Problem 22 - Essential Calculus Early Transcendentals

Question

Differentiate the function. $$ y=\frac{e^{x}}{1-e^{x}} $$

Step-by-Step Solution

Answer

The derivative is \(y' = \frac{e^x}{(1 - e^x)^2}\).

Step 1: Identify the Function Type

The given function is a rational function, meaning it is the ratio of two functions. Here, the numerator is \(e^x\) and the denominator is \(1 - e^x\). We will use the quotient rule to differentiate.

Step 2: Apply the Quotient Rule

The quotient rule states that for a function \(y = \frac{u}{v}\), the derivative \(y' = \frac{u'v - uv'}{v^2}\). For our function, let \(u = e^x\) and \(v = 1 - e^x\). Calculate the derivatives \(u'\) and \(v'\).

Step 3: Differentiate the Numerator and Denominator

The derivative of the numerator \(u = e^x\) is \(u' = e^x\). The derivative of the denominator \(v = 1 - e^x\) is \(v' = -e^x\).

Step 4: Substitute into Quotient Rule Formula

Substitute \(u\), \(u'\), \(v\), and \(v'\) into the quotient rule formula: \[ y' = \frac{e^x(1 - e^x) - e^x(-e^x)}{(1 - e^x)^2} \]

Step 5: Simplify the Expression

Simplify the expression from the previous step: - Starting with the numerator: \[ e^x(1 - e^x) + e^{2x} = e^x - e^{2x} + e^{2x} = e^x \] - The simplified derivative is: \[ y' = \frac{e^x}{(1 - e^x)^2} \]

Step 6: Finalize the Solution

We have found the derivative of the function. Thus, the derivative of \( y = \frac{e^x}{1 - e^x} \) is \( y' = \frac{e^x}{(1 - e^x)^2} \). This is the final answer.

Key Concepts

Rational Functions

Rational functions are expressions that involve the division of two polynomial functions. Think of them like fractions where both the numerator and the denominator are polynomials. In the context of differentiation, rational functions often appear because they represent a ratio of two changing quantities which is common in many real-world applications such as rates and proportions.
To spot a rational function, look for an equation of the form \( \frac{u}{v} \) where both \(u\) and \(v\) are functions of \(x\). These can be as simple as \( \frac{x}{x+1} \), or more complex, involving exponential or trigonometric terms, as seen in our example \( \frac{e^x}{1 - e^x} \).

• The numerator \(e^x\) and the denominator \(1 - e^x\) are both functions of \(x\).
• This makes the function a rational function despite the presence of exponential terms.

When differentiating rational functions, it becomes essential to understand how each part of the function contributes to the overall rate of change. This is where the quotient rule comes in handy, providing a structured approach to find the derivative.

Quotient Rule

The quotient rule is a technique for differentiating functions that are the ratio of two other functions. It is essential when you have a function that is a fraction, with both the numerator and denominator being functions of the same variable.
For a function \( y = \frac{u}{v} \), where both \(u\) and \(v\) are differentiable functions of \(x\), the derivative \( y' \) can be found using the formula: \[ y' = \frac{u'v - uv'}{v^2} \]

• \(u'\) is the derivative of the numerator \(u\).
• \(v'\) is the derivative of the denominator \(v\).

In our example, \(u = e^x\) and \(v = 1-e^x\). The application of the quotient rule involves:

• Calculating \(u' = e^x\), as the derivative of \(e^x\) is straightforwardly \(e^x\).
• Finding \(v' = -e^x\) since the derivative of \(1-e^x\) results in the negative sign due to subtraction in the function.
• Substituting these into the quotient rule formula to find the derivative of the complete rational function.

Using this method ensures an accurate representation of how the numerator and denominator contribute to the overall slope of the function.

Derivative Calculation

Once you have identified the function as rational and have the quotient rule setup, the next step is calculating the derivative. This involves plugging in the derivatives of the numerator and the denominator into the quotient rule formula from above.
Following our example, we substitute:

• \( u = e^x \) and \( u' = e^x \)
• \( v = 1-e^x \) and \( v' = -e^x \)

into:
\[ y' = \frac{e^x(1 - e^x) - e^x(-e^x)}{(1 - e^x)^2} \]
Now simplify the expression:

• The numerator becomes \( e^x(1 - e^x) + e^{2x} \), which simplifies to \( e^x \).
• Therefore, the result of the simplification leads us to the final derivative: \( y' = \frac{e^x}{(1 - e^x)^2} \)

Through this step-by-step calculation process, one can ensure that the derivative is computed correctly, capturing all subtle changes brought by exponential terms in both the numerator and denominator, allowing for precise interpretations of how the function behaves across different values of \(x\).