Problem 32 - University Calculus: Early Transcendentals

Question

Find the derivatives of the function. $$w=r e^{-r}$$

Step-by-Step Solution

Answer

The derivative is \( \frac{dw}{dr} = e^{-r}(1 - r) \).

Step 1: Recognize the Function Type

The function given is of the form \[ w = r \cdot e^{-r} \]which is a product of two functions: the linear function \( f(r) = r \) and the exponential function \( g(r) = e^{-r} \). This indicates that the product rule of differentiation should be applied.

Step 2: Apply the Product Rule

The product rule states that the derivative of two functions \( u(r) \cdot v(r) \) is given by \[ (u\cdot v)' = u'\cdot v + u \cdot v' \].Here, let \( u = r \) and \( v = e^{-r} \). Thus, we need to find \( u' \) and \( v' \).

Step 3: Differentiate Each Function

First, differentiate \( u = r \). The derivative of \( u \) with respect to \( r \) is:\[ u' = \frac{d}{dr}[r] = 1 \]The derivative of \( v = e^{-r} \) is:\[ v' = \frac{d}{dr}[e^{-r}] = -e^{-r} \]Apply the chain rule for differentiating the exponential function.

Step 4: Substitute and Simplify

Substitute the differentiated parts back into the product rule:\[\frac{dw}{dr} = u'\cdot v + u\cdot v' = 1\cdot e^{-r} + r\cdot (-e^{-r}) \]Simplify the expression:\[ \frac{dw}{dr} = e^{-r} - r \cdot e^{-r} \]Factor out \( e^{-r} \) to get:\[ \frac{dw}{dr} = e^{-r}(1 - r) \]

Key Concepts

Product Rule

The product rule is a fundamental concept in calculus, especially when dealing with derivatives. It applies when you need to differentiate a product of two functions. Let's break it down in simple steps.

• The product rule formula is: \( (u\cdot v)' = u'\cdot v + u \cdot v' \).
• "\( u \)" and "\( v \)" represent two separate functions of a variable, say \( r \).
• You need to differentiate each function separately: find \( u' \) and \( v' \).

When using the product rule in the given exercise, \( u = r \) and \( v = e^{-r} \). The derivative of \( u \), noted as \( u' \), is a straightforward calculation because it is simply \( 1 \). Meanwhile, the derivative of \( v \) is calculated using the other rules of calculus, like the chain rule.

Chain Rule

The chain rule comes into play when differentiating composite functions. A composite function is simply a function inside another function. The chain rule allows you to break down the process into manageable steps. Here's a simplified explanation:

• The chain rule formula is \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \).
• You apply the derivative of the outer function, evaluated at the inner function.
• Then, multiply by the derivative of the inner function.

In the exercise, when finding the derivative of \( e^{-r} \), the outer function is the exponential \( e^x \), while the inner function is \( -r \). The derivative of the outer function \( e^x \) remains \( e^x \), and we plug in \( -r \) for \( x \). The derivative of the inner function \( -r \) is \( -1 \). Hence, \( v' = -e^{-r} \). This showcases the utility of the chain rule in breaking down complex functions.

Exponential Functions

Exponential functions are a critical component of calculus and frequent in advanced mathematics. An exponential function generally takes the form \( e^x \), where \( e \) is Euler's number, approximately 2.718. Here’s what’s interesting about these functions:

• The derivative of \( e^x \) is itself, \( e^x \). This property holds true universally for exponential functions with base \( e \).
• However, when there's a different exponent, the chain rule helps differentiate properly.

In the task at hand, the function \( e^{-r} \) includes a negative exponent, so differentiation requires applying this unique property of exponential functions alongside the chain rule. Recognizing the structure of exponential functions makes it easier to handle their derivatives accurately. By factoring out \( e^{-r} \), you can neatly express the derivative, demonstrating both the beauty and utility of handling exponential functions.