Problem 12 - Thomas Calculus

Question

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. \begin{equation}y=(1+2 x) e^{-2 x}\end{equation}

Step-by-Step Solution

Answer

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

Step 1: Identify the function type

The given function is a product of two differentiable functions: \((1 + 2x)\) and \(e^{-2x}\). This is a product rule differentiation problem.

Step 2: Recall the product rule formula

The product rule in differentiation states that if you have two functions \(f(x)\) and \(g(x)\), then the derivative \(\dfrac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)\).

Step 3: Differentiate the first function

Differentiate \(f(x) = 1 + 2x\). The derivative is \(f'(x) = 2\) since the derivative of a constant is zero and the derivative of \(2x\) is \(2\).

Step 4: Differentiate the second function

Differentiate \(g(x) = e^{-2x}\). The derivative using the chain rule is \(g'(x) = -2e^{-2x}\), because the derivative of \(e^{u}\) with respect to \(u\) is \(e^{u}\) multiplied by the derivative of \(u = -2x\), which is \-2\.

Step 5: Apply the product rule

Using the product rule, substitute \(f'(x)\), \(g(x)\), \(f(x)\), and \(g'(x)\): \[\dfrac{dy}{dx} = (2)e^{-2x} + (1+2x)(-2e^{-2x})\]

Step 6: Simplify the expression

Simplify the expression: \[\dfrac{dy}{dx} = 2e^{-2x} - 2(1+2x)e^{-2x} = 2e^{-2x} - 2e^{-2x} - 4xe^{-2x}\] Finally, we simplify further and the answer is: \[\dfrac{dy}{dx} = -4xe^{-2x}\]

Key Concepts

Chain Rule

The chain rule is an essential technique in calculus that helps us differentiate composite functions. In this context, a composite function is a function that contains another function within itself. The chain rule states that if you have a function \(y\) that can be expressed as \(u(v(x))\), then the derivative \(\frac{dy}{dx}\) is found by multiplying the derivative of the outer function \(u\) with respect to \(v\) by the derivative of the inner function \(v\) with respect to \(x\). This can be written as:

• \(\frac{dy}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx}\)

In our exercise, when differentiating \(e^{-2x}\), we use the chain rule. Here, the outer function is the exponential \(e^u\), and the inner function is \(u = -2x\). This means the derivative is \(-2e^{-2x}\), where \(-2\) is the derivative of \(-2x\). This demonstrates how the chain rule separates complex derivatives into more manageable parts.
Understanding the chain rule makes handling complex functions much easier.

Derivative

A derivative is a fundamental concept in calculus used to measure how a function changes as its input changes. It is commonly denoted as \(\frac{dy}{dx}\), which represents the rate of change of the function \(y\) with respect to the variable \(x\). Derivatives enable us to find slopes of curves, rates of change, and optimize functions. For instance, if \(y = f(x)\), the derivative \(f'(x)\) is calculated to find the slope at any point on the curve.
In the context of our exercise, derivatives help us find how the function \(y = (1 + 2x)e^{-2x}\) changes with respect to \(x\). By differentiating each part, we can assemble the full derivative using combination rules such as the chain and product rules. On a practical note, understanding derivatives can help in various fields like physics for motion analysis, economics for cost functions, and biology for growth rates.
Therefore, derivatives are not just abstract concepts but tools used to describe and manage real-world scenarios.

Exponential Functions

Exponential functions are types of functions characterized by their exponents varying with the variable. The general form is \(f(x) = a^x\) or more commonly \(f(x) = e^x\), where \(e\) is an irrational constant approximately equal to 2.71828. These functions are widely used in scientific fields because they describe continuous growth or decay processes, such as population growth, radioactive decay, and interest calculations.

• Exponential growth occurs when the base of the exponential function is greater than one.
• Exponential decay occurs when the base is between zero and one, or more commonly when the exponent is negative, such as \(e^{-x}\).

In our exercise, \(e^{-2x}\) represents an exponential decay function. When differentiating this using the chain rule, the nature of exponential functions simplifies the derivative process since the derivative of \(e^x\) is \(e^x\), with the chain rule accounting for more complex exponents. Exponential functions' innate properties make them pivotal in modeling real-world changes over time.
Understanding their nature simplifies tackling more complex calculus problems.

Differentiable Functions

Differentiable functions are those functions which have a derivative at every point in their domain. It implies smoothness and the ability to approximate the function with a tangent line at any point. For ensuring a function is differentiable, it must be continuous across its domain, and not have any sharp corners or discontinuities.
The given function \(y = (1 + 2x)e^{-2x}\) is differentiable. This is because both components \(1 + 2x\) and \(e^{-2x}\) are differentiable on their own. By the definition, any function expressed as a product or composition of differentiable functions remains differentiable. This property is crucial, as it allows us to apply differentiation rules like the product rule with confidence.
Ensuring differentiability is an important step before any derivative calculations to guarantee valid results. Differentiable functions are foundational in calculus and aid in various analyses across science and engineering.