Problem 2 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. \(f(x)=e^{-2 x}\)

Step-by-Step Solution

Answer

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

Step 1: Identify Function and Variable

The given function is \(f(x) = e^{-2x}\), and we need to differentiate this function with respect to the independent variable \(x\).

Step 2: Apply the Derivative Rule for Exponentials

Recall that the derivative of \(e^{u}\) with respect to \(x\) is \(e^{u} \cdot \frac{du}{dx}\). In this case, \(u = -2x\), so we need to find \(\frac{du}{dx}\).

Step 3: Differentiate the Exponent

Differentiate \(-2x\) with respect to \(x\). This gives us \(\frac{d(-2x)}{dx} = -2\).

Step 4: Combine the Results

Substitute \(u = -2x\) and \(\frac{du}{dx} = -2\) back into the exponential derivative rule. Thus, the derivative of \(f(x) = e^{-2x}\) is \(e^{-2x} \cdot (-2)\).

Step 5: Simplify the Expression

Simplify the expression \(-2 \cdot e^{-2x}\). This results in \(f'(x) = -2e^{-2x}\).

Key Concepts

Exponential Functions

Exponential functions are a special type of mathematical function where the variable appears in the exponent. They have the form \(f(x) = a^{x}\), but when the base is the natural exponential number \(e\), it becomes \(f(x) = e^{x}\). Here, the function grows or decays at a constant rate, making them very useful in modeling real-world phenomena such as population growth, radioactive decay, and continuously compounded interest.

• The number \(e\) is approximately equal to 2.71828 and is the base of natural logarithms.
• In exponential functions, changes in \(x\) cause rapid increases or decreases in the function's value.
• Exponential functions are continuous and smooth, meaning they have no breaks, holes, or sharp corners.

In our exercise, \(f(x) = e^{-2x}\) is an exponential function because it has \(e\) raised to the power of \(-2x\). Understanding how to differentiate such functions is crucial, especially in calculus, as these functions frequently appear in many applications.

Chain Rule

The Chain Rule is a critical differentiation technique used when dealing with composite functions, where one function is nested inside another. It's a rule that allows us to differentiate the outer function and the inner function separately.Consider a function \(f(g(x))\), where \(f\) is the outer function and \(g\) is the inner function. The chain rule states that the derivative \(\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)\).

• Identify the outer and inner functions correctly to apply the rule.
• Differentiate the outer function leaving the inner function unchanged.
• Multiply by the derivative of the inner function.

In our original exercise, \(f(x) = e^{-2x}\), the outer function is \(e^{u}\) and the inner function is \(u = -2x\). The chain rule is employed to differentiate \(e^{u}\) with respect to \(x\), leading to \(f'(x) = e^{-2x} \cdot (-2)\). This step is vital because it allows us to differentiate more complex functions by breaking them down into simpler parts.

Differentiation Steps

Differentiation is the process of finding the derivative of a function, which represents the rate at which the function's value changes as the variable changes. It is one of the most fundamental operations in calculus. Here are the steps for differentiating a composite function with an exponential component, as seen in the exercise:1. **Identify the Function and Variable**: Locate the function to be differentiated and the variable with respect to which differentiation will occur. In our case, \(f(x) = e^{-2x}\), where the variable is \(x\).2. **Apply the Derivative Rule for Exponentials**: Use the rule \(\frac{d}{dx} e^{u} = e^{u} \cdot \frac{du}{dx}\), which requires the chain rule for differentiation.3. **Differentiate the Exponent**: Calculate \(\frac{du}{dx}\), which involves differentiating the exponent \(-2x\). This results in \(-2\).4. **Combine the Results**: Substitute back \(-2x\) and \(-2\) into the derivative expression \(e^{u} \cdot \frac{du}{dx}\).5. **Simplify the Expression**: Simplify the result; in this example, multiplying gives \(-2e^{-2x}\). These steps ensure clarity and accuracy in differentiation processes, especially when dealing with complex composite functions.