# Definition of Higher-Order Derivatives

If f(x) is a differentiable function, then its derivative f'(x) is also a function, so may have a derivative (finite or not). This function is called second derivative of f(x) because it is derivative of derivative and denoted by f'' . So, f''=(f')'.

Following notations are used for second derivative: (d^2y)/(dx^2), (d^2)/(dx^2)(f(x)), y'', f''(x), D^2 y, D^2f(x).

Example 1. Find second derivative of f(x)=1/(x+1).

We have that f'(x)=-1/(x+1)^2*(x+1)'=-1/(x+1)^2.

Now, f''(x)=(f'(x))'=(-1/(x+1)^2)'=2/(x+1)^3.

f''(x) can be interpreted as the slope of the f'(x) at (x,f'(x)). In other words, it is the rate of change of the slope of the f(x).

If s=s(t) is the position function of an object that moves in 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)=(ds)/(dt).

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)=(dv)/(dt)=s''(t)=(d^2s)/(dt^2).

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

We need to find second derivative first.

s'(t)=(3t^2+ln(t+1))'=6t+1/(t+1).

s''(t)=(s'(t))'=(6t+1/(t+1))'=6-1/(t+1)^2.

So, acceleration after 1 second is s''(1)=6-1/(1+1)^2=5.75\ m/s^2.

Similarly if function y=f(x) has finite second derivative, then its derivative (finite or not) is called third derivative of y=f(x) and denoted in following ways: (d^3y)/(dx^3), (d^3)/(dx^3)(f(x)), y''', f'''(x), D^3 y, D^3f(x).

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

We can interpret the third derivative physically in the case where 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 the jerk: j=(da)/(dt)=(d^3s)/(dt^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 fourth derivative, fifth derivative etc.

In general if (n-1)-th derivative of function y=f(x) exists and finite, then its derivative is called n-th derivative or derivative of n-th order of the function y=f(x).

So, color(red)(y^((n))=(y^((n-1)))').

Following notations are used for n-th derivative: (d^ny)/(dx^n), (d^n)/(dx^n)(f(x)), y^((n)), f^((n))(x), D^((n))y, D^((n))f(x).

In Lagrange and Cauchy notations y^((n)), f^((n)), D^((n))y, D^((n))f, if we want to explicitly state with respect to what variable we take derivative, we write y_(x^n)^((n)), f_(x^n)^((n)), D_(x^n)^((n)) y, D_(x^n)^((n))f.

For example, y''_(x^2) is second derivative of y with respect to x, f'''_(t^3) is third derivative of f with respect to t. Note, that x^2 is not x squared, it is just short record for x x, it denotes that derivative is taken two times. Similarly for t^3.

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

Example 3. Find fourth derivative of y=1/2 x^4-1/6 x^3+2x^2+4/3 x-1/2.

We have that y'=2x^3-2x^2+4x+4/3.

y''=(y')'=(2x^3-2x^2+4x+4/3)'=6x^2-4x+4.

y'''=(y'')'=(6x^2-4x+4)'=12x-4.

y^((4))=(y''')'=(12x-4)'=12.

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