Problem 36 - University Calculus: Early Transcendentals

Question

Find the derivatives of the functions in Exercises \(23-50\). $$y=(1+2 x) e^{-2 x}$$

Step-by-Step Solution

Answer

The derivative is \( y' = -4x e^{-2x} \).

Step 1: Identify Function Components

The function given is \( y = (1 + 2x) e^{-2x} \). We see that it is a product of two functions: \( u(x) = 1 + 2x \) and \( v(x) = e^{-2x} \).

Step 2: Differentiate First Component

Find the derivative of \( u(x) = 1 + 2x \). The derivative is \( u'(x) = 2 \).

Step 3: Differentiate Second Component

Find the derivative of \( v(x) = e^{-2x} \). Using the chain rule, the derivative is \( v'(x) = (-2) e^{-2x} \).

Step 4: Apply Product Rule

We use the product rule for derivatives, which states that \( (uv)' = u'v + uv' \). Substituting the found derivatives, we have \((1 + 2x)' \, e^{-2x} + (1 + 2x) \, (e^{-2x})'\).

Step 5: Substitute and Simplify

Substitute \( u'(x) = 2 \), \( v(x) = e^{-2x} \), \( u(x) = 1 + 2x \), and \( v'(x) = -2e^{-2x} \) into the product rule formula. The derivative is \( 2e^{-2x} + (1 + 2x)(-2e^{-2x}) \). Simplify it to get:\[ y' = 2e^{-2x} - 2(1 + 2x)e^{-2x} \]\[ y' = 2e^{-2x} - 2e^{-2x} - 4xe^{-2x} \]\[ y' = -4xe^{-2x} \].

Step 6: Final Derivative

The derivative of the function \( y = (1 + 2x)e^{-2x} \) is therefore \( y' = -4xe^{-2x} \).

Key Concepts

Understanding the Product Rule

The product rule is fundamental in calculus when differentiating products of two functions. If you encounter a product of the form \( y = u(x) \cdot v(x) \), the product rule helps you differentiate it. The formula to apply is \( (uv)' = u'v + uv' \). Each component, \( u(x) \) and \( v(x) \), is treated separately with respect to differentiation. This means you take the derivative of one function while keeping the other constant, and vice-versa.

• Step 1: Differentiate the first function \( u(x) \) to get \( u'(x) \).
• Step 2: Differentiate the second function \( v(x) \) to obtain \( v'(x) \).
• Step 3: Use the product rule formula to piece together these derivatives.

Remember, each term in the resulting derivative represents one of these steps. This formula is essential whenever you face a situation where both functions depend on \( x \). Recognizing when to use the product rule is key to solving such problems efficiently.

Applying the Chain Rule in Differentiation

The application of the chain rule is vital when dealing with compositions of functions, like those involving exponentials, powers, or trigonometric functions. For a function \( v(x) = e^{-2x} \) in our example, the chain rule facilitates differentiation by considering the outer and inner functions separately. Its essence lies in the idea that a change in the inner function affects the whole composition.

• Step 1: Identify the outer function \( e^x \) and the inner function, in this case \( -2x \).
• Step 2: Differentiate the outer function with respect to the inner function, leaving the inner function unchanged \( d(e^u)/du = e^u \).
• Step 3: Multiply by the derivative of the inner function \( d(-2x)/dx = -2 \).

Thus, using the chain rule gives \( v'(x) = (-2)e^{-2x} \). This rule is crucial for correctly handling complex derivatives where functions are nested within each other.

Basics of Function Differentiation

Function differentiation, the process of finding a derivative, unveils how one variable changes in relation to another. It's crucial for understanding the behavior of functions, especially in calculus. Differentiation allows you to find the slope of a curve at any given point, which is the essence of how functions evolve over an interval.

• Ensure clarity: Break down functions into simpler components, as seen with \( u(x) = 1 + 2x \) and \( v(x) = e^{-2x} \) in our example.
• Simplify calculation: Utilize established rules like the product rule and chain rule to simplify more complex derivative computations.
• Combine results: Connect derivatives of simple parts to form the derivative of more composite functions.

Understanding each step in differentiation not only helps you find derivatives but also boosts your ability to solve real-world problems by interpreting rates of change and predicting outcomes effectively.