Problem 70 - Calculus and its applications

Question

Differentiate. $$ g(x)=x^{9} \ln |2 x| $$

Step-by-Step Solution

Answer

The derivative is \( g'(x) = 9x^8 \ln |2x| + x^8 \).

Step 1: Identify the Derivative Rule

We begin by identifying that the function \(g(x) = x^9 \ln |2x|\) is a product of two separate functions: \(f(x) = x^9\) and \(h(x) = \ln |2x|\). We'll need to use the product rule to differentiate this expression. The product rule states that if \(g(x) = f(x) \cdot h(x)\), then \(g'(x) = f'(x) \cdot h(x) + f(x) \cdot h'(x)\).

Step 2: Differentiate Each Component

Next, we differentiate the two components, \(f(x) = x^9\) and \(h(x) = \ln |2x|\).1. Derivative of \(f(x) = x^9\) is \(f'(x) = 9x^8\).2. Derivative of \(h(x) = \ln |2x|\) is found using the chain rule. Begin by letting \(u = 2x\), then the derivative of \(\ln(u)\) is \(\frac{1}{u}\), and \(\frac{du}{dx} = 2\). Therefore, \(h'(x) = \frac{1}{2x} \cdot 2 = \frac{1}{x}\).

Step 3: Apply the Product Rule

Substituting the derivatives into the product rule formula, we have:\[ g'(x) = f'(x) \cdot h(x) + f(x) \cdot h'(x) \].This becomes:\[ g'(x) = (9x^8)(\ln |2x|) + (x^9)\left(\frac{1}{x}\right) \].

Step 4: Simplify the Expression

Finally, simplify the expression for \(g'(x)\):1. The first part \(9x^8 \ln |2x|\) remains unchanged.2. The second part simplifies as \(x^9 \cdot \frac{1}{x} = x^8\).Thus, the derivative of the function is:\[ g'(x) = 9x^8 \ln |2x| + x^8 \].

Key Concepts

Product Rule

The Product Rule is a powerful technique used in calculus to differentiate functions that are the product of two or more simpler functions. If you have a function, say \( g(x) = f(x) \cdot h(x) \), the Product Rule helps you find the derivative of \( g(x) \) by calculating:

• \( g'(x) = f'(x) \cdot h(x) + f(x) \cdot h'(x) \)

This rule is handy because it breaks down a potentially complex differentiation problem into simpler parts. In the case of our exercise, we applied the Product Rule to differentiate \( g(x) = x^9 \cdot \ln |2x| \), where \( f(x) = x^9 \) and \( h(x) = \ln|2x| \). By determining \( f'(x) = 9x^8 \) and \( h'(x) \) using the Chain Rule, we were able to use the Product Rule to find the derivative of the entire function.

Chain Rule

The Chain Rule is another essential principle in calculus, particularly when differentiating composite functions. A composite function is where one function is nested inside another, like \( h(x) = \ln |2x| \) in our example. The Chain Rule helps you differentiate such functions by considering the outer and inner functions separately.

• For \( h(x) = \ln |2x| \), think of it in terms of two functions: \( u = 2x \) and \( \, \ln u \), where \( u \) is inside the log function.
• First, differentiate the outer function: the derivative of \( \ln u \) is \( \frac{1}{u} \).
• Next, differentiate the inner function where \( u = 2x \); \( \frac{du}{dx} = 2 \).

The Chain Rule tells us to multiply these derivatives: \( \frac{d}{dx}\left[\ln |2x|\right] = \frac{1}{2x} \cdot 2 = \frac{1}{x} \). This process of breaking down the differentiation into manageable parts makes calculus a bit easier to tackle.

Natural Logarithm

The natural logarithm, denoted as \( \ln x \), is one of the most important functions in mathematics, particularly in calculus. It's the logarithm to the base \( e \), where \( e \approx 2.71828 \), a constant essential in many areas of math.

• In differentiation, the natural logarithm has a simple derivative formula: \( \frac{d}{dx} [\ln x] = \frac{1}{x} \).
• This makes \( \ln x \) a relatively easy function to work with when differentiating.

In our specific problem, we used an absolute value function \( \ln |2x| \). The absolute value ensures that the logarithm remains defined for both positive and negative values of \( x \). Even though the function is modified slightly, taking the derivative follows the same structure as the standard form after utilizing the Chain Rule.

Calculus

Calculus is essentially the mathematical study of change. It is foundational for many scientific and engineering applications, dealing with concepts such as limits, functions, derivatives, integrals, and infinite series.

• Differentiation, part of calculus, focuses on how a function changes smoothly – determining the rate at which this change occurs.
• In other words, it finds the slope or rate of change of a curve at any point, by applying rules such as the Product Rule and Chain Rule.

In our exercise, we used basic principles of calculus to differentiate a composite function using these rules, which is a common practice to understand and predict the behavior of dynamic systems. Embracing these techniques allows us to navigate through complex mathematical and real-world problems more effectively.