Find the following derivative. d. a ((x°+4) in x) ((x³+4) In x) =

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Problem Statement:**

Find the following derivative:

\[
\frac{d}{dx} \left( (x^3 + 4) \ln x \right)
\]

---

**Solution:**

To solve this problem, you will need to use the *product rule* for differentiation. The product rule states that if you have a product of two functions, say \( u(x) \) and \( v(x) \), their derivative is:

\[
\frac{d}{dx}[u(x) \cdot v(x)] = u'(x) v(x) + u(x) v'(x)
\]

In this case, identify:
- \( u(x) = x^3 + 4 \) and \( u'(x) = 3x^2 \)
- \( v(x) = \ln x \) and \( v'(x) = \frac{1}{x} \)

Apply the product rule:

\[
\frac{d}{dx} \left( (x^3 + 4) \ln x \right) = (3x^2) \ln x + (x^3 + 4) \cdot \frac{1}{x}
\]

Simplifying the expression:

\[
= 3x^2 \ln x + x^2 + \frac{4}{x}
\]

Thus, the derivative of \( (x^3 + 4) \ln x \) is:

\[
3x^2 \ln x + x^2 + \frac{4}{x}
\]
Transcribed Image Text:**Problem Statement:** Find the following derivative: \[ \frac{d}{dx} \left( (x^3 + 4) \ln x \right) \] --- **Solution:** To solve this problem, you will need to use the *product rule* for differentiation. The product rule states that if you have a product of two functions, say \( u(x) \) and \( v(x) \), their derivative is: \[ \frac{d}{dx}[u(x) \cdot v(x)] = u'(x) v(x) + u(x) v'(x) \] In this case, identify: - \( u(x) = x^3 + 4 \) and \( u'(x) = 3x^2 \) - \( v(x) = \ln x \) and \( v'(x) = \frac{1}{x} \) Apply the product rule: \[ \frac{d}{dx} \left( (x^3 + 4) \ln x \right) = (3x^2) \ln x + (x^3 + 4) \cdot \frac{1}{x} \] Simplifying the expression: \[ = 3x^2 \ln x + x^2 + \frac{4}{x} \] Thus, the derivative of \( (x^3 + 4) \ln x \) is: \[ 3x^2 \ln x + x^2 + \frac{4}{x} \]
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