Problem 33 - Essential Calculus Early Transcendentals

Question

Differentiate the function. $$ y=\ln \left(e^{-x}+x e^{-x}\right) $$

Step-by-Step Solution

Answer

The derivative is \(-\frac{x}{1+x}\).

Step 1: Identify the Function to Differentiate

The given function is \(y = \ln(e^{-x} + xe^{-x})\). To differentiate this function, we will use the chain rule, which is applicable because the function is composed of a natural log and a sum of two expressions inside the log.

Step 2: Differentiate the Outer Function

The outer function is the natural logarithm \(\ln(u)\), where \(u = e^{-x} + xe^{-x}\). The derivative of \(\ln(u)\) is \(\frac{1}{u}\), so we have \(\frac{1}{e^{-x} + xe^{-x}}\). This is not the final derivative because we need to apply the chain rule and differentiate \(u\) next.

Step 3: Differentiate the Inner Function

The inner function is \(u = e^{-x} + xe^{-x}\). Differentiate this expression respect to \(x\). For \(e^{-x}\), the derivative is \(-e^{-x}\), and for \(xe^{-x}\), we use the product rule: \((x)\) and \((e^{-x})\). The derivative using the product rule is \(-xe^{-x} + e^{-x}\). Thus, the derivative of the inner function is \(-e^{-x} - xe^{-x} + e^{-x} = -xe^{-x}\).

Step 4: Apply the Chain Rule

Now apply the chain rule, which states that the derivative of \(y = \ln(u)\) where \(u\) is a function of \(x\), is \(\frac{1}{u} \cdot \frac{du}{dx}\). We have \(\frac{1}{e^{-x} + xe^{-x}} \cdot (-xe^{-x})\). Simplify this expression to get \(-\frac{xe^{-x}}{e^{-x} + xe^{-x}}\).

Step 5: Simplify the Expression

Factor \(e^{-x}\) out from the numerator and denominator. We get \(-\frac{x}{1 + x}\), because \(e^{-x}\) cancels out from both the top and the bottom.

Step 6: State the Final Answer

Thus, the derivative of \(y = \ln(e^{-x} + xe^{-x})\) is \(-\frac{x}{1 + x}\).

Key Concepts

Chain Rule

When you encounter a function composed of other functions, the Chain Rule is your go-to strategy for differentiation. It's a robust method for breaking down complex expressions into manageable parts.

Suppose you have a function like \(y = \ln(u)\), where \(u\) itself is a function of \(x\). The Chain Rule helps you differentiate this by taking the derivative of the outer function first and then multiplying it by the derivative of the inner function.

• First, differentiate the outer function: For \(\ln(u)\), the derivative is \(\frac{1}{u}\).
• Next, differentiate the inner function \(u\): If \(u = e^{-x} + xe^{-x}\), you find its derivative and proceed to multiply the two results.
• Finally, combine these results: The derivative of the entire function is \(\frac{1}{u} \cdot \frac{du}{dx}\).

By following these steps, you can apply the Chain Rule to functions that might initially seem too complex to tackle. It's like peeling an onion layer by layer until you reach the core structure.

Product Rule

The Product Rule becomes essential when differentiating expressions where two functions are multiplied together. This rule is pivotal in calculus for tackling such situations with efficiency.

Imagine you have a product of two functions, \(f(x)\) and \(g(x)\). The derivative of this product isn't as straightforward as differentiating each part separately. Instead, the Product Rule provides the formula.

• The Product Rule states: \((f \cdot g)' = f' \cdot g + f \cdot g'\).
• This means you first differentiate \(f(x)\) and multiply it by \(g(x)\), and then differentiate \(g(x)\) and multiply it by the original \(f(x)\), then sum these two parts.
• For example, in the term \(xe^{-x}\), differentiate \(x\) to get \(1\), and \(e^{-x}\) to get \(-e^{-x}\).
• The result is \(-xe^{-x} + e^{-x}\).

By using the Product Rule, you can systematically break down and solve differentiations that involve product expressions. This ensures accuracy and clarity in each step.

Natural Logarithm

The Natural Logarithm, denoted as \(\ln(x)\), is an important function in mathematics and arises frequently in calculus, particularly in differentiation and integration problems.

One of its main properties is its inverse relationship with the exponential function \(e^x\). This relationship leads to some fascinating characteristics:

• The derivative of \(\ln(x)\) is \(\frac{1}{x}\). This property allows for easy integration into other compound functions using the Chain Rule.
• The natural log is especially handy in simplifying computations when working with logarithmic expressions due to its concise derivative form.
• Whenever you're differentiating a function that includes \(\ln(x)\), consider the chain rule if inside the log, there's a more complex function.

The elegance of the natural logarithm lies in its simplicity, making it a valuable tool when working through differentiation processes. Understanding its properties can greatly streamline your problem-solving process.