Question

General logarithmic and exponential derivatives Compute the following derivatives. Use logarithmic differentiation where appropriate. $$\frac{d}{d x}\left(1+\frac{1}{x}\right)^{x}$$

Short answer

Question: Find the derivative of the function \(y = (1 + \frac{1}{x})^x\). Answer: The derivative of the function is \(y' = (1 + \frac{1}{x})^x \left[\ln(1 + \frac{1}{x}) - \frac{1}{x(1+\frac{1}{x})}\right]\).

Step-by-step solution

Write down the function

Begin by writing down the given function: $$ y = (1 + \frac{1}{x})^x $$

Take the natural logarithm of both sides and rewrite the function

Take the natural logarithm of both sides and rewrite the function: $$ \ln(y) = \ln((1 + \frac{1}{x})^x) $$ Now, since \(x\) is an exponent, we can bring it as a coefficient in front of the natural logarithm, which results in: $$ \ln(y) = x\ln(1 + \frac{1}{x}) $$

Differentiate both sides with respect to \(x\)

Differentiate both sides with respect to \(x\), using the chain rule and product rule as needed. First, differentiate the left side: $$ \frac{dy}{dx}\cdot\frac{1}{y} = \frac{y'}{y} $$ Now, differentiate the right side using the product rule on \(x\) and \(\ln(1+\frac{1}{x})\): $$ =1\cdot\ln(1 + \frac{1}{x})+x\cdot\frac{1}{1+\frac{1}{x}}\cdot(-\frac{1}{x^2}) = \ln(1 + \frac{1}{x}) - \frac{1}{x(1+\frac{1}{x})} $$ We obtain: $$ \frac{y'}{y} = \ln(1 + \frac{1}{x}) - \frac{1}{x(1+\frac{1}{x})} $$

Solve for the derivative \(y'\)

Now, we solve for the derivative \(y'\) by multiplying both sides by \(y\), knowing that \(y = (1 + \frac{1}{x})^x\): $$ y' = (1 + \frac{1}{x})^x \left[\ln(1 + \frac{1}{x}) - \frac{1}{x(1+\frac{1}{x})}\right] $$ Therefore, the derivative of the given function is: $$ \frac{d}{d x}\left(1+\frac{1}{x}\right)^{x} = (1 + \frac{1}{x})^x \left[\ln(1 + \frac{1}{x}) - \frac{1}{x(1+\frac{1}{x})}\right] $$

Key concepts

Exponential Functions

Exponential functions play a significant role in calculus and other areas of mathematics. An exponential function is a function in which the variable, usually denoted as \(x\), appears in the exponent. A general form of an exponential function is \(a^x\), where \(a\) is a constant base, and \(x\) is the exponent. These functions are known for their rapid growth or decay, depending on whether the base \(a\) is greater than or less than 1.

In the context of our example, we dealt with a function \((1 + \frac{1}{x})^x\). Here, the base \((1 + \frac{1}{x})\) itself is not a constant. It changes with \(x\), making this a more complex exponential function. These kinds of functions are often seen in solving growth problems or finding complex derivatives necessitating techniques like logarithmic differentiation. Logarithmic differentiation simplifies the differentiation of exponential functions by leveraging properties of logarithms.

• Exponential functions can increase or decrease very quickly.
• Analyzing these functions involves understanding their base and how it behaves as \(x\) changes.
• The derivative of exponential functions often involves a step using natural logarithms to simplify the process.

Chain Rule

The chain rule is a foundational tool in calculus for finding derivatives of composite functions. It's crucial when a function is nested within another function. Formally, if you have two functions \(f\) and \(g\) such that \(y = f(g(x))\), the chain rule states that the derivative of \(y\) with respect to \(x\) is \(f'(g(x)) \cdot g'(x)\).

This rule was specifically applied when differentiating \(\ln(1 + \frac{1}{x})\). Given \(g(x) = 1 + \frac{1}{x}\), we see a composition where \(\ln\) is the outer function and \(1 + \frac{1}{x}\) is the inner function. Differentiating \(\ln(u)\) gives \(\frac{1}{u}\), then applying the chain rule means multiplying this by the derivative of \(g(x) = 1 + \frac{1}{x}\), which is \(-\frac{1}{x^2}\).

• The chain rule allows us to handle complex derivatives by focusing on one layer at a time.
• It’s essentially about recognizing the inner function and the outer function and working them separately.
• Often used in situations where the derivative needs to be calculated for a nested function structure.

Product Rule

The product rule is another indispensable tool in calculus, used when differentiating the product of two functions. If you have two functions, \(u(x)\) and \(v(x)\), then the derivative of their product is given by \(u'(x)v(x) + u(x)v'(x)\).

In this problem, we used the product rule to differentiate \(x \cdot \ln(1 + \frac{1}{x})\). Here, \(x\) is the first function, and \(\ln(1 + \frac{1}{x})\) is the second function. Applying the product rule:

• First, differentiate \(x\) to get \(1\), and multiply by the second function \(\ln(1 + \frac{1}{x})\).


• Next, differentiate \(\ln(1 + \frac{1}{x})\) as we found using the chain rule, which is \(-\frac{1}{x(1+\frac{1}{x})}\), and multiply by \(x\).


• Add these results to get the derivative of the product.

The product rule is particularly useful for systematically handling derivatives of complex expressions involving multiple factors interacting with each other.