Problem 17 - Thomas Calculus

Question

In Exercises \(5-36,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=\frac{x^{4}}{4} \ln x-\frac{x^{4}}{16} $$

Step-by-Step Solution

Answer

The derivative is \( x^3 \ln x \).

Step 1: Apply Derivative Rules

We need to find the derivative of the function \( y = \frac{x^4}{4} \ln x - \frac{x^4}{16} \) with respect to \( x \). Start by finding the derivative of each term separately. The derivative of a difference is the difference of the derivatives.

Step 2: Derivative of the First Term

For \( \frac{x^4}{4} \ln x \), use the product rule for derivatives, which states \((uv)' = u'v + uv'\), where \( u = \frac{x^4}{4} \) and \( v = \ln x \). First find \( u' \) and \( v' \):\[ u' = \frac{d}{dx} \left(\frac{x^4}{4}\right) = x^3,\quad v' = \frac{d}{dx}(\ln x) = \frac{1}{x}.\]Then applying the product rule:\[ \frac{d}{dx}\left(\frac{x^4}{4} \ln x\right) = x^3 \cdot \ln x + \frac{x^4}{4} \cdot \frac{1}{x} = x^3 \ln x + \frac{x^3}{4}.\]

Step 3: Derivative of the Second Term

The second term is a simple power function, \(-\frac{x^4}{16}\). The derivative is found using the power rule: \[ \frac{d}{dx}\left(-\frac{x^4}{16}\right) = -\frac{1}{16} \cdot 4x^{3} = -\frac{x^{3}}{4}.\]

Step 4: Combine Results

Combine the derivatives obtained from Step 2 and Step 3:\[ \frac{dy}{dx} = x^3 \ln x + \frac{x^3}{4} - \frac{x^3}{4}.\]Simplify by canceling terms:\[ \frac{dy}{dx} = x^3 \ln x.\]

Key Concepts

Product Rule

The product rule is a fundamental tool used in calculus for finding the derivative of a product of two functions. It's quite handy when dealing with functions like \( y = \frac{x^4}{4} \ln x \). Here, we can identify two parts: \( u = \frac{x^4}{4} \) and \( v = \ln x \).

According to the product rule, the derivative of the product \( uv \) is given by \((uv)' = u'v + uv'\). This means we need to find the derivatives of both \( u \) and \( v \) separately:

• \( u' \), the derivative of \( u \), is calculated using the power rule, which simplifies to \( x^3 \).
• \( v' \), the derivative of \( \ln x \), gives \( \frac{1}{x} \).

Now, plug these results into the product rule formula to obtain the derivative of the first term: \( x^3 \ln x + \frac{x^3}{4} \). The process highlights how the product rule efficiently breaks down complex expressions into manageable parts.

Power Rule

The power rule is one of the most widely used rules in differentiation. It's particularly useful for functions that are simple powers of \( x \). When we look at a term like \(-\frac{x^4}{16}\), it's the perfect candidate for applying the power rule.

In general, the power rule states that the derivative of \( x^n \) is \( nx^{n-1} \). Here, for \(-\frac{x^4}{16}\), we treat the coefficient separately and apply the rule to \( x^4 \).

The derivative becomes:

• First, extract the constant coefficient: \(-\frac{1}{16}\).
• Then calculate: \(-\frac{1}{16} \times 4x^3 = -\frac{x^3}{4}\).

The process is straightforward, translating the polynomial's power into its derivative form quickly and clearly. The power rule is a bedrock concept in calculus, simplifying the handling of variable powers.

Logarithmic Differentiation

Logarithmic differentiation is a strategic method used when dealing with complex expressions, particularly those involving products, quotients, or powers in logarithmic form. It gets its name from the logarithmic transformation applied to simplify these expressions.

Let's consider how we apply logarithmic differentiation to expressions involving logs: For \( y = \ln x \), the derivative \( \frac{d}{dx}(\ln x) \) is \( \frac{1}{x} \). This derivative is essential when using the product rule, offering a simpler way to differentiate terms where one part of the product is a logarithm.

Logarithmic differentiation works particularly well in tackling products where logarithms naturally appear, breaking them down into a sum of derivatives rather than a product, thus streamlining the differentiation process.

By learning these rules and their applications, students can efficiently handle a variety of derivative problems in calculus scenarios. This method becomes especially powerful when conventional differentiation proves cumbersome, highlighting its importance in advanced calculus problem-solving.