Problem 49 - Calculus of a Single Variable

Question

Finding a Derivative In Exercises \(33-54,\) find the derivative. $$ y=\frac{e^{x}+1}{e^{x}-1} $$

Step-by-Step Solution

Answer

\[\frac{dy}{dx} = \frac{-2e^{x}}{(e^{x}-1)^2}\]

Step 1: Identify u and v

The function given is in the form of a quotient \(u/v\). Here, \(u=e^{x}+1\) and \(v=e^{x}-1\).

Step 2: Find the Derivatives of u and v

Taking the derivatives of \(u\) and \(v\) with respect to \(x\), \(\frac{du}{dx} = \frac{d}{dx}(e^{x}+1) = e^{x}\) since the derivative of a constant is zero. Similarly, \(\frac{dv}{dx} = \frac{d}{dx}(e^{x}-1) = e^{x}\).

Step 3: Apply Quotient Rule

Now apply the quotient rule \[\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^{2}}\] Substituting \(u\), \(v\), \(\frac{du}{dx}\), and \(\frac{dv}{dx}\) into the quotient rule, we obtain \[\frac{dy}{dx} = \frac{(e^{x}-1) \cdot e^{x} - (e^{x}+1) \cdot e^{x}}{(e^{x}-1)^2}\] Simplifying this expression, we find that \[\frac{dy}{dx} = \frac{-2e^{x}}{(e^{x}-1)^2}\]

Key Concepts

Quotient Rule

The Quotient Rule is a handy tool for finding the derivative of a function that is the ratio of two differentiable functions. When dealing with a quotient \( \frac{u}{v} \), where \( u \) and \( v \) are both functions of \( x \), the derivative is calculated using:

• The formula \[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \]
• This requires computing the derivatives \( \frac{du}{dx} \) and \( \frac{dv}{dx} \).

When applying the quotient rule, it is crucial to correctly identify which part of the function corresponds to \( u \) and which corresponds to \( v \), as this determines the derivative terms. Once identified, you plug into the formula and simplify to find the derivative of the quotient. This rule is particularly useful when the numerator and denominator are complex or involve exponential or polynomial expressions.

Exponential Functions

Exponential functions are fundamental in calculus and are characterized by the expression \( e^x \), where \( e \) is the base of the natural logarithm, approximately equal to 2.718. They exhibit continuous growth or decay, making them applicable in many real-world phenomena.

• The derivative of an exponential function \( e^x \) is notably straightforward and unique: \( \frac{d}{dx}(e^x) = e^x \).
• This property makes exponential functions particularly easy to work with in differentiation problems.

In the given exercise, both the numerator and the denominator of the quotient involve the expression \( e^x \), which further simplifies differentiation since the exponential term's derivative is itself.``` Using exponential functions in quotients is common, hence knowing their properties and behavior is essential for efficient problem solving.

Differentiation Techniques

Differentiation techniques refer to various methods used to calculate the derivative of a function, a key concept in calculus. These include both basic rules, such as the power rule, product rule, and more advanced methods like implicit differentiation.

• The Quoteint Rule is a special differentiation technique used when dealing with quotients of functions.
• Recognizing when to use these techniques is crucial in solving complex problems efficiently.

In the exercise at hand, we specifically relied on the Quotient Rule. However, the process also exemplifies basic differentiation principles with exponential functions \( e^x \). By applying these techniques, calculating derivatives becomes systematic and manageable, leading to solutions even for intricate functions or expressions.