Problem 335 - Calculus Volume 1

Question

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: Understanding the Function

To find the derivative of the function, we first take a look at its structure. The function is given as a quotient: \( f(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} \). This consists of a numerator and a denominator that are both made up of exponential functions. We will need to use the quotient rule for differentiation to find its derivative.

Step 2: Applying the Quotient Rule

The quotient rule states that if \( u(x) = \frac{v(x)}{w(x)} \), then \( u'(x) = \frac{v'(x)w(x) - v(x)w'(x)}{(w(x))^2} \). In this case, let \( v(x) = e^x - e^{-x} \) and \( w(x) = e^x + e^{-x} \). We now need to find \( v'(x) \) and \( w'(x) \).

Step 3: Differentiating the Numerator and Denominator

Differentiate \( v(x) = e^x - e^{-x} \) to get \( v'(x) = e^x + e^{-x} \), because \( \frac{d}{dx}(e^x) = e^x \) and \( \frac{d}{dx}(e^{-x}) = -e^{-x} \).Differentiate \( w(x) = e^x + e^{-x} \) to get \( w'(x) = e^x - e^{-x} \), using the same rules as above.

Step 4: Substituting into the Quotient Rule

Substitute \( v(x) \), \( v'(x) \), \( w(x) \), and \( w'(x) \) 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: Simplifying the Result

Expand the terms in the numerator: \( (e^x + e^{-x})^2 = e^{2x} + 2 + e^{-2x} \) and \( (e^x - e^{-x})^2 = e^{2x} - 2 + e^{-2x} \).Subtract the second from the first: \( e^{2x} + 2 + e^{-2x} - (e^{2x} - 2 + e^{-2x}) = 4 \).Thus, \( f'(x) = \frac{4}{(e^x + e^{-x})^2} \).

Step 6: Final Derivative

The derivative of the function \( f(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} \) is: \[ f'(x) = \frac{4}{(e^x + e^{-x})^2} \]

Key Concepts

Quotient Rule

The quotient rule is a differential technique that is specifically used to differentiate functions that are presented as fractions, where one function is divided by another. It comes in handy when dealing with a situation like this one, where the function is in the form of a ratio of two expressions. For any function \( u(x) = \frac{v(x)}{w(x)} \), the derivative \( u'(x) \) is given by:\[ u'(x) = \frac{v'(x)w(x) - v(x)w'(x)}{(w(x))^2}\]Here’s how it works:

• Differentiate the numerator v(x) to find v'(x).
• Differentiate the denominator w(x) to get w'(x).
• Substitute these into the quotient rule formula.

The beauty of the quotient rule is that it allows us to break down complex functions into more manageable parts, enabling us to systematically find their derivatives. Remember, the order matters here, and practising the quotient rule is all about getting comfortable with differentiating the parts of the function correctly.

Exponential Functions

Exponential functions are a type of mathematical expression where the variable is in the exponent, such as \(e^x\), which defines the flow in both the numerator and the denominator of the function given. These functions are known for their growth rates which can be incredibly rapid.A key property of exponential functions is their differentiation. The derivative of the exponential function \(e^x\) is unique in that it is the same as the original function:

• \(\frac{d}{dx}(e^x) = e^x\)
• For negative exponents, such as \(e^{-x}\), the derivative is influenced by the power rule, leading to: \(\frac{d}{dx}(e^{-x}) = -e^{-x}\)

In the original function \( f(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} \), these differentiation rules play a critical role in calculating both \(v'(x)\) and \(w'(x)\). Understanding the behavior of exponential functions is crucial for tackling a wide range of problems in calculus, especially when combined with rules like the quotient rule.

Differentiation Techniques

Differentiation techniques help us to find the rate at which a function changes at any given point and are foundational in calculus. Applying these techniques correctly can reveal insightful details about the behavior and properties of functions.The exercise gives insight into several differentiation techniques:

• The quotient rule as outlined teaches us how to handle complex quotients.
• The differentiation of both simple (\(e^x\)) and complex (\(e^{-x}\)) exponential functions.
• Simplification steps that make understanding the final derivative easier.

The process begins with breaking down the function using differentiation rules, then methodically applying them to find derivatives of individual components. After differentiation comes simplification, which transforms what could be a mess of terms into a neat and interpretable expression. Understanding these steps helps students predict outcomes and explore deeper aspects of calculus, reinforcing both their problem-solving skills and theoretical knowledge.