Problem 18 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Differentiate the given function. $$f(x)=e^{-2 x} \cos 3 x$$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = e^{-2x} (-2 \text{cos}\thinspace 3x - 3 \text{sin}\thinspace 3x) \).

Step 1: Identify the necessary rule

The function is a product of two functions: \(e^{-2x}\) and \(\text{cos}\thinspace 3x\). Use the product rule for differentiation, which is: \( (u \thinspace v)' = u' \thinspace v + u \thinspace v' \).

Step 2: Differentiate \(u\)

Here, \(u = e^{-2x}\). The derivative \(u'\) is given by applying the chain rule: \( u' = \frac{d}{dx} (e^{-2x}) = e^{-2x} \frac{d}{dx} (-2x) = -2e^{-2x} \).

Step 3: Differentiate \(v\)

Here, \(v = \text{cos}\thinspace 3x\). The derivative \(v'\) is given by applying the chain rule again: \( v' = \frac{d}{dx} (\text{cos}\thinspace 3x) = \text{cos}'\thinspace 3x \frac{d}{dx}(3x) = -\text{sin}\thinspace 3x(3) = -3\text{sin}\thinspace 3x \).

Step 4: Apply the product rule

Substitute the derivatives \(u'\) and \(v'\) into the product rule formula: \( (e^{-2x} \text{cos}\thinspace 3x)' = (-2e^{-2x}) (\text{cos}\thinspace 3x) + (e^{-2x}) (-3 \text{sin}\thinspace 3x) \).

Step 5: Simplify the expression

Combine the terms to get the final result: \( f'(x) = -2e^{-2x} \text{cos}\thinspace 3x - 3 e^{-2x} \text{sin}\thinspace 3x = e^{-2x} (-2 \text{cos}\thinspace 3x - 3 \text{sin}\thinspace 3x) \).

Key Concepts

Chain Rule

The chain rule is a fundamental differentiation rule in calculus. It helps you find the derivative of a composed function. When you have one function nested inside another, you use the chain rule to differentiate them step-by-step.

For example, if you have a function of the form: \[ y = f(g(x)) \] you apply the chain rule as follows: \[ \frac{dy}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx} \]
This means you first differentiate the outer function with respect to the inner function and then multiply by the derivative of the inner function.

In our exercise, we applied the chain rule to differentiate both \( e^{-2x} \) and \( \text{cos} 3x \). Here, \( g(x) = -2x \) and \( g(x) = 3x \). We differentiated these inner functions and multiplied them by the derivatives of the outer functions, \( e^x \) and \( \text{cos }x \).

Exponential Function Differentiation

Differentiating exponential functions is straightforward when you know the rules. Exponential functions with base \( e \) (Euler's number) are particularly common in calculus.

For an exponential function of the form: \[ y = e^{g(x)} \] the derivative is given by: \[ \frac{dy}{dx} = e^{g(x)} \cdot \frac{dg(x)}{dx} \]
This is where the chain rule comes into play again, as you need to multiply by the derivative of the exponent.

In our exercise, to differentiate \( e^{-2x} \), we recognized that the derivative of \( e^{g(x)} \) is itself multiplied by the derivative of the exponent \( -2x \). This gave us \( -2 e^{-2x} \) as the final result after differentiation.

Trigonometric Function Differentiation

Trigonometric functions have their own set of differentiation rules that you should memorize. The basic derivatives are:

• \( \frac{d}{dx} \text{sin}(x) = \text{cos}(x) \)
• \( \frac{d}{dx} \text{cos}(x) = -\text{sin}(x) \)
• \( \frac{d}{dx} \text{tan}(x) = \text{sec}^2(x) \)

When trigonometric functions have a more complex argument (like \( 3x \) instead of \( x \)), you'll need to use the chain rule again.

For example, if you have \( y = \text{cos}(3x) \), the derivative is: \[ \frac{dy}{dx} = -\text{sin}(3x) \cdot \frac{d}{dx}(3x) \]
This simplifies to \( -3 \text{sin}(3x) \).

In our exercise, we used this rule to find the derivative of \( \text{cos}(3x) \), which was \( -3 \text{sin}(3x) \).