Problem 34 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the indicated derivative. \(D_{x} e^{x^{3} \ln x}\)

Step-by-Step Solution

Answer

The derivative is \( e^{x^3 \ln x} \cdot (3x^2 \ln x + x^2) \).

Step 1: Identify the Derivative Type

The derivative to be calculated is \(D_x e^{x^3 \ln x}\). We need to apply the chain rule and, potentially, other derivative rules since this is a composite function.

Step 2: Recognize the Inner Function

The function \(e^{x^3 \ln x}\) consists of the inner function \(u = x^3 \ln x\). Our goal is to first express the derivative using the chain rule: \(\frac{d}{dx} e^u = e^u \cdot \frac{du}{dx}\).

Step 3: Differentiate the Inner Function

Differentiate the inner function \(u = x^3 \ln x\). Use the product rule \( (fg)' = f'g + fg'\) where \(f = x^3\) and \(g = \ln x\).

Step 4: Apply the Product Rule

Compute \(f' = 3x^2\) and \(g' = \frac{1}{x}\).The derivative of \(u\) becomes:\[ \frac{du}{dx} = f'g + fg' = 3x^2 \ln x + x^3\left(\frac{1}{x}\right) \]Simplify it to: \( \frac{du}{dx} = 3x^2 \ln x + x^2 \).

Step 5: Apply the Chain Rule

Complete the derivative by applying the chain rule:\[D_x e^{x^3 \ln x} = e^{x^3 \ln x} \cdot (3x^2 \ln x + x^2)\]

Step 6: Final Expression

The final derivative is:\[D_x e^{x^3 \ln x} = e^{x^3 \ln x} \cdot (3x^2 \ln x + x^2)\]

Key Concepts

Chain Rule

When taking the derivative of a composite function, the chain rule is your go-to tool. It is especially handy when you are dealing with functions nested inside each other. The essence of the chain rule is to differentiate the outer function and multiply it by the derivative of the inner function. For a function expressed as both an outer function and an inner function, such as in this exercise with \(e^{u}\) and \(u=x^3 \ln x\), the rule simplifies the process.
To use the chain rule effectively:

• Identify the outer and inner functions.
• Differentiate the outer function while considering the inner function constant.
• Multiply this result by the derivative of the inner function.

This two-layer approach ensures you capture all the nuances of the composite nature. In the given problem, recognizing that you need to multiply \(e^{u}\) by \(\frac{du}{dx}\) encapsulates the chain rule beautifully.

Product Rule

The product rule is a derivative rule used when differentiating the product of two functions. It is essential when tackling problems with multiple variable components multiplying each other. The key formula for the product rule is \((fg)' = f'g + fg'\), where \(f\) and \(g\) are both functions of \(x\).
Utilizing the product rule involves these steps:

• Determine the two functions involved in the product.
• Calculate the derivative of each function individually (f' and g').
• Apply the rule formula by multiplying each function by the derivative of the other and then summing the results.

In this exercise, \(f = x^3\) and \(g = \ln x\), making their derivatives \(f' = 3x^2\) and \(g' = \frac{1}{x}\) respectively. Applying the product rule delivers the derivative of \(u = x^3 \ln x\), critical for the chain rule application.

Composite Functions

Composite functions involve one function nested inside another, thus forming a chain of operations. Recognizing a composite function is the first crucial step for correctly applying calculus techniques like the chain or product rule. In simpler terms, a composite function like \(e^{x^3 \ln x}\) can be seen as a function within a function.
The process to handle composite functions includes:

• Identifying the outer and inner functions.
• Applying the appropriate differentiation rules for each part: the chain rule for the composite nature and potentially additional rules for any products or sums.
• Combining all these differentiation steps into a complete derivative expression.

Understanding composite functions simplify complex problems into manageable segments, aiding in theoretical and practical calculus applications.