Problem 17 - Brief Applied Calculus

Question

Find the second derivative of each function. $$ f(x)=\frac{x}{x-1} $$

Step-by-Step Solution

Answer

The second derivative is \( f''(x) = \frac{2}{(x-1)^3} \).

Step 1: Find the First Derivative

To find the first derivative of \( f(x) = \frac{x}{x-1} \), we use the quotient rule. The quotient rule states that if we have a function \( g(x) = \frac{u(x)}{v(x)} \), then \( g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \). Here, \( u(x) = x \) and \( v(x) = x - 1 \). Thus, \( u'(x) = 1 \) and \( v'(x) = 1 \). Applying the quotient rule, we have:\[f'(x) = \frac{1(x-1) - x(1)}{(x-1)^2} = \frac{x-1-x}{(x-1)^2} = \frac{-1}{(x-1)^2} \]

Step 2: Differentiate Again for the Second Derivative

Next, we find the second derivative \( f''(x) \) by differentiating \( f'(x) = \frac{-1}{(x-1)^2} \). We can rewrite \( f'(x) = -(x-1)^{-2} \) to apply the power rule. The power rule says if \( y = x^n \), then \( y' = nx^{n-1} \). So, differentiating \( -(x-1)^{-2} \) gives:\[\f''(x) = -(-2)(x-1)^{-3} \cdot (1) = \frac{2}{(x-1)^3} \]

Step 3: Simplify the Expression

The second derivative is \( f''(x) = \frac{2}{(x-1)^3} \). This is already in its simplest form. Thus, we conclude our process.

Key Concepts

Quotient Rule

In calculus, the quotient rule is an essential tool for differentiating functions that are written as fractions. When you have a function, say \( g(x) = \frac{u(x)}{v(x)} \), the quotient rule helps find its derivative. The formula for the derivative using the quotient rule is:

• \( g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \)

In this exercise, we worked with a function \( f(x) = \frac{x}{x-1} \), where the numerator \( u(x) = x \) and the denominator \( v(x) = x - 1 \). Let’s break this down:

• The derivative of the numerator \( u'(x) = 1 \)
• The derivative of the denominator \( v'(x) = 1 \)

Applying these into the quotient rule formula, you perform the calculation by subtracting the product of the numerator's derivative and the denominator from the product of the numerator and the derivative of the denominator. Don’t forget to square the denominator in the final result. This leads to the first derivative \( f'(x) = \frac{-1}{(x-1)^2} \). This approach employs the quotient rule, ensuring each term is properly differentiated according to function position.

Power Rule

The power rule is a fundamental technique in differentiation used when a function is in the form of a power of \( x \). When you have a function like \( y = x^n \), the power rule tells you that its derivative is \( y' = nx^{n-1} \). It simplifies the process of taking derivatives by allowing you to handle exponents directly.In our exercise, after finding the first derivative using the quotient rule, we had \( f'(x) = \frac{-1}{(x-1)^2} \), which we rewrote as \( f'(x) = -(x-1)^{-2} \). To find the second derivative, we applied the power rule. Here’s how it works:

• We took the expression \( -(x-1)^{-2} \), a simple power rule application on \( (x-1)^{-2} \).
• By applying the power rule, the exponent \( -2 \) gets multiplied in front, so it becomes \( -(-2) \).
• The power of the term is reduced by one, giving \( (x-1)^{-3} \).

This results in the expression \( f''(x) = \frac{2}{(x-1)^3} \). The power rule helps us handle negative exponents and apply differentiation efficiently.

Differentiation

Differentiation is a core concept in calculus, involving the process of finding a derivative. This derivative represents the rate of change or the slope of a function at a given point. In our exercise, we first applied the quotient rule to derive the first derivative, \( f'(x) \). Differentiation can involve various rules and techniques depending on the form of the function being differentiated. Here's why it’s important:

• It helps in determining the rate at which a function is changing at any given point, essential in physics and engineering for calculating speed, acceleration, etc.
• Different rules are applied based on function types, such as product rule, chain rule, power rule, and quotient rule, providing flexibility in solving complex problems.

Once we had the first derivative, we differentiated once more to find the second derivative \( f''(x) \), which represents the rate at which the rate of change is changing. This is crucial in understanding the concavity and points of inflection of the original function. Each step in differentiation unpacks more information about the behavior of the function.