Problem 42 - Brief Applied Calculus

Question

Use the Generalized Power Rule to find the derivative of each function. $$ f(x)=\left(\frac{x-1}{x+1}\right)^{5} $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = \frac{10(x-1)^4}{(x+1)^6} \).

Step 1: Identify the Functions

To apply the Generalized Power Rule, we need to identify the outer and inner functions in the expression.The given function is \( f(x) = \left(\frac{x-1}{x+1}\right)^{5} \), where the outer function \( g(u) = u^{5} \) and the inner function \( u = \frac{x-1}{x+1} \).

Step 2: Derivative of the Outer Function Using the Power Rule

Find the derivative of the outer function, which is \( g(u) = u^{5} \). According to the power rule, the derivative is:\[ g'(u) = 5u^{4} \].

Step 3: Derivative of the Inner Function Using the Quotient Rule

Find the derivative of the inner function \( u = \frac{x-1}{x+1} \) using the quotient rule, which states that if \( u = \frac{v}{w} \), then:\[ u' = \frac{v'w - vw'}{w^2} \]Here, \( v = x-1 \) and \( w = x+1 \), so:- \( v' = 1 \)- \( w' = 1 \)Thus, the derivative is:\[ u' = \frac{1(x+1) - (x-1)(1)}{(x+1)^2} = \frac{x+1 - x + 1}{(x+1)^2} = \frac{2}{(x+1)^2} \].

Step 4: Apply the Chain Rule

Using the chain rule, multiply the derivative of the outer function by the derivative of the inner function:\[ f'(x) = g'(u) \cdot u' = 5\left(\frac{x-1}{x+1}\right)^4 \cdot \frac{2}{(x+1)^2} \].

Step 5: Simplify the Derivative Expression

Combine and simplify the expression for the derivative:\[ f'(x) = \frac{10(x-1)^4}{(x+1)^{4}} \cdot \frac{1}{(x+1)^2} = \frac{10(x-1)^4}{(x+1)^6} \].Thus, the derivative of \( f(x) \) simplifies to:\[ f'(x) = \frac{10(x-1)^4}{(x+1)^6} \].

Key Concepts

Generalized Power Rule

The Generalized Power Rule is an advanced form of the traditional power rule used in calculus. It enables us to find derivatives of functions raised to a power, even when the base of the exponent is not a simple variable. Instead, the base can be a more complex function of a variable. In these scenarios, the power rule alone is not sufficient.

The rule can be expressed simply as: If you have a function defined as \[ f(x) = (u(x))^n \] where \( u(x) \) is any differentiable function and \( n \) is a constant, the derivative is: \[ f'(x) = n(u(x))^{n-1}u'(x) \].
Here's how this applies in practice:

• First, identify the outer function, which is the exponent operation, and the inner function, which is the base function.
• Take the derivative of the outer function according to the power, then multiply it by the derivative of the inner function.

For example, in \( f(x) = \left( \frac{x-1}{x+1} \right)^{5} \), we apply the power rule to \( u^{5} \) with \( u = \frac{x-1}{x+1} \) and proceed with further steps involving the chain rule.

Quotient Rule

The Quotient Rule is essential for differentiating functions that are expressed as a division of two simpler functions. When you have a function \( u(x) = \frac{v(x)}{w(x)} \), its derivative is found using:\[ u'(x) = \frac{v'(x)w(x) - v(x)w'(x)}{(w(x))^2} \].

This formula is pivotal when differentiating rational functions to keep the process systematic.

Steps to use the quotient rule:

• Identify the 'top' function, \( v(x) \), and the 'bottom' function, \( w(x) \).
• Compute the derivatives of both these functions.
• Plug these into the quotient rule formula to find \( u'(x) \).

For example, in the expression \( \frac{x-1}{x+1} \), applying the quotient rule involves:

• Taking \( v(x) = x-1 \), thus \( v'(x) = 1. \)
• Taking \( w(x) = x+1 \), thus \( w'(x) = 1. \)
• Using these, the derivative \( u'(x) \) becomes \( \frac{2}{(x+1)^2} \), forming a crucial part in finding the derivative of our original function when combined with the chain rule.

Chain Rule

The Chain Rule is a core technique in calculus, allowing us to differentiate composite functions. It is especially helpful when dealing with nested functions, where one function is inside another. The rule states:\[ \text{if } f(x) = g(h(x)), \text{ then } f'(x) = g'(h(x)) \, \times \, h'(x) \].

The rule can be broken down into simple steps:

• Identify the 'outer' function \( g \) and the 'inner' function \( h \).
• Differentiate both functions separately.
• Multiply these derivatives to get the derivative of the composite function.

In utilizing the chain rule for \( f(x) = \left( \frac{x-1}{x+1} \right)^{5} \), the outer function \( g(u) = u^{5} \) was differentiated to \( 5u^4 \), while the inner function \( u = \frac{x-1}{x+1} \) was dealt with using the quotient rule.

Combining these insights, you achieve the derivative: \( f'(x) = 5 \left( \frac{x-1}{x+1} \right)^4 \cdot \frac{2}{(x+1)^2} \), which simplifies further. This showcases the power and necessity of the chain rule in calculus.