Problem 15 - Calculus

Question

Find the derivatives of the functions $$ \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{4}{(e^x + e^{-x})^2} \).

Step 1: Identify the Function and Apply Quotient Rule

We are tasked with finding the derivative of the function \( \frac{e^{x} - e^{-x}}{ e^{x} + e^{-x} } \). This is a quotient of two functions. To differentiate it, we will use the Quotient Rule, which states \( \left( \frac{f}{g} \right)' = \frac{f'g - fg'}{g^2} \), where \( f(x) = e^x - e^{-x} \) and \( g(x) = e^x + e^{-x} \).

Step 2: Differentiate the Numerator and the Denominator

Differentiate the numerator \( f(x) = e^x - e^{-x} \):\[ f'(x) = e^x + e^{-x} \]Differentiate the denominator \( g(x) = e^x + e^{-x} \):\[ g'(x) = e^x - e^{-x} \]

Step 3: Apply the Quotient Rule Formula

Substitute into the Quotient Rule formula:\[ \left( \frac{e^x - e^{-x}}{e^x + e^{-x}} \right)' = \frac{(e^x + e^{-x})(e^x + e^{-x}) - (e^x - e^{-x})(e^x - e^{-x})}{(e^x + e^{-x})^2} \]

Step 4: Simplify the Expression

Expand both products in the numerator:\[ (e^x + e^{-x})^2 = e^{2x} + 2 + e^{-2x} \]\[ (e^x - e^{-x})^2 = e^{2x} - 2 + e^{-2x} \]Now calculate the difference:\[ (e^{2x} + 2 + e^{-2x}) - (e^{2x} - 2 + e^{-2x}) = 4 \]Thus, the numerator simplifies to 4.

Step 5: Write Down the Final Derivative

Substitute back into the formula to get the derivative:\[ \frac{4}{(e^x + e^{-x})^2} \]

Key Concepts

Understanding Derivatives

When we talk about derivatives in calculus, we're referring to the rate at which a function is changing at any given point. You can think of it as the slope of the tangent line to the graph of the function at that point. This concept is fundamental, as it shows how variables change in relation to each other. For a function \( f(x) \), the derivative, denoted \( f'(x) \) or \( \frac{d}{dx}f(x) \), details how \( f \) changes when \( x \) changes.

Derivatives have many rules for computation: the product rule, chain rule, and the quotient rule, to name a few. These rules help break down complex problems into manageable steps. In this specific exercise, we used the Quotient Rule due to the function's format as a fraction of two expressions. Differentiation allows us to analyze the behavior of functions more deeply, making it a cornerstone topic in calculus.

Exponential Functions Explained

Exponential functions are a special class of functions where a constant base is raised to a variable exponent. The most common base is Euler's number \( e \), approximately equal to 2.718. An exponential function with this base is written as \( e^x \). Exponentials grow quickly and have unique properties, such as their derivatives.

When differentiating exponential functions, something interesting happens: the derivative of \( e^x \) is \( e^x \), and for \( e^{-x} \), it's \( -e^{-x} \). This is quite handy because exponentials remain the same or only differ by a sign when differentiated, which simplifies calculations significantly. This exercise involved these exponential functions, making it necessary to leverage their properties to find the derivative efficiently.

Differentiation Techniques: The Quotient Rule

The Quotient Rule is a specific technique used in calculus when you need to differentiate a function that is the ratio of two other functions. It provides a systematic way to handle such functions and is expressed as:

• \( \left( \frac{f}{g} \right)' = \frac{f'g - fg'}{g^2} \)

This formula might seem complex at first, but it follows a straightforward process:

• Find the derivative of the numerator (top function), \( f(x) \).
• Find the derivative of the denominator (bottom function), \( g(x) \).
• Substitute these derivatives into the Quotient Rule formula.

The rule helps to systematically solve for the derivative of a fraction without getting lost in the steps. In the exercise, we applied this rule to the function \( \frac{e^{x} - e^{-x}}{e^{x} + e^{-x}} \), and by using this clear method, we arrived at the final derivative with ease.