Problem 22 - Calculus Single Variable

Question

An expression for \(f(x)\) is given. Compute the first, second, and third derivatives of \(f(x)\) with respect to \(x\). \(x / 3^{x}\)

Step-by-Step Solution

Answer

First derivative: \(f'(x) = \frac{1 - x \ln(3)}{3^x}\). Second derivative: \(f''(x) = \frac{-\ln(3) + x \ln^2(3)}{3^x}\). Third derivative: \(f'''(x) = \frac{\ln^2(3) - \ln^3(3) - x \ln^3(3)}{3^x}\).

Step 1: Write down the function

The function given is \( f(x) = \frac{x}{3^x} \). We will differentiate this function with respect to \( x \).

Step 2: Use the Quotient Rule for First Derivative

To find the first derivative \( f'(x) \), use the Quotient Rule: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \) where \( u = x \) and \( v = 3^x \). First, compute \( u' = 1 \) and \( v' = 3^x \ln(3) \). Apply the rule:\[ f'(x) = \frac{1 \cdot 3^x - x \cdot 3^x \ln(3)}{(3^x)^2} = \frac{3^x - x 3^x \ln(3)}{9^x} = \frac{1 - x \ln(3)}{3^x} \].

Step 3: Simplify the First Derivative

Simplify the expression by factoring out \( \frac{1}{3^x} \):\[ f'(x) = \frac{1 - x \ln(3)}{3^x} \].

Step 4: Find the Second Derivative

Differentiate \( f'(x) = \frac{1 - x \ln(3)}{3^x} \) using the Quotient Rule again. Assign \( u = 1 - x \ln(3) \) and \( v = 3^x \). Calculate \( u' = -\ln(3) \) and \( v' = 3^x \ln(3) \). Use the rule:\[ f''(x) = \frac{(-\ln(3))3^x - (1-x \ln(3))3^x \ln(3)}{(3^x)^2} \].

Step 5: Simplify the Second Derivative

Combine terms for simplicity:\[ f''(x) = \frac{-\ln(3) + x \ln^2(3)}{3^x} \].

Step 6: Find the Third Derivative

Apply the Quotient Rule once more to \( f''(x) \). Set \( u = -\ln(3) + x \ln^2(3) \) and \( v = 3^x \). Then compute \( u' = \ln^2(3) \) and \( v' = 3^x \ln(3) \). Use the rule:\[ f'''(x) = \frac{\ln^2(3)3^x - (-\ln(3) + x \ln^2(3))3^x \ln(3)}{(3^x)^2} \].

Step 7: Simplify the Third Derivative

After combining like terms, the expression simplifies to:\[ f'''(x) = \frac{\ln^2(3) - \ln^3(3) - x \ln^3(3)}{3^x} \].

Key Concepts

Mastering the Quotient Rule

When calculating the derivative of a function that is a ratio of two functions, the Quotient Rule is your go-to technique. It's like the Product Rule's twin brother, specifically designed for fractions. If you have a function like \( \frac{u}{v} \), where \( u \) and \( v \) are functions of \( x \), then its derivative is given by:\[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \]This formula helps you differentiate when you have two functions being divided. Here are the key steps when applying this:- Differentiate the numerator \( u \) to get \( u' \).- Differentiate the denominator \( v \) to get \( v' \).- Plug these derivatives into the formula.In our function \( f(x) = \frac{x}{3^x} \), \( u = x \) and \( v = 3^x \) lead us to get \( f'(x) = \frac{1 - x \ln(3)}{3^x} \) after simplification. Practice using the Quotient Rule until it becomes second nature. Remember, it’s a powerful tool for functions involving division.

Exploring Higher-Order Derivatives

Higher-order derivatives involve taking the derivative of a derivative. So, the first derivative \( f'(x) \) tells us the rate at which \( f(x) \) changes. The second derivative, \( f''(x) \), tells us how the rate itself changes. Similarly, the third derivative gives insights into the change of the change of the change!In the exercise, we've used the Quotient Rule multiple times to find:

• First Derivative \( f'(x) \)
• Second Derivative \( f''(x) = \frac{-\ln(3) + x \ln^2(3)}{3^x} \)
• Third Derivative \( f'''(x) = \frac{\ln^2(3) - \ln^3(3) - x \ln^3(3)}{3^x} \)

By examining these higher-order derivatives, we gain deeper insight into the function's behavior over its domain. Each derivative level provides more detailed information, like acceleration compared to speed in Physics