Problem 14 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the Function Type

The given function is a rational function of the form \( f(x) = \frac{g(x)}{h(x)} \), where \( g(x) = x - e^{-x} \) and \( h(x) = 1 + x e^{-x} \). We'll use the quotient rule for differentiation.

Step 2: Recall the Quotient Rule

The quotient rule states that for a function \( f(x) = \frac{u(x)}{v(x)} \), its derivative is given by \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \). Here, \( u(x) = x - e^{-x} \) and \( v(x) = 1 + x e^{-x} \).

Step 3: Differentiation of the Numerator

Differentiate \( u(x) = x - e^{-x} \) to get \( u'(x) = 1 + e^{-x} \), since the derivative of \( x \) is 1, and the derivative of \( -e^{-x} \) is \( e^{-x} \).

Step 4: Differentiation of the Denominator

Differentiate \( v(x) = 1 + x e^{-x} \). This requires the product rule for \( x e^{-x} \), giving \( v'(x) = e^{-x} - x e^{-x} \). The derivative of \( 1 \) is 0.

Step 5: Apply the Quotient Rule

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

Step 6: Simplify Each Part

Expand the terms in the numerator:- Expand \( (1 + e^{-x})(1 + x e^{-x}) \) to get \( 1 + x e^{-x} + e^{-x} + x e^{-2x} \).- Expand \( (x - e^{-x})(e^{-x} - x e^{-x}) \) to get \( x e^{-x} - x^2 e^{-2x} - e^{-2x} + x e^{-x} \).

Step 7: Combine Like Terms

Combine all terms in the numerator to simplify:- Combine and simplify to get \( 1 + 2x e^{-x} + 2e^{-x} \) after canceling and combining terms. This gives the final expression for the simplified derivative.

Step 8: Final Expression for the Derivative

The simplified derivative of \( f(x) \) is then:\[ f'(x) = \frac{1 + 2x e^{-x} + 2e^{-x}}{(1 + x e^{-x})^2} \].

Key Concepts

Rational Functions

Rational functions are one of the building blocks of calculus, helping us understand complex relationships between variables. A rational function is expressed as the ratio of two polynomials. In our case, the function is \( f(x) = \frac{x - e^{-x}}{1 + x e^{-x}} \). The numerator and denominator are functions themselves: \( g(x) = x - e^{-x} \) and \( h(x) = 1 + x e^{-x} \). When dealing with a rational function, especially in differentiation, it's essential to recognize this structure. This allows us to apply specific rules like the quotient rule to find derivatives. This awareness helps take apart the problem step by step, making complex differentiation much more manageable.
To differentiate rational functions effectively, we rely on distinguishing the form \( \frac{g(x)}{h(x)} \), and then carefully applying rules that suit their structure. Mastery in identifying these forms simplifies the calculus process significantly, especially when preparing for higher-level mathematical challenges.

Differentiation

Differentiation is a core concept in calculus that involves finding the derivative of a function, which represents the rate of change. For the function \( f(x) = \frac{x - e^{-x}}{1 + x e^{-x}} \), differentiation entails breaking it down into smaller parts that can be handled with well-defined rules.

• The derivative of a sum or difference is simply the sum or difference of their derivatives.
• When encountering exponential functions such as \( e^{-x} \), we use their specific derivatives: the derivative of \( e^{-x} \) is \( -e^{-x} \).
• Differentiation of simple terms, like \( x \), results in 1.

By understanding the key principles of differentiation, you can tackle most mathematically expressed problems involving rate of change, slopes, speeds, and much more. It’s especially pivotal when working with complex functions, offering the foundational skills needed for more advanced calculus applications.

Product Rule

The product rule is a crucial tool in differentiation, helping when dealing with functions that are multiplied by each other. When differentiating \( v(x) = 1 + x e^{-x} \), the product rule becomes indispensable, as it is required for the component \( x e^{-x} \). Here's how the product rule works:
For two functions \( u(x) \) and \( v(x) \), the derivative of their product is given by:

• \( (uv)' = u'v + uv' \).

Applying this to \( x e^{-x} \):

• The derivative of \( x \) is 1.
• The derivative of \( e^{-x} \) is \( -e^{-x} \).

Thus, when applying the product rule, the derivative of \( x e^{-x} \) becomes \( 1 \cdot e^{-x} + x(-e^{-x}) \), which simplifies to \( e^{-x} - x e^{-x} \).
The product rule is indispensable when addressing combinations of different types of functions, allowing us to systematically find their derivative while maintaining an organized structure.