivative of y{y} with respect to x{x}, ft3{f{'''}}_{{{{t}}^{{3}}}} is the third derivative of f{f{}} with respect to t{t}. Note that x2{{x}}^{{2}} is not x{x} squared, it is just the short record for xx{x}{x}, it denotes that the derivative is taken two times. The same is true for t3{{t}}^{{3}}.

So, acceleration can be written as a=st2{a}={s}''_{{{{t}}^{{2}}}}, and jerk is j=st3{j}={s}'''_{{{{t}}^{{3}}}}.

Example 3. Find the fourth derivative of y=12x416x3+2x2+43x12{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=2x32x2+4x+43{y}'={2}{{x}}^{{3}}-{2}{{x}}^{{2}}+{4}{x}+\frac{{4}}{{3}}.

y=(y)=(2x32x2+4x+43)=6x24x+4{y}''={\left({y}'\right)}'={\left({2}{{x}}^{{3}}-{2}{{x}}^{{2}}+{4}{x}+\frac{{4}}{{3}}\right)}'={6}{{x}}^{{2}}-{4}{x}+{4}.

y=(y)=(6x24x+4)=12x4{y}'''={\left({y}''\right)}'={\left({6}{{x}}^{{2}}-{4}{x}+{4}\right)}'={12}{x}-{4}.

y(4)=(y)=(12x4)=12{{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.