Problem 26 - Single Variable Calculus

Question

Find the derivative of the function. $$ g(x)=e^{-2 x} \cos 3 x $$

Step-by-Step Solution

Answer

The derivative of the function \(g(x) = e^{-2x} \cos 3x\) is \(g'(x) = -2e^{-2x}\cos{3x} - 3e^{-2x}\sin{3x}\).

Step 1: Apply the product rule

Let's denote \(u(x) = e^{-2x}\) and \(v(x) = \cos{3x}\). We can now rewrite our function as \(g(x) = u(x) \cdot v(x)\). To find the derivative of the function \(g(x)\), we will apply the product rule: \[ \frac{d}{dx}(u(x) \cdot v(x)) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \]

Step 2: Find the derivatives of u(x) and v(x)

We will now find the derivatives of the functions u(x) and v(x). For \(u(x) = e^{-2x}\), apply the chain rule: \[ u'(x) = \frac{d}{dx}(e^{-2x}) = -2e^{-2x} \] For \(v(x) = \cos{3x}\), apply the chain rule: \[ v'(x) = \frac{d}{dx}(\cos{3x}) = -3\sin{3x} \]

Step 3: Use the product rule with the derivatives

Now we will substitute the derivatives u'(x) and v'(x) into the product rule formula: \[ g'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) = (-2e^{-2x})(\cos{3x}) + (e^{-2x})(-3\sin{3x}) \]

Step 4: Simplify the expression

Now, let's simplify the expression: \[ g'(x) = -2e^{-2x}\cos{3x} - 3e^{-2x}\sin{3x} \] So, the derivative of the function \(g(x) = e^{-2 x} \cos 3 x\) is: \[ g'(x) = -2e^{-2x}\cos{3x} - 3e^{-2x}\sin{3x} \]

Key Concepts

Understanding the Product Rule

When you need to find the derivative of a function that is the product of two other functions, the product rule comes into play. This rule states that if you have a function \(g(x) = u(x) \cdot v(x)\), then the derivative \(g'(x)\) is computed as:

• First derivative of \(u(x)\), multiplied by \(v(x)\)
• Plus \(u(x)\), multiplied by the derivative of \(v(x)\)

Mathematically, it's written as:\[\frac{d}{dx}(u(x) \cdot v(x)) = u'(x) \cdot v(x) + u(x) \cdot v'(x)\]This means you need to differentiate each function separately and then combine them using this formula. It’s an essential rule for tackling complex functions, especially when dealing with combinations like trigonometric and exponential functions.

Making Sense of the Chain Rule

The chain rule is used when you have a function within a function—commonly called a composition of functions. If you have a function \(f(g(x))\), the chain rule helps you find its derivative. The rule tells us to differentiate the outer function first and then multiply by the derivative of the inner function.

• Differentiate the outer function while keeping the inner function unchanged.
• Multiply by the derivative of the inner function.

This technique is simple but very powerful. For example, with \(u(x) = e^{-2x}\), the outer function is the exponential function \(e^u\), and the inner function is \(-2x\). Differentiating gives \(-2e^{-2x}\) because you multiply the derivative of the outer by the derivative of the inner.

The Role of Trigonometric Functions

Trigonometric functions, like sine and cosine, have specific derivatives that are important in calculus. For a function like \(\cos{3x}\), the derivative involves the chain rule since it's a composition function. The derivative of \(\cos{u}\) is \(-\sin{u}\). Thus, finding \(v'(x)\) for \(v(x) = \cos{3x}\) means:

• Firstly, differentiate \(\cos{u}\) to get \(-\sin{u}\).
• Secondly, multiply by the derivative of the inner, which is the \(3x\) in this case, giving \(-3\sin{3x}\).

Understanding these derivatives helps you solve a wide range of problems and find the rate of change of trigonometric functions quickly.

Understanding Exponential Functions

Exponential functions often appear in calculus problems due to their constant rate of change characteristics. When differentiating an exponential function like \(e^{-2x}\), we utilize the chain rule.

• The derivative of \(e^x\) is itself, \(e^x\).
• For \(e^{-2x}\), apply the chain rule: the outer function is \(e^x\) and inner function is \(-2x\), thus the derivative becomes \(-2e^{-2x}\).

This concept is pivotal in both natural processes and mathematical modeling. Knowing how to differentiate exponential functions allows you to tackle problems in calculus with greater confidence.