Problem 21 - Essential Calculus Early Transcendentals

Question

Differentiate the function. $$ y=\frac{x}{e^{x}} $$

Step-by-Step Solution

Answer

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

Step 1: Apply the Quotient Rule

The given function is a quotient of two functions: the numerator is \( x \) and the denominator is \( e^x \). To differentiate the function \( y = \frac{x}{e^x} \), use the quotient rule, which is \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \). Identify \( u = x \) and \( v = e^x \).

Step 2: Differentiate the Numerator and Denominator

Differentiate the numerator \( u = x \), which is \( u' = 1 \). Differentiate the denominator \( v = e^x \), which is \( v' = e^x \).

Step 3: Plug Derivatives into the Quotient Rule

Substitute \( u' = 1 \), \( v = e^x \), \( u = x \), and \( v' = e^x \) into the quotient rule formula: \[ \frac{d}{dx} \left( \frac{x}{e^x} \right) = \frac{1 \cdot e^x - x \cdot e^x}{(e^x)^2}. \]

Step 4: Simplify the Expression

Simplify the expression from Step 3: \[ = \frac{e^x - xe^x}{e^{2x}}. \] Factor out \( e^x \) in the numerator: \[ = \frac{e^x(1 - x)}{e^{2x}}. \] Simplify further by canceling \( e^x \) in the numerator and denominator: \[ = \frac{1 - x}{e^x}. \]

Step 5: Final Answer

The derivative of the function \( y = \frac{x}{e^x} \) is \( \frac{1-x}{e^x} \).

Key Concepts

Quotient Rule

The Quotient Rule is a fundamental technique in calculus used for differentiating functions that are expressed as a quotient. When you have a function expressed as \( y = \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), the derivative \( \frac{dy}{dx} \) is calculated using the formula: \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]. This rule is crucial because it helps us handle situations where the function isn't in a simple, single-variable form.

• Step 1: Identify the top function \( u \) and the bottom function \( v \).
• Step 2: Differentiate \( u \) with respect to \( x \) to get \( u' \).
• Step 3: Differentiate \( v \) with respect to \( x \) to obtain \( v' \).
• Step 4: Substitute \( u, u', v, \) and \( v' \) into the Quotient Rule formula.

Using the Quotient Rule correctly will yield the derivative of the quotient of two differentiable functions.

Exponential Functions

Exponential functions are a class of functions where the variable appears in the exponent, with a constant base. The most common and important form of an exponential function in calculus is \( e^x \), where \( e \) is Euler's number, approximately 2.718.
These functions have a unique and convenient property: their derivatives are themselves. This means that \( \frac{d}{dx} (e^x) = e^x \). This characteristic simplifies the process of differentiation, especially when applying rules like the Quotient Rule, since the derivative of \( e^x \) is straightforward.
Exponential functions are not only mathematically interesting but also incredibly useful in modeling real-world situations such as population growth, radioactive decay, and continuous interest calculations. Their self-derivative nature means they grow at a rate proportional to their current value, making them natural models for continuous change in many scenarios.

Derivative Simplification

Once you've obtained the derivative of a function using rules like the Quotient Rule, it's essential to simplify the expression as much as possible. Simplification involves reducing complex fractions and expressions to their simplest form, which makes them easier to interpret and use.
In the context of the derivative from the example, we initially found the derivative to be \( \frac{e^x - xe^x}{e^{2x}} \). This expression can be unwieldy, so the next step is to look for common factors that can be simplified. Here, notice that \( e^x \) is common in the numerator, allowing us to factor it out: \( e^x(1 - x) \). This step makes it possible to cancel \( e^x \) in both the numerator and the denominator, resulting in a significantly simpler form: \( \frac{1-x}{e^x} \).
Simplified expressions are not only pleasing to look at, they also provide more insights, reduce errors in further calculation, and often are exactly what you need as a final cleaned-up answer in calculus.