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In Exercises 1 through \(8,\) use the product rule to find the derivative. $$ x^{4} \ln x $$

Short Answer

Expert verified
The derivative is \(4x^3 \ln x + x^3\).

Step by step solution

01

Identify Functions

The expression given is a product of two functions: \(f(x) = x^4\) and \(g(x) = \ln x\). These are the functions we need to differentiate using the product rule.
02

Recall the Product Rule

The product rule states that if you have two functions \(u(x)\) and \(v(x)\), then the derivative of their product is \(\frac{d}{dx}[u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x)\).
03

Differentiate Each Function

First, compute the derivative of \(f(x) = x^4\), which is \(f'(x) = 4x^3\).Next, compute the derivative of \(g(x) = \ln x\), which is \(g'(x) = \frac{1}{x}\).
04

Apply the Product Rule

Substitute \(f(x)\), \(f'(x)\), \(g(x)\), and \(g'(x)\) into the product rule formula: \[\frac{d}{dx}[x^4 \ln x] = 4x^3 \cdot \ln x + x^4 \cdot \frac{1}{x}.\]
05

Simplify the Expression

Simplify the expression obtained from the product rule application: \[\frac{d}{dx}[x^4 \ln x] = 4x^3 \ln x + x^3.\] The final simplified derivative is: \[4x^3 \ln x + x^3.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calculus
Calculus is a fundamental branch of mathematics that deals with the study of how things change. It provides tools to analyze and understand the behavior of functions, particularly how they vary. At its core, calculus is all about two main operations: differentiation and integration.

  • **Differentiation:** This is the process of finding the rate of change or the slope of a curve. It helps us understand how one quantity changes in relation to another.
  • **Integration:** This is essentially the reverse operation of differentiation. It's about finding the total amount accumulated, like finding the area under a curve.
In this exercise, we focus on differentiation using one specific technique: the product rule. By applying this rule, we find the derivative of functions composed of multiple other functions. Calculus lets us solve complex real-world problems such as optimizing systems, predicting trends, and analyzing dynamic systems, making it an essential study for science and engineering fields.
Derivative
A derivative is a central concept in calculus representing the rate at which a function is changing at any given point. Imagine watching a car's speedometer—the reading at any instant is analogous to the derivative of the position of the car. It's a powerful tool because it tells you how steep or flat a curve is at a specific point.

In mathematical terms, the derivative of a function \( f(x) \) is represented as \( f'(x) \) or \( \frac{df}{dx} \). This notation means we are determining the limit of the average rate of change as the change in the x-value approaches zero.

When given composite expressions, such as a product of two functions, we need to employ specific strategies, like the product rule, to find their derivatives effectively. This allows us to handle more complex relationships between variables and better understand their dynamics.
Differentiation Techniques
Differentiation techniques are strategies used to find the derivative of a function. These techniques allow us to handle various forms of functions, from simple to complex. Here is an overview of some common differentiation techniques:

  • **Power Rule:** Used when differentiating powers of x. If \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \).
  • **Product Rule:** Applied to find the derivative of a product of two functions, as seen in this problem. For functions \( u(x) \) and \( v(x) \), it's \( u'(x) \cdot v(x) + u(x) \cdot v'(x) \).
  • **Chain Rule:** Used for composite functions, helping us differentiate functions nested within each other.
Understanding and mastering these techniques allows us to efficiently calculate derivatives of more complex functions by strategically breaking them down into simpler parts. Each method provides a different way to approach and solve problems, ensuring we can handle a variety of situations in calculus.

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