Problem 29 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate the indicated derivative. $$ f^{\prime}(3) \text { if } f(x)=\left(\frac{x^{2}+1}{x+2}\right)^{3} $$

Step-by-Step Solution

Answer

The derivative at \( x = 3 \) is 9.6.

Step 1: Understand the Function

The function given is \( f(x) = \left(\frac{x^2 + 1}{x + 2}\right)^3 \). Our task is to find the derivative of this function and evaluate it at \( x = 3 \).

Step 2: Apply the Chain Rule

Use the chain rule to differentiate the function. Let \( u = \frac{x^2 + 1}{x + 2} \), then \( f(x) = u^3 \). The derivative of \( u^3 \) with respect to \( x \) is \( 3u^2 \cdot \frac{du}{dx} \).

Step 3: Differentiate \( u \) Using the Quotient Rule

The function \( u \) is \( \frac{x^2 + 1}{x + 2} \). Differentiate using the quotient rule: if \( u = \frac{v}{w} \), then \( \frac{du}{dx} = \frac{v' w - vw'}{w^2} \). Here, \( v = x^2+1 \), \( v' = 2x \), \( w = x+2 \), \( w' = 1 \).

Step 4: Perform the Quotient Rule Calculation

Substitute into the quotient rule: \( \frac{du}{dx} = \frac{(2x)(x+2) - (x^2+1)(1)}{(x+2)^2} \). Simplify this to get \( \frac{du}{dx} = \frac{2x^2 + 4x - x^2 - 1}{(x+2)^2} = \frac{x^2 + 4x - 1}{(x+2)^2} \).

Step 5: Substitute Back into the Chain Rule Derivative

Combining our results, \( f'(x) = 3u^2 \cdot \frac{du}{dx} = 3 \left(\frac{x^2+1}{x+2}\right)^2 \cdot \frac{x^2 + 4x - 1}{(x+2)^2} \).

Step 6: Evaluate \( f'(3) \)

Substitute \( x = 3 \) into the derivative: \( f'(x) = 3 \left(\frac{3^2+1}{3+2}\right)^2 \cdot \frac{3^2 + 4(3) - 1}{(3+2)^2} \). Simplify the expression to find the value of the derivative at \( x = 3 \).

Step 7: Simplification

Calculate the quantities: \( u = \frac{10}{5} = 2 \), then \( u^2 = 4 \). For \( \frac{du}{dx} \), calculate the numerator: \( 9 + 12 - 1 = 20 \), and denominator \( 25 \), so \( \frac{20}{25} = \frac{4}{5} \). Therefore, \( f'(3) = 3 \times 4 \times \frac{4}{5} = \frac{48}{5} = 9.6 \).

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus that enables you to compute the derivative of a composite function. This means it helps find how fast a function is changing, even when it has functions inside it or made up of other functions.
When you have a function like \( f(x) = (u)^n \) where \( u \) itself is another function (for example, \( u = \frac{x^2 + 1}{x + 2} \) in our case), the chain rule allows you to break it down into easier parts for differentiation.
The rule is defined as: if \( y = (u)^n \), then the derivative \( \frac{dy}{dx} = n(u)^{n-1} \cdot \frac{du}{dx} \).

• We first differentiate the outer function with respect to \( u \), treating \( u \) as a single variable. For example, \( 3u^2 \) when \( y = u^3 \).
• Then we multiply the result by the derivative of the inner function \( u \) with respect to \( x \), or \( \frac{du}{dx} \).

This cascading process makes it easier to handle even the most complex functions by breaking them down into manageable components.

Quotient Rule

The quotient rule is a technique for finding the derivative of a fraction of two functions. In simpler terms, it helps us differentiate expressions where one function divides another.
For a function \( u(x) = \frac{v(x)}{w(x)} \), the rule is mathematically expressed as: \[ \frac{du}{dx} = \frac{v'(x)w(x) - v(x)w'(x)}{[w(x)]^2} \]
Here's how it works practically:

• Identify the top and bottom parts of the fraction. Here, \( v = x^2 + 1 \) and \( w = x + 2 \).
• Calculate their derivatives, \( v'(x) = 2x \) and \( w'(x) = 1 \).
• Plug into the formula to find \( \frac{du}{dx} \), and simplify.

The quotient rule deftly handles the division of functions, providing an elegant solution even when both numerator and denominator are complex expressions.

Derivative Evaluation

Once we find the derivative of a function, the next step is often to evaluate it at a specific point. This involves plugging a number into the derivative to find the rate of change at that specific value.
In the given exercise, we determine \( f'(x) \) and then evaluate it at \( x = 3 \). Steps include:

• Substitute the number into each part of the derivative expression. For instance, where \( x = 3 \) in \( f'(x) = 3 \left(\frac{x^2+1}{x+2}\right)^2 \cdot \frac{x^2 + 4x - 1}{(x+2)^2} \).
• Simplify each component separately. Start with the inside functions and work outwards to streamline calculations.
• Multiply through to find the final value of \( f'(3) \).

Derivative evaluation is particularly useful in practical scenarios, such as calculating the instantaneous speed of a moving object at a particular time, ensuring understanding goes beyond just theoretical exercise.