Title: Default for 11.3 Higher Derivatives
Course: HELM Workbooks 2023-2024

The derivative of a derivative

You have already learnt how to calculate the derivative of a function using a table of derivatives and applying some basic rules. By differentiating the function, \(y(x)\), we obtain the derivative,\( \dfrac{dy}{dx}\). By repeating the process we can obtain higher derivatives.

Example

Calculate the first, second and third derivatives of \(y = x^4 + 6x^2\).

Solution

The first derivative is \[ \text{first derivative} =4x^3 + 12x\] To obtain the second derivative we differentiate the first derivative. \[ \text{second derivative} = 12x^2 + 12 \] The third derivative is found by differentiating the second derivative. \[ \text{third derivative} = 24x + 0 = 24x \]

Exercise

Calculate \( \dfrac{d^2y}{dx^2}\) and \( \dfrac{d^3y}{dx^3}\) given \(y = [value]\).
1. First find \( \dfrac{dy}{dx}\)
\( \dfrac{dy}{dx} =\)
2. Now obtain the second derivative:
\( \dfrac{d^2y}{dx^2} =\)
3. Finally, obtain the third derivative:
\( \dfrac{d^3y}{dx^3} =\)
Feedback: We are given the function \( y= [value].\) 1. \( \dfrac{dy}{dx} = [value], \) 2. \( \dfrac{d^2y}{dx^2} = [value] ,\) 3. \( \dfrac{d^3y}{dx^3} = [value] .\)

Notation for derivatives

Just as there is a notation for the first derivative so there is a similar notation for higher derivatives. Consider the function, \(y(x)\). We know that the first derivative is \( \dfrac{dy}{dx} \) or \( \dfrac{d}{dx}(y)\) which is the instruction to differentiate the function \(y(x)\). The second derivative is calculated by differentiating the first derivative, that is \[ \text{second derivative} = \dfrac{d}{dx} \left( \dfrac{dy}{dx} \right) \] So, using a fairly obvious adaptation of our derivative notation, the second derivative is denoted by \( \dfrac{d^2y}{dx^2} \) and is read as `dee two \(y\) by dee \(x\) squared'. This is often written more concisely as \(y''\).

In similar manner, the third derivative is denoted by \( \dfrac{d^3y}{dx^3}\) or \(y'''\) and so on. So, referring to the previous example, we could have written \[ \text{first derivative} = \dfrac{dy}{dx} =4x^3 + 12x\] \[ \text{second derivative} =  \dfrac{d^2y}{dx^2} = 12x^2 + 12 \] \[ \text{third derivative} =  \dfrac{d^3y}{dx^3} = 24x + 0 = 24x \]

Key Point

If \(y = y(x)\) then its first, second and third derivatives are denoted by:

\[ \dfrac{dy}{dx} \text{ or } y' \] \[ \dfrac{d^2y}{dx^2} \text{ or } y'' \] \[ \dfrac{d^3y}{dx^3} \text{ or } y''' \]

In most examples we use \(x\) to denote the independent variable and \(y\) the dependent variable. However, in many applications, time \(t\) is the independent variable. In this case a special notation is used for derivatives. Derivatives with respect to \(t\) are often indicated using a dot notation, so \( \dfrac{dy}{dt}\) can be written as \(\dot{y}\), pronounced `y dot'. Similarly, a second derivative with respect to \(t\) can be written as \(\ddot{y}\), pronounced `y double dot'.

Key Point

If \(y = y(t)\) then

\(\dot{y}\) stands for \( \dfrac{dy}{dt}\), and

\(\ddot{y}\) stands for \( \dfrac{d^2y}{dt^2}\).

Evaluating higher derivatives

We may need to evaluate higher derivatives at specific points. We use an obvious notation. The second derivative of \(y(x)\), evaluated at say, \(x = 2\), is written as \( \dfrac{d^2y}{dx^2} (2)\), or more simply as \(y''(2)\). The third derivative evaluated at \(x = -1\) is written as \( \dfrac{d^3y}{dx^3} (-1)\) or \(y'''(-1)\).

Exercise

Given \(y = [value]\), find to 3 decimal places
1. \( \dfrac{dy}{dx}([value])\)
\( \dfrac{dy}{dx}([value]) =\)
2. \( \dfrac{d^2y}{dx^2}([value])\)
\( \dfrac{d^2y}{dx^2}([value]) =\)
3. \( \dfrac{d^3y}{dx^3}([value]) \)
\( \dfrac{d^3y}{dx^3}([value]) =\)
Feedback: We are given the function \( y= [value]\). 1. \( \dfrac{dy}{dx} = [value] \), which when evaluated at \( x = [value]\) is \( [value] \). 2. \( \dfrac{d^2y}{dx^2} = [value] \), which when evaluated at \( x = [value]\) is \( [value] \). 3. \( \dfrac{d^3y}{dx^3} = [value] \), which when evaluated at \( x = [value]\) is \( [value] \).

Exercise

Calculate \( \dfrac{dy}{dx} \), \( \dfrac{d^2y}{dx^2}\) and \( \dfrac{d^3y}{dx^3}\) given \(y = [value]\).
1. \( \dfrac{dy}{dx} =\)
2. \( \dfrac{d^2y}{dx^2} =\)
3. \( \dfrac{d^3y}{dx^3} =\)
Feedback: We have the function \( y = [value].\) 1. \( \dfrac{dy}{dx} = [value], \) 2. \( \dfrac{d^2y}{dx^2} = [value], \) 3. \( \dfrac{d^3y}{dx^3} = [value] .\)

Exercise

Calculate \( \dfrac{dy}{dx} \), \( \dfrac{d^2y}{dx^2}\) and \( \dfrac{d^3y}{dx^3}\) given \(y = [value] .\)
1. \( \dfrac{dy}{dx} =\)
2. \( \dfrac{d^2y}{dx^2} =\)
3. \( \dfrac{d^3y}{dx^3} =\)
Feedback: We are given the function \( y=[value].\) 1. \( \dfrac{dy}{dx} = [value], \) 2. \( \dfrac{d^2y}{dx^2} = [value] ,\) 3. \( \dfrac{d^3y}{dx^3} = [value]. \)

Exercise

Calculate \( \dfrac{dy}{dx} \), \( \dfrac{d^2y}{dx^2}\) and \( \dfrac{d^3y}{dx^3}\) given \(y = [value] \).
1. \( \dfrac{dy}{dx} =\)
2. \( \dfrac{d^2y}{dx^2} =\)
3. \( \dfrac{d^3y}{dx^3} =\)
Feedback: We have the function \(y= [value].\) 1. \( \dfrac{dy}{dx} = \displaystyle{[value]}, \) 2. \( \dfrac{d^2y}{dx^2} = \displaystyle{[value]}, \) 3. \( \dfrac{d^3y}{dx^3} = \displaystyle{[value]}. \)

Exercise

Calculate \( \dfrac{dy}{dx} \), \( \dfrac{d^2y}{dx^2}\) and \( \dfrac{d^3y}{dx^3}\) given \(y = [value] \).
1. \( \dfrac{dy}{dx} =\)
2. \( \dfrac{d^2y}{dx^2} =\)
3. \( \dfrac{d^3y}{dx^3} =\)
Feedback: We have the function \(y=[value].\) 1. \( \dfrac{dy}{dx} = \displaystyle{[value]} ,\) 2. \( \dfrac{d^2y}{dx^2} = \displaystyle{[value]}, \) 3. \( \dfrac{d^3y}{dx^3} = \displaystyle{[value]}. \)

Exercise

Given \(y = [value] \), find \(y''\), \(y'''\) at \(x = [value]\).
1. \( y''([value]) =\)
2. \( y'''([value]) =\)
Feedback: We are given the function \( y = [value] = [value].\) We start by calculating the first derivative: \( \displaystyle \dfrac{dy}{dx} = [value] .\) 1. \( \dfrac{d^2y}{dx^2} = [value] \), which when evaluated at \( x = [value]\) is \( [value]. \) 2. \( \dfrac{d^3y}{dx^3} = [value] \), which when evaluated at \( x = [value]\) is \( [value] .\)

Exercise

Given \(y = \sin ([value]x) + \cos(x) \), find \(y''\), \(y'''\) at \(x = [value]\) to 3 decimal places.
1. \( y''([value]) =\)
2. \( y'''([value]) =\)
Feedback: We are given the function \( y=[value].\) We start by calculating the first derivative \(\dfrac{dy}{dx} = [value].\) 1. \( \dfrac{d^2y}{dx^2} = [value] \), which when evaluated at \( x = [value]\) is \( [value]. \) 2. \( \dfrac{d^3y}{dx^3} = [value] \), which when evaluated at \( x = [value]\) is \( [value] .\)

Exercise

Given \(y = [value] e^{x} + e^{ [value] x} \), find \(y''\), \(y'''\) at \(x = [value]\) to 3 decimal places.
1. \( y''([value]) =\)
2. \( y'''([value]) =\)
Feedback: We are given the function \( y = [value].\) We start by calculating the first derivative: \( \displaystyle \dfrac{dy}{dx} = [value] .\) 1. \( \dfrac{d^2y}{dx^2} = [value] \), which when evaluated at \( x = [value]\) is \( [value]. \) 2. \( \dfrac{d^3y}{dx^3} = [value] \), which when evaluated at \( x = [value]\) is \( [value] .\)

Exercise

Given \(y = \cos\left( \dfrac{x}{[value]}\right) \), find \(y''\), \(y'''\) at \(x = [value]\) to 3 decimal places.
1. \( y''([value]) =\)
2. \( y'''([value]) =\)
Feedback: We are given the function \( y = [value].\) We start by calculating the first derivative: \( \displaystyle \dfrac{dy}{dx} = [value] .\) 1. \( \dfrac{d^2y}{dx^2} = [value] \), which when evaluated at \( x = [value]\) is \( [value]. \) 2. \( \dfrac{d^3y}{dx^3} = [value] \), which when evaluated at \( x = [value]\) is \( [value] .\)