Problem 53 - Calculus of a Single Variable

Question

Finding a Derivative In Exercises \(37-58\) , find the derivative of the function. (Hint: In some exercises, you may find it helpful to apply logarithmic properties before differentiating.) $$ f(x)=\log _{2} \frac{x^{2}}{x-1} $$

Step-by-Step Solution

Answer

The derivative of the function \(f(x)=\log _{2} \frac{x^{2}}{x-1}\) is given by \(f'(x) = \frac{x - 2}{x(x-1) \ln 2}\)

Step 1: Apply Logarithmic Properties

First, use the logarithmic property \(\log_b(a/c) = \log_b a - \log_b c\) to simplify the given function. Hence, the function \(f(x)=\log _{2} \frac{x^{2}}{x-1}\) becomes \(f(x) = \log_2 x^2 - \log_2 (x-1)\). Further simplify it using another logarithmic property \(\log_b(a^c) = c \log_b a\) to write it as \(f(x) = 2 \log_2 x - \log_2 (x-1)\).

Step 2: Differentiate the Simplified Function

Now that it's in a simpler form, take the derivative. Recall the derivative of the natural logarithm function: \(d/dx( \ln(u)) = u' / u\). By analogy, the derivative of \(\log_2 x\) is \(\frac{1}{\ln(2) \cdot x}\). Apply the chain rule where necessary, the derivative becomes \(f'(x) = \frac{2}{x \ln 2} - \frac{1}{(x-1) \ln 2}\).

Step 3: Simplify the Derivative

Let's express the derivative in a simpler form by combining the two fractions in the result obtained into one fraction. This turns out to be \(f'(x) = \frac{2(x-1) - x}{x(x-1) \ln 2} = \frac{x - 2}{x(x-1) \ln 2}\).

Key Concepts

Logarithmic Properties

Understanding logarithmic properties is crucial for simplifying complex expressions before taking derivatives. Logarithms, by definition, are the inverses of exponentiation. They have unique properties that can simplify multiplication into addition, division into subtraction, and exponentiation into multiplication.

For instance, one key property is \[ \log_b\left(\frac{a}{c}\right) = \log_b a - \log_b c \] This property can separate a single log expression with a fraction into two separate log terms. Similarly, the property \[ \log_b(a^c) = c \log_b a \] helps in pulling out exponents as coefficients. These properties enable us to rewrite \[ f(x)=\log _{2} \frac{x^{2}}{x-1} \] as \[ f(x) = 2 \log_2 x - \log_2 (x-1) \] This simplification is the stepping stone for finding derivatives more easily and forms an essential foundation for calculus students.

Chain Rule

The chain rule is a fundamental rule in calculus used for finding the derivative of a composite function. It's essentially a formula to differentiate a function of a function. In mathematical terms, if you have a function \( h(x) = g(f(x)) \), the chain rule states that the derivative \( h'(x) \) can be found using \( h'(x) = g'(f(x)) \cdot f'(x) \).

This rule is particularly useful when dealing with functions inside logarithms, as you often need to take the derivative of an inner function within the log function. For example, if we have a function \( f(x) = \log_2 (g(x)) \), we would find the derivative by first taking the derivative of \( g(x) \) and then applying the chain rule to multiply it by the outer derivative.

Derivative of Logarithmic Function

The derivative of logarithmic functions is a frequent task in calculus. If we have a natural log function of a variable \(u\), represented as \(\ln(u)\), its derivative is \(\frac{d}{dx}[\ln(u)] = \frac{u'}{u}\), where \(u'\) is the derivative of \(u\) with respect to \(x\).

For other logarithmic bases, like \(\log_2(x)\), we consider the change of base formula. The derivative of a log function with base \(b\) is \(\frac{1}{\ln(b) \cdot u} \cdot u'\). Applying this to our simplified function, we find the derivatives of \(\log_2 x^2\) and \(\log_2 (x-1)\) by invoking this rule along with the chain rule for any inner function \(g(x)\).

Simplifying Expressions

In calculus, simplifying expressions before and after differentiation can make the process of finding derivatives much easier and the results more interpretable. The objective is to break down complex fractions, combine like terms, and factor wherever possible.

After applying the properties of logarithms and the chain rule to find the derivative, we often obtain an expression that may still be complex. In our case, the derivative is \(f'(x) = \frac{2}{x \ln 2} - \frac{1}{(x-1) \ln 2}\). To simplify, we look for a common denominator and combine these terms into a single fraction. This leads to the simpler form \(f'(x) = \frac{x - 2}{x(x-1) \ln 2}\), which is easier to understand and use for further calculus operations, like evaluating the behavior of the original function or finding its extrema.