Problem 16 - Calculus for Biology and Medicine

Question

Differentiate the functions in Problems 1-52 with respect to the independent variable. $$ f(x)=\frac{x}{e^{x}+e^{-x}} $$

Step-by-Step Solution

Answer

The derivative of the function is \( f'(x) = \frac{(1-x)e^x + (1+x)e^{-x}}{(e^x + e^{-x})^2} \).

Step 1: Identify the Functions and the Form

The given function is \( f(x) = \frac{x}{e^x + e^{-x}} \). This can be identified as a quotient of two functions, which means we need to use the quotient rule for differentiation.

Step 2: Recall the Quotient Rule

The quotient rule is used to differentiate functions that are divided by each other. The rule is stated as: if \( u(x) \) and \( v(x) \) are functions of \( x \), then the derivative of \( \frac{u(x)}{v(x)} \) with respect to \( x \) is \( \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \).

Step 3: Assign Functions as \( u(x) \) and \( v(x) \)

Set \( u(x) = x \) and \( v(x) = e^x + e^{-x} \). Compute the derivatives: \( u'(x) = 1 \) and \( v'(x) = e^x - e^{-x} \). This is because the derivative of \( e^x \) is \( e^x \) and the derivative of \( e^{-x} \) is \( -e^{-x} \).

Step 4: Apply the Quotient Rule

Using the quotient rule, substitute \( u(x) \), \( v(x) \), \( u'(x) \), and \( v'(x) \):\[f'(x) = \frac{1 \cdot (e^x + e^{-x}) - x \cdot (e^x - e^{-x})}{(e^x + e^{-x})^2}\]

Step 5: Simplify the Expression

Simplify the expression from Step 4:\[f'(x) = \frac{e^x + e^{-x} - x(e^x - e^{-x})}{(e^x + e^{-x})^2}\]Expanding and simplifying further gives:\[ f'(x) = \frac{e^x + e^{-x} - xe^x + xe^{-x}}{(e^x + e^{-x})^2}\]Group like terms:\[f'(x) = \frac{(1-x)e^x + (1+x)e^{-x}}{(e^x + e^{-x})^2}\]

Step 6: Check the Final Answer

The final derivative of the function \( f(x) = \frac{x}{e^x + e^{-x}} \) is \[f'(x) = \frac{(1-x)e^x + (1+x)e^{-x}}{(e^x + e^{-x})^2}\]

Key Concepts

Quotient Rule

The quotient rule is a critical tool in calculus, especially when differentiating fractions where both the numerator and denominator are functions of the variable \( x \). It helps in finding the derivative of a function that is the division of two differentiable functions. Imagine having two functions \( u(x) \) and \( v(x) \). The quotient rule gives us a formula to differentiate the function \( \frac{u(x)}{v(x)} \). This is especially helpful because directly applying simple derivative rules isn't possible when dealing with such division.Here's the formula: if you have \( f(x) = \frac{u(x)}{v(x)} \), then the derivative \( f'(x) \) is given by:

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

This formula tells us that we first find the derivative of the numerator \( u(x) \) and multiply it by the denominator \( v(x) \), and from this product, subtract the result of multiplying the numerator \( u(x) \) by the derivative of the denominator \( v'(x) \). Finally, we divide everything by the square of the denominator \( (v(x))^2 \).Remember, to use the quotient rule, it is essential both \( u(x) \) and \( v(x) \) are differentiable. Applying the rule correctly ensures precise results when differentiating quotients.

Functions of x

Within calculus, understanding functions of \( x \) is pivotal. Functions of \( x \) form the basis of most calculus problems. A function outputs a result based on the input value \( x \). For example, in our provided exercise, the function is \( f(x) = \frac{x}{e^x + e^{-x}} \). This is a perfect illustration of a ratio where both the numerator \( x \) and the denominator \( e^x + e^{-x} \) are functions of \( x \).In such composite functions, where the components involve different operations (like addition, subtraction, exponentials), the main task is usually differentiating them effectively and efficiently. Differentiating such functions often requires a mix of basic derivative rules and special rules, like the quotient rule, to manage their complexity.Recognizing each part of the function and its role can simplify the process of applying differentiation rules. For our example, identifying \( u(x) = x \) and \( v(x) = e^x + e^{-x} \) allows us to directly apply the quotient rule.

Derivative Calculation

Derivative calculation is the core of many calculus problems. Calculating derivatives means finding the rate at which a function is changing at any given point. For the exercise provided, we focused on the function \( f(x) = \frac{x}{e^x + e^{-x}} \). To differentiate this function, we applied the quotient rule.First, we identified the components \( u(x) \) and \( v(x) \):

• \( u(x) = x \)
• \( v(x) = e^x + e^{-x} \)

Calculating the derivatives of these functions gives us:

• \( u'(x) = 1 \)
• \( v'(x) = e^x - e^{-x} \)

Next, substitute these into the quotient rule formula to find \( f'(x) \): \[ f'(x) = \frac{1 \cdot (e^x + e^{-x}) - x \cdot (e^x - e^{-x})}{(e^x + e^{-x})^2}\]After simplifying, we arrive at:\[ f'(x) = \frac{(1-x)e^x + (1+x)e^{-x}}{(e^x + e^{-x})^2}\]The result gives the derivative of the function, showing how the function changes with respect to \( x \). This powerful technique allows us to understand the behavior of complex functions in calculus.