Problem 18 - Thomas Calculus

Question

In Exercises \(5-36,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=\frac{x^{3}}{3} \ln x-\frac{x^{3}}{9} $$

Step-by-Step Solution

Answer

The derivative is \(y' = x^2 \ln x\).

Step 1: Identify the function components

The given function is \(y = \frac{x^3}{3} \ln x - \frac{x^3}{9}\). This consists of two terms: \(\frac{x^3}{3} \ln x\) and \(\frac{x^3}{9}\). We will find the derivative of each term separately and then combine them.

Step 2: Differentiate the first term

To differentiate the first term \(\frac{x^3}{3} \ln x\), we use the product rule, since it is the product of \(\frac{x^3}{3}\) and \(\ln x\). The product rule states that \((uv)' = u'v + uv'\). Let \(u = \frac{x^3}{3}\) and \(v = \ln x\).- Find \(u'\): \(u' = \frac{d}{dx}\left(\frac{x^3}{3}\right) = x^2\).- Find \(v'\): \(v' = \frac{d}{dx}\left(\ln x\right) = \frac{1}{x}\).Apply the product rule:\[\frac{d}{dx}\left(\frac{x^3}{3} \ln x\right) = \left(x^2\right)(\ln x) + \left(\frac{x^3}{3}\right)\left(\frac{1}{x}\right) = x^2 \ln x + \frac{x^2}{3}.\]

Step 3: Differentiate the second term

Differentiate the second term \(\frac{x^3}{9}\) using the power rule. The power rule states that \(\frac{d}{dx}x^n = nx^{n-1}\).\[\frac{d}{dx}\left(\frac{x^3}{9}\right) = \frac{3}{9}x^{3-1} = \frac{1}{3}x^2.\]

Step 4: Combine the derivatives

Combine the derivatives from Steps 2 and 3 to find the derivative of \(y\). These two steps give the following:\[y' = \left(x^2 \ln x + \frac{x^2}{3}\right) - \frac{x^2}{3}\].The term \(\frac{x^2}{3}\) cancels out, leaving us with:\[y' = x^2 \ln x.\]

Key Concepts

Product Rule

The product rule is a crucial method in calculus for finding the derivative of a product of two functions. In essence, it provides a means to differentiate expressions where two functions are multiplied together. The product rule formula is as follows:

• \((uv)' = u'v + uv'\)

To apply the product rule effectively, follow these simple steps:- Identify the two functions involved in the product. These functions are often denoted as \(u\) and \(v\).- Differentiate each function independently to get \(u'\) and \(v'\).- Substitute \(u'\), \(v\), \(u\), and \(v'\) into the product rule formula. For example, in the function \(\frac{x^3}{3} \ln x\),

• let \(u = \frac{x^3}{3}\) and \(v = \ln x\).

This leads to:- \(u' = x^2\) since the derivative of \(\frac{x^3}{3}\) is \(x^2\), using the power rule. - \(v' = \frac{1}{x}\) since the derivative of \(\ln x\) is \(\frac{1}{x}\).Plug these derivative values into the product rule formula:\[\frac{d}{dx}\left(\frac{x^3}{3} \ln x\right) = x^2 \ln x + \frac{x^3}{3} \cdot \frac{1}{x} = x^2 \ln x + \frac{x^2}{3}. \] This example showcases how the product rule can simplify finding derivatives for multi-part functions.

Power Rule

The power rule is one of the most straightforward and frequently used techniques for differentiation in calculus. When you deal with polynomial expressions, the power rule helps to quickly find the derivative of terms in the form of \(x^n\). The power rule can be expressed as:

• \(\frac{d}{dx}x^n = nx^{n-1}\)

To apply the power rule, follow these key points:- Identify the exponent \(n\) in the expression.- Multiply by the exponent.- Subtract one from the original exponent.Consider the second term in the given function, \(\frac{x^3}{9}\). Here's how the power rule applies:1. Identify \(n\) as \(3\). 2. Multiply the coefficient \(\frac{1}{9}\) by \(3\) (from \(3 \cdot \frac{1}{9}\)), which results in \(\frac{1}{3}\).3. Reduce the power of \(x\) by one, giving \(x^{2}\).Resulting in:\[\frac{d}{dx}\left(\frac{x^3}{9}\right) = \frac{1}{3}x^2.\]Utilizing the power rule allows you to easily differentiate polynomial terms, a frequent requirement in calculus.

Logarithmic Differentiation

Logarithmic differentiation is a powerful tool, especially useful for complex functions involving products, divisions, or powers where traditional methods can become cumbersome. This technique helps simplify differentiation by taking advantage of logarithmic properties. Here's a step-by-step guide to using logarithmic differentiation:1. **Take the natural logarithm of both sides of the function.**
This process simplifies the differentiation by converting multiplication into addition and division into subtraction. It is especially useful with functions that are products or quotients of expressions. 2. **Differentiate using implicit differentiation.**
Apply the derivative to both sides of the equation. The derivative of \(\ln y\) with respect to \(x\) is \(\frac{1}{y} \cdot \frac{dy}{dx}\).3. **Solve for \(\frac{dy}{dx}\).**
Multiply both sides by \(y\) to isolate the derivative \(\frac{dy}{dx}\).In practice, if faced with a complex product like \(y = \frac{x^3}{3} \ln x - \frac{x^3}{9}\), which breaks down into simple products using the product rule, logarithmic differentiation might not be necessary. However, for more dense functions, it simplifies the differentiation and helps avoid extensive algebra. Logarithmic differentiation shines in functions where power rules or product rules alone may not be efficient, but in this exercise, it isn't explicitly required but remains useful in broader calculus applications.