Problem 335 - Calculus Volume 1

Question

For the following exercises, find \(f^{\prime}(x)\) for each function. $$f(x)=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$$

Step-by-Step Solution

Answer

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

Step 1: Identify the Function Type

The function given is a quotient of two expressions. This suggests the use of the quotient rule for differentiation.

Step 2: Apply the Quotient Rule

The quotient rule states that if you have a function \( f(x) = \frac{g(x)}{h(x)} \), then its derivative is given by \( f'(x) = \frac{g'(x) \, h(x) - g(x) \, h'(x)}{(h(x))^2} \). Here, \( g(x) = e^x - e^{-x} \) and \( h(x) = e^x + e^{-x} \).

Step 3: Differentiate the Numerator

The derivative of \( g(x) = e^x - e^{-x} \) is \( g'(x) = e^x + e^{-x} \). This is calculated using the fact that the derivative of \( e^{ax} \) is \( a \, e^{ax} \).

Step 4: Differentiate the Denominator

The derivative of \( h(x) = e^x + e^{-x} \) is \( h'(x) = e^x - e^{-x} \). This is also obtained by applying the exponential differentiation rule.

Step 5: Substitute into the Quotient Rule Formula

Plugging the derivatives into the quotient rule formula, we have:\[f'(x) = \frac{(e^x + e^{-x})(e^x + e^{-x}) - (e^x - e^{-x})(e^x - e^{-x})}{(e^x + e^{-x})^2}\]

Step 6: Simplify the Expression

Simplify each part of the numerator: \[(e^x + e^{-x})^2 = e^{2x} + 2 + e^{-2x}\]\[(e^x - e^{-x})^2 = e^{2x} - 2 + e^{-2x}\]Subtract these to get:\[2(e^{2x} + e^{-2x} - (e^{2x} - e^{-2x})) = 4\]Thus, the expression simplifies to:\[f'(x) = \frac{4}{(e^x + e^{-x})^2}\]

Step 7: Review the Final Expression

The final expression for the derivative is \( f'(x) = \frac{4}{(e^x + e^{-x})^2} \). This is a much simplified form.

Key Concepts

Quotient Rule

When a function is the quotient of two other functions, we use the quotient rule to find its derivative. The quotient rule comes in handy because it allows us to differentiate a ratio neatly. Consider a function of the form \( f(x) = \frac{g(x)}{h(x)} \). The derivative \( f'(x) \) is given by:

• \( f'(x) = \frac{g'(x) h(x) - g(x) h'(x)}{(h(x))^2} \)

This rule requires finding the derivatives of both the numerator \( g(x) \) and the denominator \( h(x) \). Once these derivatives are determined, you substitute them into the quotient rule formula.

It’s crucial to keep the order of terms in mind, as mistakenly reversing them can lead to incorrect solutions. The quotient rule helps handle fractions efficiently, making it easier to simplify after differentiating.

Exponential Functions

Exponential functions are significant in calculus due to their unique properties. They have the form \( e^{kx} \), where \( e \) is the base of natural logarithms, approximately equal to 2.71828. One fascinating characteristic of exponential functions is that their rate of change matches their value.

When differentiating exponential functions, recall:

• The derivative of \( e^{x} \) is \( e^{x} \)
• For \( e^{-x} \), the derivative is \( -e^{-x} \) because of the chain rule

This feature is what makes exponential functions realistically model processes like population growth or radioactive decay.
They are simple yet powerful, especially since they remain self-similar under differentiation, a property that significantly simplifies calculus operations.

Differentiation

Differentiation is the mathematical technique of finding the derivative of a function. The derivative represents the function's rate of change or slope at any point. Understanding how to differentiate correctly is essential in calculus as it forms the foundation for understanding changes and trends.

Here are some differentiation fundamentals:

• Basic rules include the power rule, the chain rule, and the product rule
• For polynomial functions, apply the power rule: \( \frac{d}{dx} x^n = nx^{n-1} \)
• For products or quotients of functions, use the product rule or quotient rule respectively

Differentiation is a powerful tool that can calculate concepts such as velocity, acceleration, and even the slope of a tangent line on a curve.
Mastering differentiation paves the way to more complex mathematical concepts like integration and differential equations, making it an indispensable part of mathematical and scientific studies.