Problem 22 - Thomas Calculus

Question

In Exercises \(17-36,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=(1+2 x) e^{-2 x} $$

Step-by-Step Solution

Answer

The derivative is \(-4xe^{-2x}\).

Step 1: Identify the Function Type

The given function is the product of two functions: \((1+2x)\) and \(e^{-2x}\). This means we will need to use the product rule to find the derivative.

Step 2: Apply the Product Rule

The product rule for differentiation is given by \((u \cdot v)' = u' \cdot v + u \cdot v'\), where \(u = 1 + 2x\) and \(v = e^{-2x}\). We need to find \(u'\) and \(v'\).

Step 3: Differentiate Each Function Separately

Differentiate \(u = 1 + 2x\) with respect to \(x\). The derivative \(u' = \frac{d}{dx}(1 + 2x) = 2\). For \(v = e^{-2x}\), differentiate with respect to \(x\) using the chain rule. \(v' = \frac{d}{dx}(e^{-2x}) = -2e^{-2x}\).

Step 4: Plug into the Product Rule Formula

Substitute \(u\), \(u'\), \(v\), and \(v'\) into the product rule formula: \[(1+2x)' \cdot e^{-2x} + (1+2x) \cdot (e^{-2x})' = 2 \cdot e^{-2x} + (1+2x) \cdot (-2e^{-2x})\].

Step 5: Simplify the Expression

Compute the expression: \[2e^{-2x} - 2(1+2x)e^{-2x} = 2e^{-2x} - 2e^{-2x} - 4xe^{-2x}\].This simplifies to \[-4xe^{-2x}\].

Key Concepts

Understanding the Product Rule

When faced with functions that are products of two separate functions, the product rule is your go-to tool for differentiation. This rule helps break down the complexity of differentiating such combinations by considering each function's individual derivative and their interaction. For any two functions, say \( f(x) \) and \( g(x) \), the derivative of their product is given by:
\[(f \cdot g)' = f' \cdot g + f \cdot g'\]
This formula implies that you need to differentiate each function separately and then apply the given formula. This allows us to handle products of functions effectively.

• Differentiate the first function and multiply by the second function as it is.
• Then, differentiate the second function and multiply by the first function as it is.

The sum of these results gives you the product's derivative. With practice, the product rule becomes a straightforward process.

Exploring the Chain Rule

The chain rule is essential when dealing with composite functions, which are functions of a function. It allows you to differentiate outer layers by considering the derivatives of inner layers. An example of a composite function could be \( e^{-2x} \).
To apply the chain rule, follow these steps:

• Identify the inner function and differentiate it.
• Differentiate the outer function with respect to the inner function.
• Multiply these derivatives to get the final result.

For example, in \( e^{-2x} \), if we consider \(-2x\) as \( g(x) \) and \( e^{g(x)} \) as the outer function, the derivative becomes:
\[\frac{d}{dx} e^{-2x} = \frac{d}{dg}(e^g) \cdot \frac{dg}{dx} = e^{-2x} \cdot (-2) = -2e^{-2x}\]
The chain rule is a powerful tool that is widely applicable in calculus.

Basic Differentiation Principles

Differentiation is a cornerstone of calculus, used to determine how a function changes as its input changes. It's essentially about finding the rate of change or slope of a function at any point. Differentiation applies rules systematically to simplify these calculations.
Some basic rules of differentiation include:

• Constant Rule: The derivative of a constant is zero.
• Power Rule: For \(x^n\), the derivative is \(nx^{n-1}\).
• Sum Rule: The derivative of a sum is the sum of the derivatives.

Differentiation utilizes these rules in combination with others like product and chain rules to handle more complex functions effectively.

Working with Exponential Functions

Exponential functions, such as \( e^x \), are pervasive in calculus because of their unique property where the derivative is proportional to the function itself. This makes them easier to differentiate, especially when combined with the chain rule.
Consider the function \( e^{-2x} \). Its derivative involves the chain rule because of the inner function \(-2x\). The derivative results in:
\[\frac{d}{dx} e^{-2x} = -2e^{-2x}\]
In essence, the process involves:

• Recognizing that the base of the exponential function remains unchanged.
• Applying the chain rule for the exponent's derivative.

Understanding exponential functions and their differentiability is crucial for solving various application problems in calculus.