Problem 10 - Thomas Calculus in SI Units

Question

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 \( \frac{dy}{dx} = -4xe^{-2x} \).

Step 1: Identify the Function Structure

The function given is a product of two terms: \( (1+2x) \) and \( e^{-2x} \). This indicates that we'll need to use the Product Rule for differentiation.

Step 2: Recall the Product Rule

The Product Rule states that if you have a function \( y = u(x) \, v(x) \), the derivative \( \frac{dy}{dx} \) is given by \( u'(x)v(x) + u(x)v'(x) \).

Step 3: Define Components for Product Rule

Let \( u(x) = 1 + 2x \) and \( v(x) = e^{-2x} \). We need to find \( u'(x) \) and \( v'(x) \).

Step 4: Differentiate \( u(x) \)

Differentiate \( u(x) = 1 + 2x \) with respect to \( x \) to get \( u'(x) = 2 \).

Step 5: Differentiate \( v(x) \)

Differentiate \( v(x) = e^{-2x} \) with respect to \( x \). Using the chain rule, \( v'(x) = -2e^{-2x} \).

Step 6: Apply the Product Rule

Use the product rule to find \( \frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) \). Substitute \( u'(x) = 2 \), \( v(x) = e^{-2x} \), \( u(x) = 1+2x \), and \( v'(x) = -2e^{-2x} \).

Step 7: Substitute and Simplify

Substitute into the product rule: \( \frac{dy}{dx} = 2e^{-2x} + (1+2x)(-2e^{-2x}) \). Simplify to get \( \frac{dy}{dx} = 2e^{-2x} - 2e^{-2x} - 4xe^{-2x} = -4xe^{-2x} \).

Step 8: Present the Final Derivative

The derivative of \( y \) with respect to \( x \) is \( \frac{dy}{dx} = -4xe^{-2x} \).

Key Concepts

Product Rule

Understanding the product rule is essential when dealing with differentiation, especially when a function is the product of two smaller functions. The rule helps you find the derivative of a product of two functions, say \( u(x) \) and \( v(x) \). According to the product rule, the derivative of \( y = u(x) \cdot v(x) \) is given by:\[\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)\]Here, you need to differentiate each of the functions individually and then apply the product rule. Let’s break it down:

• Differentiate \( u(x) \): Find \( u'(x) \).
• Use original \( v(x) \): Don’t change \( v(x) \).
• Add \( u(x) \) times \( v'(x) \): Differentiate \( v(x) \) to get \( v'(x) \) and multiply it by the original \( u(x) \).

The product rule results in a new function that shows how the rate of change is affected by both original functions.

Chain Rule

The chain rule is another crucial rule in differentiation which is particularly useful for composite functions—functions within another function. It allows you to find the derivative of a composite function by focusing on each layer of the function separately.The chain rule states that if a function \( y \) is composed of a function \( f(g(x)) \), then the derivative \( \frac{dy}{dx} \) is:\[\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\]Here's how this works:

• Differentiate the outer function: Treat the inside function as a single variable and differentiate the outer part (\( f'(g(x)) \)).
• Multiply by the derivative of the inner function: Differentiate the inside function \( g(x) \) to get \( g'(x) \), then multiply this result by the derivative of the outer function.

In our exercise, we applied the chain rule to differentiate \( e^{-2x} \), recognizing "\(-2x\)" as the inner function.

Derivative of Exponential Functions

Derivatives of exponential functions have their own set of rules, but they aren't too tricky once you get the hang of it. For functions in the form of \( e^{u(x)} \), the differentiation process involves using the chain rule, especially when \( u(x) \) isn’t a simple \( x \).If you are differentiating \( e^{u(x)} \), you will follow these steps:

• Exponential stays the same: The \( e^{u(x)} \) part remains unchanged in the differentiation.
• Multiply by the derivative of the exponent: Differentiate \( u(x) \) to get \( u'(x) \) and multiply it by \( e^{u(x)} \).

In the given exercise, with \( e^{-2x} \), you use the chain rule: the derivative is \(-2 \times e^{-2x}\). Exponential derivatives maintain the exponential part while multiplying by the derivative of the exponent, thus linking closely with chain rule techniques.