Question

In Exercises, find the derivative of the function. $$ f(x)=\log _{2} x $$

Short answer

The derivative of the function \(f(x) = \log_2 x\) is \(f'(x) = 1/(x*0.693)\)

Step-by-step solution

Transform to natural logarithm

First, transform the logarithm base 2 into a natural logarithm, as the derivative rules are easier with natural logarithms. This can be done using the following equation \(\log_b a = \ln a / \ln b\). After transformation, the function becomes \(f(x) = ln(x) / ln(2)\).

Apply the differentiation rule

Use the differentiation rule for natural logarithms. The derivative of natural logarithm function \(\ln(u)\) is \(1/u\) times the derivative of \(u\). Thus, the derivative of \(f(x) = ln(x) / ln(2)\) becomes \(f'(x) = 1/(x*ln(2))\). Note that \(ln(2)\) is a constant.

Simplify the derivative

After finding the derivative, simplify it. Here, the derivative after simplification becomes \(f'(x) = 1/(x*ln(2)) = 1/(x*0.693)\).

Key concepts

Logarithm Differentiation

Differentiating logarithmic functions requires a solid understanding of the logarithm rules and how they relate to derivatives. Let's discuss how to tackle derivatives of logarithmic functions with different bases. Consider a function like \( f(x) = \log_{2}x \). Differentiating it directly might not seem straightforward, but we can use a conversion formula: \( \log_b a = \frac{\ln a}{\ln b} \), where \( b \) is the base of the logarithm and \( a \) is the argument. This transformation is essential because the properties and rules for the natural logarithm (denoted as \( \ln \) and having \( e \) as its base) are well-established and easier to apply in differentiation.

Once the logarithm is expressed in terms of the natural logarithm, we can proceed to apply the derivative rules. The fundamental rule for differentiating \( \ln(u) \) is that the derivative is \( \frac{1}{u} \) times the derivative of \( u \) with respect to \( x \). Therefore, for our example, after converting the function to a natural logarithm, we differentiate it to obtain \( f'(x) = \frac{1}{x \cdot \ln(2)} \). To streamline the process of logarithm differentiation, remember this key strategy: convert, differentiate using natural logarithm rules, and then simplify.

Natural Logarithm Properties

Understanding the properties of the natural logarithm (\(\ln\)) is crucial when dealing with calculus problems. The natural logarithm is the inverse function of the exponential function with the base \( e \) (Euler's number), which is approximately 2.71828. Some of the fundamental properties of natural logarithms that are particularly useful in calculus are:

• \(\ln(1) = 0\): The natural log of one is always zero.
• \(\ln(e) = 1\): Since \( e \) is the base of the natural logarithm, its logarithm is one.
• \(\ln(ab) = \ln(a) + \ln(b)\): The logarithm of a product is equal to the sum of the logarithms of the individual factors.
• \(\ln(\frac{a}{b}) = \ln(a) - \ln(b)\): The logarithm of a quotient is the difference of the logarithms.
• The derivative of \( \ln(x) \) is \( \frac{1}{x} \) when \( x \) is the variable.

These properties are particularly useful when simplifying expressions before differentiating, or when integrating. For instance, when we come across \( \log_{2}x \), we use the change-of-base formula to rewrite it as \( \frac{\ln(x)}{\ln(2)} \) before applying differentiation techniques. This relies on the understanding that \( \ln(2) \) is a constant and that the derivative of \( \ln(x) \) is \( \frac{1}{x} \).

Calculus in Algebra

When applied to algebraic functions, calculus becomes a powerful tool for analyzing and understanding their behavior. Consider differentiating logarithmic functions within the context of algebra. What we are essentially doing is applying the principles of calculus—specifically derivatives—to algebraic expressions involving logarithms. The process often involves manipulating expressions into a form that is more amenable to the application of derivative rules.

As seen in our example for logarithm differentiation, the algebraic manipulation of changing the base of a logarithm to natural logarithm using algebraic properties is a critical step in calculus. These skills not only simplify the differentiation process but also highlight the interconnectedness between algebra and calculus. In essence, algebra provides the language and the structure, while calculus offers the tools to analyze change. By mastering algebraic manipulation and understanding the principles of calculus, students can tackle a wide range of mathematical problems, making this cross-disciplinary knowledge invaluable.