# 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.