Problem 18 - Brief Applied Calculus

Question

Find the second derivative of each function. $$ f(x)=\frac{x}{x-2} $$

Step-by-Step Solution

Answer

The second derivative is \( f''(x) = \frac{4}{(x-2)^3} \).

Step 1: Identify First Derivative

We need to find the first derivative of the function \( f(x) = \frac{x}{x-2} \). This is a quotient of two functions, so we will use the quotient rule, which states \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]. Let \( u = x \) and \( v = x - 2 \). Then \( u' = 1 \) and \( v' = 1 \).

Step 2: Apply Quotient Rule for First Derivative

Substituting the values into the quotient rule formula, the first derivative \( f'(x) \) is: \[ f'(x) = \frac{(1)(x-2) - (x)(1)}{(x-2)^2} = \frac{x-2-x}{(x-2)^2} = \frac{-2}{(x-2)^2}. \]

Step 3: Differentiate Again to Find Second Derivative

Now, we differentiate \( f'(x) = \frac{-2}{(x-2)^2} \) to find \( f''(x) \). This can be done using the chain rule. Let \( g(x) = -2 \) and \( h(x) = (x-2)^{-2} \). The derivative of \( g(x)h(x) \) is \( g'(x)h(x) + g(x)h'(x) \).

Step 4: Apply Chain Rule for Second Derivative

The function \( h(x) = (x-2)^{-2} \) has derivative \( h'(x) = -2(x-2)^{-3}(1) = -2(x-2)^{-3} \). This means \( f''(x) = 0 \cdot h(x) + (-2) \cdot h'(x) = -2(-2(x-2)^{-3}) = 4(x-2)^{-3} \).

Step 5: Simplify the Second Derivative

Simplify \( 4(x-2)^{-3} \), which is \( \frac{4}{(x-2)^3} \). This is the second derivative of \( f(x) \).

Key Concepts

Quotient Rule

The quotient rule is an essential concept for differentiation, particularly when dealing with fractions of two functions. When you have a function that is the quotient of two other functions, like \( \frac{u}{v} \), you use the quotient rule to find its derivative. The rule states that the derivative is \( \frac{u'v - uv'}{v^2} \).
The steps are to first identify the functions \( u \) and \( v \) in your quotient:

• Let \( u = x \) and \( v = x - 2 \) for the original function \( f(x) = \frac{x}{x-2} \).
• Differentiate both \( u \) and \( v \): \( u' = 1 \) and \( v' = 1 \).

Next, plug these derivatives back into the quotient rule formula to find the derivative of the entire function. This will give you \( f'(x) = \frac{(1)(x-2) - (x)(1)}{(x-2)^2} = \frac{-2}{(x-2)^2} \). The quotient rule is instrumental because it systematically addresses the complexity that arises when dividing two functions.

Chain Rule

The chain rule is crucial for differentiating compositions of functions. It helps when a function involves another function, like \( h(x) = (x-2)^{-2} \) seen in steps to find the second derivative.
The chain rule formula is \( (f(g(x)))' = f'(g(x))g'(x) \). To differentiate our function after applying the quotient rule, we treat \( (x-2)^{-2} \) as a composite function where:

• Outer function: \( y = x^{-2} \)
• Inner function: \( u = x-2 \)

Derive the inner function: \( du/dx = 1 \). The derivative of the outer function is calculated by the power rule: \( y' = -2x^{-3} \). Combine them to get the derivative for \( (x-2)^{-2} \), resulting in \( -2(x-2)^{-3} \). This step ensures that you serve the complexity of function composition in derivatives.

Differentiation Steps

Differentiation involves finding how a function changes at any point, which involves several precise steps. After computing the first derivative using the quotient rule, the second derivative requires further differentiation.
The specific steps are:

• Identify the type of function you're differentiating: quotient, product, chain, etc.
• For \( f(x) = \frac{x}{x-2} \), compute the first derivative using the quotient rule: \( f'(x) = \frac{-2}{(x-2)^2} \).
• Apply the chain rule to differentiate \( f'(x) \) by identifying inner/outer components, resulting in \( f''(x) = \frac{4}{(x-2)^3} \).

Each step in differentiation is carefully executed to ensure accurate results. This approach reveals how derivatives help analyze the rate and manner in which functions change, foundational for calculus and essential for complex problem-solving.