Higher Order Derivatives: Concepts, Notation, and Applications

Higher Order Derivatives

Introduction to Higher Order Derivatives

Higher order derivatives extend the concept of the derivative to describe rates of change beyond the first derivative. These derivatives are essential in analyzing the behavior of functions, especially in applications involving motion, economics, and optimization.

• First Derivative: Represents the rate of change of a function, often interpreted as the slope of the tangent line.

• Second Derivative: Represents the rate of change of the first derivative, describing how the slope itself changes.

• Higher Order Derivatives: Successive derivatives, each describing the rate of change of the previous derivative.

Notation and Calculation of Higher Order Derivatives

Higher order derivatives are denoted using prime notation, Leibniz notation, or parentheses for the nth derivative.

• Prime Notation: $f'(x)$, $f''(x)$, $f^{(n)}(x)$

• Leibniz Notation: $\frac{d^n y}{dx^n}$

• Calculation: Each derivative is found by differentiating the previous derivative.

Example: For $y = x^5 - 3x^4 + x$:

• First derivative: $y' = 5x^4 - 12x^3 + 1$

• Second derivative: $y'' = 20x^3 - 36x^2$

• Third derivative: $y^{(3)} = 60x^2 - 72x$

• Fourth derivative: $y^{(4)} = 120x - 72$

• Fifth derivative: $y^{(5)} = 120$

• For $n \geq 6$, $y^{(n)} = 0$

Leibniz Notation for Higher Order Derivatives

Leibniz's notation is commonly used for expressing higher order derivatives, especially in physics and engineering.

• Second derivative: $\frac{d^2 y}{dx^2}$ or $\frac{d}{dx}\left(\frac{dy}{dx}\right)$

• Read as: "the second derivative of y with respect to x"

Worked Examples

Examples illustrate the process of finding higher order derivatives for various functions.

• Example 1: For $y = \frac{1}{x}$, find $\frac{d^2 y}{dx^2}$ $y = x^{-1}$ $\frac{dy}{dx} = -x^{-2}$ $\frac{d^2 y}{dx^2} = 2x^{-3} = \frac{2}{x^3}$

• Example 2: For $y = (x^2 + 10x)^{20}$, find $y'$ and $y''$ By the Extended Chain Rule: $y' = 20(x^2 + 10x)^{19}(2x + 10)$ Using the Product Rule and Extended Chain Rule: $y'' = 40(x^2 + 10x)^{18}(39x + 3900x + 950)$

Quick Check Problems and Solutions

Practice problems reinforce the calculation of higher order derivatives.

• Find $y''$ for:

  • (i) $y = -6x^4 + 3x^2$ $y' = -24x^3 + 6x$ $y'' = -72x^2 + 6$

  • (ii) $y = \frac{2}{x}$ $y' = -\frac{2}{x^2}$ $y'' = \frac{4}{x^3}$

  • (iii) $y = (3x^2 + 1)^2$ $y' = 6x(3x^2 + 1)$ $y'' = 36x^2 + 12$

• Find $\frac{d^4}{dx^4}\left[\frac{1}{x}\right]$: $\frac{d^4}{dx^4}\left[\frac{1}{x}\right] = \frac{24}{x^5}$

Applications: Velocity and Acceleration

Higher order derivatives have important applications in physics, particularly in describing motion.

• Velocity: The first derivative of position with respect to time. $v(t) = s'(t) = \lim_{h \to 0} \frac{s(t + h) - s(t)}{h}$

• Acceleration: The second derivative of position, or the first derivative of velocity. $a(t) = v'(t) = s''(t)$

Example: Motion Analysis

• Given $s(t) = 10t^2$, find $v(t)$ and $a(t)$: $v(t) = s'(t) = 20t$ $a(t) = v'(t) = s''(t) = 20$

• At $t = 4$ hr: $s(4) = 160$ mi $v(4) = 80$ mi/hr $a(4) = 20$ mi/hr2

Example: Free Fall

• Let $s(t) = 16t^2$ (distance in feet, time in seconds): After $t = 3.5$ s: Distance: $s(3.5) = 196$ ft Velocity: $v(3.5) = 112$ ft/s Acceleration: $a(3.5) = 32$ ft/s2

Summary of Key Concepts

• The second derivative is the derivative of the first derivative: $f''(x) = [f'(x)]'$

• The second derivative describes the rate of change of the rate of change.

• Acceleration is a real-life example of a second derivative.

• Common notation for the nth derivative: $f^{(n)}(x)$ or $\frac{d^n}{dx^n}f(x)$