Economics at Zero Marginal Cost: Differentiation of Higher Order

If $y=f(x)$ then $f'(x)\equiv \frac{dy}{dx}$ is the first derivative of the function.

Since the first derivative $f'(x)$ is itself a function (i.e., a derived function from a primitive function $y=f(x)$ , it too should be differentiable with respect to , provided it is continuous and smooth.

The first derivative can, therefore, be differentiated again and the result of this differentiation known as the second derivative of the function is denoted by:

${f}''(x)\equiv \frac{d^2y}{dx^2}\left ( \equiv \frac{d}{dx}\left ( \frac{dy}{dx} \right ) \right )$

Again, since the second derivative is a function of $x$ , it can be differentiated with respect to $x$ again to produced a third derivative, which in turn can be the source of a fourth derivative, and so on as long as differentiability condition is met.

Given a function $y=f(x)$

First Derivative: $f'(x)\equiv \frac{dy}{dx}$

Derivatives of higher order can be represented as:

Second Derivative: $f''(x)\equiv \frac{d^2y}{dx^2}\left ( \equiv \frac{d}{dx}\left ( \frac{dy}{dx} \right ) \right )$

Third Derivative: $f''(x)\equiv \frac{d^3y}{dx^3}\left ( \equiv \frac{d}{dx}\left ( \frac{d^2y}{d^2x} \right ) \right )$

Fourth Derivative: $f''(x)\equiv \frac{d^4y}{dx^4}\left ( \equiv \frac{d}{dx}\left ( \frac{d^3y}{d^3x} \right ) \right )$

Exercise:

Find the second and third derivatives of the following:

$\\1.\;y=ax^2+bx+c\\ 2.\;y=4x^4-3x-4$

1. $y=ax^2+bx+c$

Solution:

First derivatives:

$\\\frac{dy}{dx}=\frac{d}{dx}y=\frac{d}{dx}(ax^2+bx+c)\\ \frac{dy}{dx}=\frac{d}{dx}ax^2+\frac{d}{dx}bx+\frac{d}{dx}c\\ \frac{dy}{dx}=a\frac{d}{dx}x^2+b\frac{d}{dx}x+0\\ \frac{dy}{dx}=a2x^{2-1}+b.1 \frac{dy}{dx}=2ax+b$

Second derivative:

$\\\frac{d^2y}{dx^2}=\frac{d}{dx}\left ( \frac{dy}{dx} \right )\\ \frac{d^2y}{dx^2}=\frac{d}{dx}(2ax+b)\\ \frac{d^2y}{dx^2}=\frac{d}{dx}2ax+\frac{d}{dx}b\\ \frac{d^2y}{dx^2}=2a\frac{d}{dx}x+0\\ \frac{d^2y}{dx^2}=2a.1\\ \frac{d^2y}{dx^2}=2a$

Third derivative:

$\\\frac{d^3y}{dx^3}=\frac{d}{dx}\left ( \frac{d^2y}{dx^2} \right )\\ \frac{d^3y}{dx^3}=\frac{d}{dx}(2a)\\ \frac{d^3y}{dx^3}=0$

$\therefore \frac{d^y}{dx^2}=2a \;\;and\;\; \frac{d^3y}{dx^3}=0$

2. $y=4x^4-3x-4$

Solution:

First derivative:

$\\\frac{dy}{dx}=\frac{d}{dx}y=\frac{d}{dx}(y=4x^4-3x-4)\\ \frac{dy}{dx}=\frac{d}{dx}4x^4-\frac{d}{dx}3x-\frac{d}{dx}4\\ \frac{dy}{dx}=4\frac{d}{dx}x^4-3\frac{d}{dx}x-\frac{d}{dx}\\ \frac{dy}{dx}=4.4x^{4-1}-3.1\\ \frac{dy}{dx}=16x^3-3$

Second derivative:

$\\\frac{d^2y}{dx^2}=\frac{d}{dx}\left ( \frac{dy}{dx} \right )\\ \frac{d^2y}{dx^2}=\frac{d}{dx}(16x^3-3)\\ \frac{d^2y}{dx^2}=\frac{d}{dx}16x^3-\frac{d}{dx}3\\ \frac{d^2y}{dx^2}=16\frac{d}{dx}x^{3}-0\\ \frac{d^2y}{dx^2}=16.3x^{3-1}\\ \frac{d^2y}{dx^2}=48x^2\\$

Third derivative:

$\\\frac{d^3y}{dx^3}=\frac{d}{dx}\left ( \frac{d^2y}{dx^2} \right )\\ \frac{d^3y}{dx^3}=\frac{d}{dx}(48x^2)\\ \frac{d^3y}{dx^3}=48\frac{d}{dx}x^2\\ \frac{d^3y}{dx^3}=48.2x^{2-1}\\ \frac{d^3y}{dx^3}=96x$