Problem 91 - Precalculus

Question

In calculus we prove that the derivative of \(f+g\) is \(f^{\prime}+g^{\prime}\) and that the derivative of \(f-g\) is \(f^{\prime}-g^{\prime} .\) It is also shown in calculus that if \(f(x)=\ln x\) then \(f^{\prime}(x)=\frac{1}{x}\) Use these properties to find the derivative of \(f(x)=\ln x^{2}\)

Step-by-Step Solution

Answer

The derivative of \( f(x) = \ln x^2 \) is \( f'(x) = \frac{2}{x} \).

Step 1: Rewrite the Function Using Logarithm Properties

The given function is \( f(x) = \ln x^2 \). We can use the property of logarithms that states \( \ln(a^b) = b \cdot \ln(a) \). Apply this property to rewrite the function as \( f(x) = 2 \cdot \ln x \).

Step 2: Identify the Constant Multiplier

The rewritten function is \( f(x) = 2 \cdot \ln x \). Recognize that the number \( 2 \) is a constant multiplier of the function \( \ln x \).

Step 3: Differentiate Using the Constant Multiple Rule

Apply the constant multiple rule of differentiation, which states that the derivative of \( c \cdot g(x) \) is \( c \cdot g'(x) \). Here, \( c = 2 \) and \( g(x) = \ln x \) with \( g'(x) = \frac{1}{x} \). So, the derivative is \( f'(x) = 2 \cdot \frac{1}{x} \).

Step 4: Simplify the Expression

Simplify the derived expression \( f'(x) = 2 \cdot \frac{1}{x} \) to \( f'(x) = \frac{2}{x} \).

Key Concepts

Logarithm Properties

Logarithms come with a set of handy properties that simplify complex expressions. One such property is the power rule, which states that the logarithm of an exponentiated number can be expressed as a multiple. In equation form, it is

• \( \ln(a^b) = b \cdot \ln(a) \)

This means if you have \( \ln(x^2) \), you can rewrite it as \( 2 \cdot \ln(x) \). This simplification makes differentiation much more straightforward. By using these properties, we break down complex logarithmic forms into simpler linear forms, easing the process of finding derivatives.
This property is crucial when dealing with logarithmic differentiation. It turns products and powers into sums and products that are more manageable under differentiation rules.

Constant Multiple Rule

The constant multiple rule is a fundamental principle in calculus differentiation. It simplifies the act of taking derivatives when a constant is multiplied by a function. The rule states:

• The derivative of \( c \cdot g(x) \) is \( c \cdot g'(x) \), where \( c \) is a constant and \( g(x) \) is a function of \( x \).

This essentially means that when you have a constant in front of a function, you can take the constant out of the differentiation process, differentiate the remaining function, and then multiply back by the constant. For instance, if we have \( f(x) = 2 \cdot \ln(x) \), the derivative \( f'(x) \) becomes \( 2 \cdot \frac{1}{x} \).
This rule is particularly useful in calculus as it allows differentiation to be broken down into simpler steps, making the process both efficient and straightforward.

Calculus Differentiation

Differentiation is a core concept in calculus that's used to find the rate at which things change. For the function \( f(x) = \ln(x) \), it's well-proven that the derivative is \( \frac{1}{x} \). This means that for every increment in \( x \), the function \( \ln(x) \) increases by \( \frac{1}{x} \), illustrating how sharply or gently the curve rises.
When we differentiate more complex functions, this principle remains at the heart of the process. We use foundational rules like the sum and difference rules (which state, for instance, the derivative of \( f+g \) is \( f' + g' \)) and apply properties like the constant multiple rule.
Understanding differentiation involves mastering these basic ideas, enabling you to tackle much more complicated calculus problems with confidence. Mastery of differentiation not only aids in academic pursuits but also empowers problem-solving in fields such as physics, engineering, and economics.