Problem 19 - Thomas Calculus

Question

In Exercises \(7-38,\) 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: Differentiate the first term

The first term is \(\frac{x^4}{4} \ln x\). Use the product rule of differentiation, where if \(u = \frac{x^4}{4}\) and \(v = \ln x\), then the derivative \(\frac{d}{dx}(uv) = u'v + uv'\). Also, remember that \(\frac{d}{dx}(x^4) = 4x^3\) and \(\frac{d}{dx}(\ln x) = \frac{1}{x}\). Calculate the derivative:\[\frac{d}{dx}\left(\frac{x^4}{4} \ln x\right) = \left(\frac{d}{dx}\frac{x^4}{4}\right) \ln x + \frac{x^4}{4} \left(\frac{d}{dx}\ln x\right)\]\[= x^3 \ln x + \frac{x^3}{4}\]

Step 2: Differentiate the second term

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

Step 3: Combine the derivatives

Combine the derivatives of both terms from Step 1 and Step 2:\[\frac{dy}{dx} = \left(x^3 \ln x + \frac{x^3}{4}\right) - \frac{x^3}{4}\]Notice that \(-\frac{x^3}{4}\) and \(+\frac{x^3}{4}\) will cancel each other out. Therefore, the derivative simplifies to:\[\frac{dy}{dx} = x^3 \ln x\]

Key Concepts

Product Rule

When finding the derivative of a product of two functions, we use the product rule. This rule helps us differentiate expressions where two functions are multiplied together. Here, the product rule formula is given by

• If you have a function given by the product of two functions, say u(x) and v(x), then the derivative of their product u(x) v(x) is \[\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\]

In simpler terms, this means:

• Differentiate the first function and multiply it by the second function as it is.
• Then, add the first function as it is, multiplied by the derivative of the second function.
• Combine these two products to get your resultant derivative.

In our exercise, the product rule was applied to the first term \(\frac{x^4}{4} \ln x\). By treating \(\frac{x^4}{4}\) as u(x) and \(\ln x\) as v(x), we were able to find the derivative of the term step by step.

Power Rule

The power rule is a fundamental tool in differentiation, simplifying how we find the derivative of a function that involves a power of \(x\). The rule states:

• If \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\)

This is extremely useful for quick differentiation of polynomial terms. You keep the coefficient in front constant, bring down the exponent as a multiplier, and then reduce the exponent by one.
This was used in the exercise to find the derivative of the term \(-\frac{x^4}{16}\) by applying it as follows:

• Multiply 4 (the exponent) by \(-\frac{1}{16}\) (the coefficient), which gives \(-\frac{1}{4}\).
• Then, reduce the power of \(x\) from 4 to 3, resulting in \(-\frac{x^3}{4}\).

Combining terms is straightforward once you've applied the power rule. Just remember to apply subtraction, as shown in the steps of the exercise.

Logarithmic Differentiation

Logarithmic differentiation steps in when dealing especially with products, quotients, or complex expressions where direct application of other rules is cumbersome. This technique leverages natural logs to simplify and differentiate complicated expressions.

• The key idea is to take the natural log of both sides of the equation and then use the properties of logs to simplify before differentiating.

In the original problem, although logarithmic differentiation wasn't explicitly used, the understanding of the log derivative was essential. Remember, when differentiating \(\ln x\), the result is \(\frac{1}{x}\), a fact used notably within the product rule application.
This strategy allows tricky expressions to be broken down into simpler parts, often turning multiplication into addition and exponents into multipliers by using the property \(\ln(a \cdot b) = \ln a + \ln b\) and \(\ln(a^b) = b \ln a\). While not directly applied in this solution, it's a helpful tool when dealing with logs in differentiation.