Problem 329 - Calculus Volume 1

Question

For the following exercises, compute \(d y / d x\) by differentiating \(\ln y\). $$y=e^{-\ln x}$$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = -\frac{1}{x^2}\).

Step 1: Express y in Terms of Natural Logarithm

Given that \(y = e^{-\ln x}\), we can simplify this expression. Using the property that \(e^{\ln a} = a\), we have: \(y = e^{-\ln x} = \frac{1}{x}\).

Step 2: Apply Natural Logarithm to Both Sides

To differentiate \(y\), we first take the natural logarithm of both sides of the equation: \(\ln y = \ln \left(\frac{1}{x}\right) = -\ln x\).

Step 3: Differentiate the Logarithmic Expression

Differentiate both sides with respect to \(x\). The derivative of \(\ln y\) with respect to \(x\) is \(\frac{1}{y}\frac{dy}{dx}\), and the derivative of \(-\ln x\) is \(-\frac{1}{x}\): \(\frac{1}{y} \frac{dy}{dx} = -\frac{1}{x}\).

Step 4: Solve for \(\frac{dy}{dx}\)

To find \(\frac{dy}{dx}\), multiply both sides by \(y\): \(\frac{dy}{dx} = -\frac{y}{x}\). Since we found in Step 1 that \(y = \frac{1}{x}\), substitute \(\frac{1}{x}\) for \(y\): \(\frac{dy}{dx} = -\frac{1}{x} \cdot \frac{1}{x} = -\frac{1}{x^2}\).

Key Concepts

Natural Logarithm

The natural logarithm, often written as \(\ln\), is a logarithm to the base \(e\), where \(e\) is approximately 2.71828. It's fundamental in calculus because it simplifies the process of differentiation and integration involving exponential functions.

Some important properties of the natural logarithm include:

• \(\ln(e) = 1\)
• \(\ln(1) = 0\)
• \(\ln(a \cdot b) = \ln(a) + \ln(b)\)
• \(\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)\)
• \(\ln(a^b) = b \cdot \ln(a)\)

These properties make logarithms particularly useful in simplifying complex expressions, especially when dealing with exponential functions. In the exercise, we use the property \(\ln(a^b) = b \cdot \ln(a)\) to work with \(y = e^{-\ln x}\). This converts a potentially difficult function into an easier, more manageable one.

Derivative of Logarithmic Functions

Differentiating logarithmic functions is a crucial part of calculus as it enables us to understand how a function changes at any given point. The derivative of a natural logarithm \(\ln(x)\) is \(\frac{1}{x}\).

When differentiating a function of the form \(y = \ln(f(x))\), we use the chain rule. The derivative is \(\frac{1}{f(x)} \cdot f'(x)\), which can be condensed to \(\frac{f'(x)}{f(x)}\). For example, if \(y = \ln(g(x))\), the differentiation becomes straightforward by using the chain rule.

In the exercise provided, the function \(\ln y = -\ln x\) is differentiated. We take the derivative of both sides, which for \(\ln y\) results in \(\frac{1}{y}\frac{dy}{dx}\) and for \(-\ln x\) results in \(-\frac{1}{x}\). Solving for \(\frac{dy}{dx}\) involves straightforward algebraic manipulation, which provides further insight into the function’s dynamics.

Exponential Functions

Exponential functions are expressions where the variable is in the exponent, such as \(y = e^x\). They are ubiquitous in fields ranging from finance to biological sciences, often used to model growth and decay processes.

One interesting property of exponential functions involving \(e\) (the base of the natural logarithm) is that they are their own derivatives. This property simplifies differentiation since the derivative of \(e^{g(x)}\) is \(e^{g(x)}g'(x)\) using the chain rule.

In the exercise, we dealt with an exponential function in logarithmic disguise: \(y = e^{-\ln x}\). By recognizing the form \(e^{\ln a} = a\), we can simplify \(y\) to \(\frac{1}{x}\), clarifying the differentiation process. This type of transformation showcases how exponential functions can be flexibly manipulated and simplifies the handling of complex expressions.