Problem 62 - Calculus and its applications

Question

Differentiate. $$ y=x^{4} \ln x-\frac{1}{2} x^{2} $$

Step-by-Step Solution

Answer

The derivative is \( y' = 4x^3 \ln x + x^3 - x \).

Step 1: Identify Terms

The function to differentiate is given by \[ y = x^4 \ln x - \frac{1}{2} x^2. \]It consists of two separate terms: \( x^4 \ln x \) and \(-\frac{1}{2} x^2.\)

Step 2: Differentiate the First Term

To find the derivative of the first term \( x^4 \ln x \), use the product rule which states that if you have two functions \( u(x) \) and \( v(x) \), then the derivative \( (uv)' = u'v + uv' \). Set \( u = x^4 \) and \( v = \ln x \). First, calculate the derivatives: - The derivative of \( u = x^4 \) is \( u' = 4x^3 \).- The derivative of \( v = \ln x \) is \( v' = \frac{1}{x} \).Now apply the product rule:\[ (x^4 \ln x)' = (4x^3)(\ln x) + (x^4)\left(\frac{1}{x}\right) = 4x^3 \ln x + x^3. \]

Step 3: Differentiate the Second Term

Differentiate the second term \(-\frac{1}{2}x^2\). The derivative of \(-\frac{1}{2}x^2\) is obtained by applying the power rule:\[ \left(-\frac{1}{2}x^2\right)' = -\frac{1}{2}(2)x^{2-1} = -x. \]

Step 4: Combine the Derivatives

Now add the derivatives of the two terms together. Sum the derivative of the first term \(4x^3 \ln x + x^3\) and the derivative of the second term \(-x\):\[ y' = 4x^3 \ln x + x^3 - x. \]

Key Concepts

Product Rule

The Product Rule is essential in calculus when differentiating products of two functions. It states: if you have two functions, say, \( u(x) \) and \( v(x) \), the derivative of their product \( (uv)' \) can be found using \( (uv)' = u'v + uv' \).
This rule helps tackle more complex functions that are not just a simple product, such as \( x^4 \ln x \) in our original exercise.

In the example, we identify \( u = x^4 \) and \( v = \ln x \). Differentiating separately, we find \( u' = 4x^3 \) and \( v' = \frac{1}{x} \).

Plugging these into the Product Rule formula gives:

• \( (x^4 \ln x)' = (4x^3)(\ln x) + (x^4)\left(\frac{1}{x}\right) \)
• Simplifying this: \( 4x^3 \ln x + x^3 \).

Understanding how to apply the Product Rule equips you with a powerful tool for a variety of differentiation problems.

Power Rule

The Power Rule is one of the fundamental techniques for differentiation. It makes finding derivatives of terms involving a variable raised to a power straightforward.
The general form of the rule is: if \( y = x^n \), then its derivative \( y' = nx^{n-1} \). This is useful for any term like \( x^n \).
In the original problem, we applied the Power Rule to the term \(-\frac{1}{2}x^2\). Here's how:

• The exponent \( n \) is 2.
• Multiply by the exponent: \(-\frac{1}{2} \times 2 \).
• Reduce the exponent by 1: \( x^{2-1} \).

This results in the derivative: \(-x\).
The Power Rule simplifies the process of differentiation, especially for polynomial terms, and is constantly applied in calculus differentiation tasks.

Logarithmic Differentiation

Logarithmic Differentiation is a technique particularly helpful when differentiating products or quotients more complex than typical functions.
Although it wasn't directly used in the original exercise, understanding it can broaden your differentiation skills when encountering functions involving logarithms.

To use Logarithmic Differentiation, you typically follow these steps:

• Take the natural log (\( \ln \)) of both sides of the equation \( y = f(x) \).
• Use properties of logs to simplify the equation.
• Differentiate both sides with respect to \( x \).
• Simplify and solve for \( y' \).

This method is advantageous when dealing with powers and products, as it transforms multiplicative processes into additive ones, simplifying the process of differentiation.
Even if it's not applied in every scenario, having it in your toolkit enhances your ability to tackle intricate functions that feature logs.