Problem 99 - Calculus and its applications

Question

Differentiate. $$ f(x)=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} $$

Step-by-Step Solution

Answer

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

Step 1: Identify the function to differentiate

We need to find the derivative of the function \( f(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} \). This is a rational function, which suggests the use of the quotient rule.

Step 2: Recall the quotient rule formula

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

Step 3: Differentiate the numerator and the denominator

Calculate \( u'(x) \) and \( v'(x) \). For \( u(x) = e^x - e^{-x} \), the derivative \( u'(x) = e^x + e^{-x} \). For \( v(x) = e^x + e^{-x} \), the derivative \( v'(x) = e^x - e^{-x} \).

Step 4: Apply the quotient rule formula

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

Step 5: Simplify the numerator

Expand the numerator: \[(e^x + e^{-x})^2 - (e^x - e^{-x})^2 = (e^{2x} + 2 + e^{-2x}) - (e^{2x} - 2 + e^{-2x})\].Simplify further: \[(e^{2x} + 2 + e^{-2x}) - (e^{2x} - 2 + e^{-2x}) = 4\].

Step 6: Write the final expression for the derivative

Substituting back, we obtain: \[f'(x) = \frac{4}{(e^x + e^{-x})^2}\]. This is the simplified derivative of the given function.

Key Concepts

Quotient Rule

The Quotient Rule is an essential tool in calculus that helps differentiate functions expressed as a quotient, meaning one function divided by another. It applies when you have a function written in the form \( f(x) = \frac{u(x)}{v(x)} \), where both \( u \) and \( v \) are differentiable functions of \( x \).

To apply the Quotient Rule, you need to know the derivatives of the numerator and the denominator. The formula for the derivative of the quotient is:

• \[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \]

Essentially, the derivative \( f'(x) \) results from the difference between the product of the derivative of the numerator and the original denominator, and the product of the numerator and the derivative of the denominator, all divided by the square of the denominator.

When using the Quotient Rule, as seen in the original exercise, it's crucial to:

• Correctly identify \( u(x) \) and \( v(x) \).
• Calculate their derivatives, \( u'(x) \) and \( v'(x) \).
• Carefully substitute these into the formula to avoid errors in signs and factors.

Derivative of Exponential Functions

Exponential functions, such as those involving \( e^x \), exhibit unique characteristics in calculus due to the constant base \( e \), an irrational number approximately equal to 2.718. The derivative of an exponential function is notably straightforward:

• The derivative of \( e^x \) is simply \( e^x \).
• For \( e^{-x} \), the derivative is \(-e^{-x} \) because of the chain rule, which accounts for the negative sign.

In the context of our original exercise, where we deal with the function \( u(x) = e^x - e^{-x} \), applying these principles yields \( u'(x) = e^x + e^{-x} \). Similarly, for \( v(x) = e^x + e^{-x} \), the derivative is \( v'(x) = e^x - e^{-x} \). It's this understanding of exponential derivatives that makes applying the Quotient Rule in our exercise possible.

Remember: For exponential functions, differentiation often mirrors the function itself, but be vigilant with signs when negative exponents come into play.

Simplification of Expressions

Simplification is the process of reducing a mathematical expression to its simplest form, making it easier to interpret and use for further calculations. After applying the Quotient Rule in our exercise, simplification of the resulting expression allows us to find a clean and clear final answer.

In the original solution, you were tasked to expand the numerator after applying the Quotient Rule. Starting with:

• \[ (e^x + e^{-x})^2 - (e^x - e^{-x})^2 \]

Placing the components in their expanded forms means distributing and combining like terms:

• \[ e^{2x} + 2 + e^{-2x} - (e^{2x} - 2 + e^{-2x}) = 4 \]

Any equivalent terms that cancel out are removed, leaving a simplified result of \( 4 \) in the numerator.

This simplification is essential as it yields a simple final derivative: \[ f'(x) = \frac{4}{(e^x + e^{-x})^2} \]. With a clean equation, the function's behavior at various points is easier to analyze or integrate if needed. Simplification helps make complex derivatives more manageable for future tasks.