Problem 92 - 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 \(f(x)=\ln \frac{1}{x}\)

Step-by-Step Solution

Answer

The derivative is \( f'(x) = -\frac{1}{x} \).

Step 1: Rewrite the Function Using Logarithm Properties

Rewrite the function \(f(x)=\ln \frac{1}{x}\) using the property of logarithms: \( \ln \frac{1}{x} = \ln x^{-1} = -\ln x \). Therefore, \( f(x) = -\ln x \).

Step 2: Apply the Derivative Property

Differentiate \( f(x) = -\ln x \) using the linearity of differentiation. This can be expressed as: \( f'(x) = -(\ln x)' \).

Step 3: Differentiate by Applying Known Derivative

Use the known derivative \( (\ln x)' = \frac{1}{x} \). Hence, differentiating \( -\ln x \) yields: \( f'(x) = -\frac{1}{x} \).

Key Concepts

Logarithm Properties

When dealing with logarithms, understanding their properties can significantly simplify equations. A common property used in calculus is that of rewriting logarithms. For example, the formula for the logarithm of a reciprocal helps us transform expressions into a more workable format.

• If you have an expression like \( \ln \frac{1}{x} \), you can use the property \( \ln \frac{1}{x} = \ln x^{-1} = -\ln x \). This allows you to express the function as \( f(x) = -\ln x \).

Breaking down expressions using these properties simplifies differentiation and integration later on. By rewriting \( \ln \frac{1}{x} \) to \( -\ln x \), we've already made substantial progress towards finding its derivative.

Linearity of Differentiation

The linearity of differentiation is a crucial tool in calculus, especially for dealing with sums and scalar multiples. This principle states that the derivative of a sum is the sum of the derivatives, and constants can be factored out.

• In mathematical terms, this is expressed as \( (a \cdot f(x) + b \cdot g(x))' = a \cdot f'(x) + b \cdot g'(x) \).
• Similarly, for subtraction, \( (f(x) - g(x))' = f'(x) - g'(x) \).

Using these properties, you can easily differentiate functions like \( f(x) = -\ln x \) by recognizing that the derivative of a constant multiplied by a function is just that constant times the derivative of the function. So, we have \( f'(x) = -1 \cdot (\ln x)' \).

Chain Rule

The chain rule is a fundamental differentiation technique used when dealing with composed functions. Although it's not explicitly necessary for the differentiation of \( f(x) = -\ln x \), understanding it is essential for more complex scenarios.

• The chain rule can be described as: if you have a composite function \( h(x) = g(f(x)) \), then the derivative is \( h'(x) = g'(f(x)) \cdot f'(x) \).
• In practice, this means differentiating the outer function and multiplying by the derivative of the inner function.

While simple logarithmic functions rely more on basic properties and linearity, the chain rule becomes invaluable for nested functions and helps manage complex derivatives efficiently.