Problem 29 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 29-32, 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 \( f'(3) = \frac{48}{5} \).

Step 1: Identify the Derivative Method

Since the function is a composition of functions, we should use the chain rule to differentiate it. Specifically, we'll use a combination of the chain rule and the quotient rule.

Step 2: Apply the Chain Rule

Let \( u(x) = \frac{x^2 + 1}{x + 2} \). The function can then be rewritten as \( f(x) = u(x)^3 \). According to the chain rule, \( f'(x) = 3u(x)^2 \, u'(x) \).

Step 3: Differentiate the Inner Function Using the Quotient Rule

To find \( u'(x) \), use the quotient rule: \[ u'(x) = \frac{(x+2)(2x) - (x^2+1)(1)}{(x+2)^2} = \frac{2x^2 + 4x - x^2 - 1}{(x+2)^2} = \frac{x^2 + 4x - 1}{(x+2)^2}. \]

Step 4: Substitute and Simplify

Now substitute \( u(x) \) and \( u'(x) \) back into the chain rule expression for \( f'(x) \):\[ f'(x) = 3 \left(\frac{x^2 + 1}{x + 2}\right)^2 \cdot \frac{x^2 + 4x - 1}{(x+2)^2}. \]Simplify further if necessary, but for evaluation, this expression is sufficient.

Step 5: Evaluate at \( x = 3 \)

Substitute \( x = 3 \) into the equation:First, find \( u(3) = \frac{3^2 + 1}{3 + 2} = \frac{10}{5} = 2 \).Next, \( u'(3) = \frac{3^2 + 4(3) - 1}{(3+2)^2} = \frac{9 + 12 - 1}{25} = \frac{20}{25} = \frac{4}{5} \).So, \( f'(3) = 3(2)^2 \cdot \frac{4}{5} = 3 \cdot 4 \cdot \frac{4}{5} = \frac{48}{5}.\)

Key Concepts

Understanding the Chain Rule

The chain rule is a fundamental principle in calculus for finding the derivative of a composition of functions. Imagine you have a function that is wrapped inside another function. To find the derivative of this composite function, the chain rule helps us by breaking it into manageable pieces.

Here's the chain rule formula: If you have a function composed as \( f(u(x)) \), where \( u \) is another function of \( x \), the derivative \( f'(x) \) is found by differentiating \( f \) with respect to \( u \) (denoted as \( f'(u) \)) and then multiplying by the derivative of \( u \) with respect to \( x \) (denoted as \( u'(x) \)).

This can be more simply written as:

• \( f'(x) = f'(u(x)) \cdot u'(x) \)

Using the chain rule allows us to action this process effectively step by step, especially useful in functions such as powers or exponentials involving more complex inner functions.

Applying the Quotient Rule

The quotient rule is another essential calculus tool for differentiating functions, but specifically those expressed as quotients. When a function is divided by another, the quotient rule offers a systematic approach to find derivatives.

The formula for the quotient rule is: if \( u(x) \) and \( v(x) \) are two functions, then the derivative of \( \frac{u}{v} \) is given by:

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

Using the quotient rule, we orderly differentiate the numerator \( u(x) \) and the denominator \( v(x) \), taking the products and then dividing over the square of the original denominator.

In practical scenarios, such as exercises where multiple rules combine, understanding both the numerator and denominator's derivatives, and applying the rule precisely facilitates reaching the correct derivative expression.

Evaluating Derivatives

After calculating the derivative formula using rules like the chain or quotient rules, the next step often involves evaluating it at a specific point. This final step, known as evaluating derivatives, allows us to find the actual slope or rate of change at a particular value of \( x \).

For instance, if you are tasked with finding \( f'(3) \), it means plugging the value \( x = 3 \) into your derivative expression. Here’s what to do:

• First, substitute \( x \) into any auxiliary functions like \( u(x) \).
• Next, calculate the derivative of these functions at this \( x \) value.
• Finally, substitute these values back into your derivative formula to compute the result.

This step is crucial in practical applications, providing exact numerical values for derivatives that convey how fast or slow a function is changing at the specified point.