Problem 14 - Calculus - AP Edition

Question

Find \(d y / d x\). $$y=x^{3} \ln x$$

Step-by-Step Solution

Answer

The derivative is \( \frac{d y}{d x} = 3x^2 \ln x + x^2 \).

Step 1: Identify the Differentiation Technique

The function given is in the form of a product: \( y = x^3 \ln x \). It contains two functions, \( u = x^3 \) and \( v = \ln x \). We will use the product rule for differentiation, which states that if \( y = uv \), then \( \frac{d y}{d x} = u'v + uv' \).

Step 2: Differentiate the First Function

Differentiate the first function, \( u = x^3 \), with respect to \( x \). The derivative of \( u \) with respect to \( x \) is \[ u' = \frac{d}{dx}(x^3) = 3x^2 \].

Step 3: Differentiate the Second Function

Differentiate the second function, \( v = \ln x \), with respect to \( x \). The derivative of \( v \) with respect to \( x \) is \[ v' = \frac{d}{dx}(\ln x) = \frac{1}{x} \].

Step 4: Apply the Product Rule

Substitute \( u, u', v, \) and \( v' \) into the product rule formula: \[ \frac{d y}{d x} = u'v + uv' \]. - \( u' = 3x^2 \) - \( v = \ln x \) - \( u = x^3 \) - \( v' = \frac{1}{x} \)Plug these into:\[ \frac{d y}{d x} = (3x^2)(\ln x) + (x^3)\left(\frac{1}{x}\right) \].

Step 5: Simplify the Expression

Simplify the expression obtained from the product rule.1. Calculate \( (3x^2)(\ln x) = 3x^2 \ln x \).2. Calculate \( (x^3)(\frac{1}{x}) = x^2 \).Combine the terms:\[ \frac{d y}{d x} = 3x^2 \ln x + x^2 \].

Key Concepts

Derivative of Logarithmic Functions

When finding the derivative of logarithmic functions, it's essential to understand their basic differentiation rules. One of the key rules is that the derivative of the natural logarithm, \( \ln x \), is \( \frac{1}{x} \). This stems from the fact that the logarithm function is the inverse of the exponential function.
Another point to consider is the domain of the logarithmic function, which is only defined for positive values of \( x \). This means that when differentiating, we must ensure \( x \) is positive to maintain the validity of the function.
Differentiating logarithmic functions is a common requirement in calculus, as they appear frequently in various mathematical models, especially those involving growth and decay.
In our given exercise, the function \( \ln x \) is paired with polynomial \( x^3 \), showcasing both logarithmic and polynomial differentiation, a common scenario in calculus problems.

Differentiation Techniques

In calculus, it's important to choose the appropriate differentiation technique for the function you are working with. A variety of techniques exist, and using the right one simplifies the solution process.

• The Product Rule: This is used when you need to differentiate a product of two functions. The formula is \( \frac{d}{dx}(uv) = u'v + uv' \).
• The Chain Rule: Useful when the function is a composition of two functions.
• The Power Rule: Ideal for functions in the form \( x^n \), where the derivative is \( nx^{n-1} \).

In the context of our problem, since \( y = x^3 \ln x \) is a product of \( x^3 \) and \( \ln x \), the product rule was suitable. We identified each part of the product as separate functions, differentiated them accordingly, and applied the product rule. This highlights the importance of classifying the function correctly before selecting the appropriate technique to use.

Calculus Problem-Solving

Solving calculus problems involves a blend of understanding theoretical concepts and applying them through calculations. A structured approach typically enhances success, as showcased in our exercise.
The first step is identifying the type of operation or method needed—deciding whether it's a product, chain, or another rule. This identification guides the calculation process, as it did for the product rule in our problem.
Once the correct method is selected, the next step is detailed calculation. Each function in the product was individually differentiated, and the derivatives were substituted back into the product rule formula.
Lastly, simplifying the result is vital. By combining and reducing terms, the derivative becomes clearer, as seen when we simplified \( 3x^2 \ln x + x^2 \).

• Identify the rule or technique
• Carry out detailed calculations
• Simplify your result for clarity

Mastering these steps with patience and practice improves problem-solving skills in calculus.