If f(x) is a differentiable function, its derivative f′(x) is also a function, so it may have a derivative (either finite or not). This function is called the second derivative of f(x), because it is the derivative of a derivative, and is denoted by f′′. So, f′′=(f′)′.
The following notations are used for the second derivative: dx2d2y, dx2d2(f(x)), y′′, f′′(x), D2y, D2f(x).
Example 1. Find the second derivative of f(x)=x+11.
We have that f′(x)=−(x+1)21⋅(x+1)′=−(x+1)21.
Now, f′′(x)=(f′(x))′=(−(x+1)21)′=(x+1)32.
f′′(x) can be interpreted as the slope of f′(x) at (x,f′(x)). In other words, it is the rate of change of the slope of f(x).
If s=s(t) is the position function of an object that moves along a straight line, we know that its first derivative represents the velocity v(t) of the object as a function of time: v(t)=s′(t)=dtds.
The instantaneous rate of change of velocity with respect to time is called the acceleration a(t) 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)=dtdv=s′′(t)=dt2d2s.
Example 2. If the position function of the object is s(t)=3t2+ln(t+1), find the acceleration of the object after 1 second.
We need to find the second derivative first.
s′(t)=(3t2+ln(t+1))′=6t+t+11.
s′′(t)=(s′(t))′=(6t+t+11)′=6−(t+1)21.
So, the acceleration after 1 second is s′′(1)=6−(1+1)21=5.75 s2m.
Similarly, if the function y=f(x) has a finite second derivative, then its derivative (finite or not) is called the third derivative of y=f(x) and is denoted in the following ways: dx3d3y, dx3d3(f(x)), y′′′, f′′′(x), D3y, D3f(x).
The third derivative is the derivative of the second derivative: f′′′=(f′′)′. So, f′′′(x) can be interpreted as the slope of f′′(x) or as the rate of change of f′′(x).
We can interpret the third derivative physically in the case when the function is the position function s=s(t) of an object that moves along a straight line. Because s′′′=(s′′)′=a′, the third derivative of the position function is the derivative of the acceleration function and is called jerk: j=dtda=dt3d3s.
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 (n−1)−th derivative of the function y=f(x) 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).
So, y(n)=(y(n−1))′.
The following notations are used for the n-th derivative: dxndny, dxndn(f(x)), y(n), f(n)(x), D(n)y, D(n)f(x).
In Lagrange's and Cauchy's notations y(n), f(n), D(n)y, D(n)f, if we want to explicitly state with respect to what variable we take the derivative, we write yxn(n), fxn(n), Dxn(n)y, Dxn(n)f.
For example, yx2′′ is the second derivative of y with respect to x, f′′′t3 is the third derivative of f with respect to t. Note that x2 is not x squared, it is just the short record for xx, it denotes that the derivative is taken two times. The same is true for t3.
So, acceleration can be written as a=st2′′, and jerk is j=st3′′′.
Example 3. Find the fourth derivative of y=21x4−61x3+2x2+34x−21.
We have that y′=2x3−2x2+4x+34.
y′′=(y′)′=(2x3−2x2+4x+34)′=6x2−4x+4.
y′′′=(y′′)′=(6x2−4x+4)′=12x−4.
y(4)=(y′′′)′=(12x−4)′=12.
Note that the derivatives of an order higher than 4 will be zero in this case.