Problem 14 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The derivative is \( \frac{1 + x^2 e^{-x} + x e^{-2x}}{(1 + x e^{-x})^2} \).

Step 1: Identify the Differentiation Rule

The function \( f(x) = \frac{x - e^{-x}}{1 + x e^{-x}} \) is a rational function, which means we need to use the quotient rule for differentiation. Recall that the quotient rule states \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \), where \( u \) and \( v \) are functions of \( x \).

Step 2: Define the Top and Bottom Functions

Identify the top function as \( u = x - e^{-x} \) and the bottom function as \( v = 1 + x e^{-x} \). We'll differentiate these separately to use in the quotient rule.

Step 3: Differentiate the Top Function

Differentiate \( u \) with respect to \( x \): \( u = x - e^{-x} \)\( u' = 1 + e^{-x} \)

Step 4: Differentiate the Bottom Function

Differentiate \( v \) with respect to \( x \):\( v = 1 + x e^{-x} \)Use the product rule on \( x e^{-x} \):\( \frac{d}{dx}(x e^{-x}) = x(-e^{-x}) + e^{-x}(1) = -x e^{-x} + e^{-x} \)Thus, \( v' = -x e^{-x} + e^{-x} \).

Step 5: Apply the Quotient Rule

Substitute \( u \), \( u' \), \( v \), and \( v' \) into the quotient rule formula:\[\left( \frac{x - e^{-x}}{1 + x e^{-x}} \right)' = \frac{(1 + e^{-x})(1 + x e^{-x}) - (x - e^{-x})(-x e^{-x} + e^{-x})}{(1 + x e^{-x})^2}\]

Step 6: Simplify the Expression

Simplify the expression in the derivative:1. Calculate the numerator: - Expand the products: - First term: \( 1 \times (1 + x e^{-x}) + e^{-x} \times (1 + x e^{-x}) = 1 + x e^{-x} + e^{-x} + x e^{-2x}\) - Second term: \( -(x \times -x e^{-x} + e^{-x}) = x^2 e^{-x} - x e^{-x} \)2. Combine like terms: - Collect all terms: \[ 1 + x e^{-x} + e^{-x} + x e^{-2x} + x^2 e^{-x} - x e^{-x} - e^{-2x} \] - Final simplified numerator: \[ 1 + x^2 e^{-x} + x e^{-2x} \]3. The expression becomes: \[\frac{1 + x^2 e^{-x} + x e^{-2x}}{(1 + x e^{-x})^2}\]

Step 7: Conclusion

The derivative of \( f(x) = \frac{x - e^{-x}}{1 + x e^{-x}} \) with respect to \( x \) is:\[\frac{1 + x^2 e^{-x} + x e^{-2x}}{(1 + x e^{-x})^2}\]

Key Concepts

Quotient rule

The quotient rule is an essential technique in calculus for finding the derivative of a function that is expressed as the division or quotient of two other functions. Whenever we have a function in the form \( \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), the quotient rule helps us find the derivative of this expression efficiently. The rule is defined as: \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]This formula tells us that we need to:

• Differentiate the numerator \( u \) to get \( u' \).
• Differentiate the denominator \( v \) to get \( v' \).
• Apply the formula using the derivatives and functions themselves.

Then you calculate \( u'v - uv' \) and finally divide by \( v^2 \), where \( v \) is the original denominator.
Remember, it's crucial that the denominator \( v \) is not zero, otherwise, the function or its derivative is not defined at those points. With the quotient rule, you can tackle a wide array of rational functions smoothly.

Product rule

When it comes to the differentiation of products of two functions, the product rule is our go-to tool. If you have a function \( y = f(x)g(x) \), where both \( f \) and \( g \) are differentiable, the product rule provides a straightforward path to its derivative. It states:\[ ( f(x)g(x) )' = f'(x)g(x) + f(x)g'(x) \]This means that to differentiate the product of two functions, you:

• Find the derivative of the first function \( f \), call it \( f' \), and multiply it by the second function \( g \).
• Find the derivative of the second function \( g \), call it \( g' \), and multiply it by the first function \( f \).
• Add the two results together to get the final derivative.

This rule is particularly helpful because it breaks down differentiation into manageable steps. Unlike simply multiplying the derivatives, the product rule ensures that all interactions between the two functions are taken into account. In the differentiation processes involving quotients or complex interactions as in the given exercise, product rule assists in differentiating parts of the function separately, such as when a term in the quotient involves a product.

Rational function differentiation

Rational functions, which are expressions forming a ratio of two polynomials, require special consideration when differentiating. Generally, rational function differentiation employs the quotient rule, as these functions consist of a numerator and a denominator. To differentiate a rational function properly:

• Identify the numerator and denominator polynomials individually.
• Apply the quotient rule, where each part is carefully differentiated. This is often essential since each component might involve their own differentiation rules.
• Simplify the resulting derivative expression, merging terms where possible for clarity.

In the context of differentiating functions like \( f(x) = \frac{x - e^{-x}}{1 + x e^{-x}} \), rational functions reveal subtle complexities.
To differentiate, first recognize that both the numerator \( x - e^{-x} \) and the denominator \( 1 + x e^{-x} \) need their derivatives via product and basic differentiation rules.
After implementing the quotient rule, simplify the expression by organizing like terms to make the function easier to interpret and use. Understanding and applying these methods offers a robust way to handle rational expressions in calculus.