Problem 41 - University Calculus: Early Transcendentals

Question

Find the derivatives of all orders of the functions. $$y=\frac{x^{2}}{2}-\frac{3}{2} x^{2}-x$$

Step-by-Step Solution

Answer

First derivative: \(-2x - 1\), second derivative: \(-2\), higher derivatives: 0 for \(n \geq 3\).

Step 1: Simplify the Expression

First, simplify the given expression to make the differentiation easier.\[ y = \frac{x^2}{2} - \frac{3}{2}x^2 - x = \left(\frac{1}{2} - \frac{3}{2}\right)x^2 - x = -x^2 - x \] Thus, the simplified expression is \[ y = -x^2 - x \].

Step 2: First Derivative

Differentiate the simplified expression with respect to x to find the first derivative.\[ y' = \frac{d}{dx}(-x^2 - x) \] Using the power rule \( \frac{d}{dx}(x^n) = nx^{n-1} \), we have: \[ y' = -2x - 1 \].

Step 3: Second Derivative

Differentiate the first derivative to find the second derivative.\[ y'' = \frac{d}{dx}(-2x - 1) \] The derivative of \( -2x \) is \( -2 \) and the derivative of the constant \( -1 \) is \( 0 \). Thus, \[ y'' = -2 \].

Step 4: Higher Order Derivatives

Since \( y'' = -2 \) is a constant, all higher derivatives (third, fourth, etc.) will be zero. Therefore: \[ y^{(n)} = 0 \text{ for } n \geq 3 \].

Key Concepts

Power Rule

The power rule is a fundamental concept in calculus used to find the derivative of a function of the form \( x^n \). Derivatives are essential for understanding how a function changes and are widely used in various fields such as physics and engineering. The power rule states that if you have a function \( y = x^n \), then its derivative can be found using:

• \( \frac{d}{dx}(x^n) = nx^{n-1} \)

This rule simplifies the process of differentiation, so you don't have to calculate from scratch each time.
For example, in the original exercise, we started with a simplified expression \( y = -x^2 - x \). By applying the power rule:

• For \( -x^2 \), treat it as \( (-1)x^2 \), and apply the power rule to get \( -2x \).
• For \( -x \), treat it as \( (-1)x^1 \), and applying the power rule gives \( -1 \).

Together, these produce the derivative \( y' = -2x - 1 \). The simplicity of the power rule makes differentiation straightforward for most polynomial functions.

Higher Order Derivatives

Higher order derivatives are derivatives of a function that have been differentiated more than once. They provide information about the curvature and concavity of the original function, offering deeper insight into its behavior.

• The first derivative \( y' = -2x - 1 \) reflects the slope of the tangent line to the curve at any point \( x \).
• The second derivative \( y'' = -2 \) shows the acceleration or change in the slope and is especially useful for determining concavity. In our exercise, this constant \( -2 \) indicates that the curve is concave down.

If you continue differentiating to find higher derivatives, you may find certain functions eventually yield constants or zero values.
In this exercise, after the second derivative, every succeeding derivative becomes zero, as they are derivatives of the constant \( -2 \). Thus, for \( n \geq 3 \), \( y^{(n)} = 0 \). This implies no further change in the rate of change, indicating the function has reached a point of stability in its behavior.

Differentiation Techniques

Differentiation techniques are methods used to find the derivative of a function. Understanding these techniques is crucial for tackling a wide variety of problems in calculus. One of the most commonly used techniques is the simplification of expressions before applying the power rule.

• In the original solution, simplifying the expression \( y = \frac{x^2}{2} - \frac{3}{2}x^2 - x \) to \( y = -x^2 - x \) made the differentiation straightforward.
• Once simplified, you can easily apply the power rule to each term to find the derivative.

Beyond simplification, other techniques such as product rule, chain rule, and quotient rule can be utilized for more complex differentiation tasks.
In this exercise, using the power rule on the simplified terms was sufficient, but in more complex situations, knowledge of these other techniques will be invaluable for effective problem-solving in calculus. Mastery of these techniques allows one to differentiate a wide array of functions efficiently and accurately.