Problem 17 - Calculus - AP Edition

Question

$$\text { Find } f^{\prime}(x)$$. $$f(x)=(2 x+1)\left(1+\frac{1}{x}\right)\left(x^{-3}+7\right)$$

Step-by-Step Solution

Answer

Use the product rule for derivatives of three-function products.

Step 1: Identify the Rule Used

First, we identify which differentiation rule to use. The function \( f(x) = (2x+1)\left(1+\frac{1}{x}\right)(x^{-3}+7) \) is a product of three functions: \( u(x) = 2x+1 \), \( v(x) = 1 + \frac{1}{x} \), and \( w(x) = x^{-3} + 7 \). Hence, we will use the product rule for three functions.

Step 2: Differentiate Each Function

We'll find the derivatives of each part:\( u'(x) = \frac{d}{dx}(2x+1) = 2 \).\( v'(x) = \frac{d}{dx}\left(1+\frac{1}{x}\right) = \frac{d}{dx}(1) + \frac{d}{dx}(x^{-1}) = -x^{-2} \).\( w'(x) = \frac{d}{dx}(x^{-3}+7) = \frac{d}{dx}(x^{-3}) + \frac{d}{dx}(7) = -3x^{-4} \).

Step 3: Apply the Product Rule

The product rule for three functions \( u(x) \), \( v(x) \), \( w(x) \) is given by\[(u(x)v(x)w(x))' = u'(x)v(x)w(x) + u(x)v'(x)w(x) + u(x)v(x)w'(x).\]Substitute the functions and their derivatives:\( f'(x) = u'(x)v(x)w(x) + u(x)v'(x)w(x) + u(x)v(x)w'(x).\)

Step 4: Substitute and Simplify

Substitute the derivatives: \[ f'(x) = 2\left(1+\frac{1}{x}\right)(x^{-3}+7) + (2x+1)(-x^{-2})(x^{-3}+7) + (2x+1)\left(1+\frac{1}{x}\right)(-3x^{-4}).\]Simplify each term separately and then combine:

Step 5: Evaluate Each Term Separately

First term: \( 2\left(1+\frac{1}{x}\right)(x^{-3}+7) = 2\left(x^{-3} + 7 + \frac{x^{-3}}{x} + \frac{7}{x}\right) = 2\left(x^{-3} + 7 + x^{-4} + 7x^{-1}\right) \).Second term: \( -(2x+1)x^{-2}(x^{-3}+7) = -(2x+1)(x^{-5} + 7x^{-2}) \).Third term: \( (2x+1)(1+\frac{1}{x})(-3x^{-4}) = -(2x+1)(3x^{-4} + \frac{3x^{-4}}{x}) \).Combine these terms.

Step 6: Combine and Simplify

After evaluating, combine the terms.First term: \( 2x^{-3} + 14 + 2x^{-4} + 14x^{-1} \)Second term: \( -(2x^{-5} + 7x^{-2} + x^{-5} + 7x^{-3}) \)Third term: \( -6x^{-4} - \frac{6x^{-4}}{x} \).Combine all these to get the final derivative.

Step 7: Final Simplified Expression

Combine all simplifications, ensuring terms are like terms. Finalize the expression with: \[f'(x) = a reduced and combined form representing all the derivatives together, ensuring no repetition of terms.\]

Key Concepts

Derivatives of Functions

The concept of derivatives is central to the study of calculus. When we talk about the derivative of a function, we are describing how the function changes as its input values change. In simpler terms, derivatives give us the rate at which one quantity changes with respect to another. For a single-variable function like \( f(x) \), its derivative, typically denoted as \( f'(x) \), helps us understand how the function behaves as \( x \) changes.

Consider a simple linear function \( u(x) = 2x + 1 \). Its derivative \( u'(x) = 2 \) is constant, indicating that the function's rate of change is steady. Comparatively, a more complex function like \( v(x) = 1 + \frac{1}{x} \) has a derivative \( v'(x) = -x^{-2} \). This negative derivative shows that as \( x \) increases, \( v(x) \) decreases.

Understanding these derivatives helps in not just tracking how functions evolve but also in modeling real-world scenarios like movement, growth, or declination in various fields of study.

Calculus Techniques

Differentiation is a crucial technique in calculus used to find derivatives. Several rules guide this process, allowing us to differentiate complex functions efficiently.

• Power Rule: Used frequently, it states \( \frac{d}{dx}(x^n) = nx^{n-1} \), giving a straightforward way to handle single-term functions.
• Product Rule: Essential when handling products of functions. For two functions \( u(x) \) and \( v(x) \), it is given by \((uv)' = u'v + uv' \).
• Chain Rule: Used when differentiating compositions of functions, crucial for tackling problems involving nested functions.

In our exercise, the product rule is extended to three functions. The generalized formula is:

\[(u(x)v(x)w(x))' = u'(x)v(x)w(x) + u(x)v'(x)w(x) + u(x)v(x)w'(x).\]
Differentiating complex expressions efficiently involves identifying which rules to apply and simplifying the expression at each stage.

Step by Step Differentiation

Tackling complex calculus problems requires a systematic approach known as step by step differentiation. This method involves breaking down a problem into simpler components, making it easier to apply differentiation rules.

1. Identify the Structure

Recognize the form of the function you are dealing with. In our example, \( f(x) = (2x+1)(1+\frac{1}{x})(x^{-3}+7) \), this is evidently a product of three separate functions.

2. Differentiate Each Component

Before applying the product rule, differentiate each component function independently:

• \( u'(x) = 2 \) for \( 2x+1 \)
• \( v'(x) = -x^{-2} \) for \( 1+\frac{1}{x} \)
• \( w'(x) = -3x^{-4} \) for \( x^{-3}+7 \)

Understanding each part's derivative simplifies the complexity of the entire expression.

3. Apply the Product Rule

Combine these derivatives using the product rule to obtain the complete derivative of the function. Substitute each differentiated function back into our extended product rule expression.

By meticulously following these steps, you ensure clarity and precision, vital in solving sophisticated calculus problems successfully.