Problem 18 - Calculus Early Transcendentals

Question

Find \(f^{\prime}(x)\). \(f(x)=x^{-5}\left(x^{2}+2 x\right)(4-3 x)\left(2 x^{9}+1\right)\)

Step-by-Step Solution

Answer

Use the product rule and simplify after finding derivatives of each part.

Step 1: Understand the Problem

We need to find the derivative of the function \(f(x)\) which is a product of four functions: \(x^{-5}, (x^{2}+2x), (4-3x),\) and \((2x^{9}+1)\).

Step 2: Apply the Product Rule

The product rule states that if you have multiple functions multiplied, \((u \, v \, w \, t)\), their derivative is \(u'vwt + uv'wt + uvw't + uvwt'\). Thus, we'll differentiate each part individually and apply the product rule.

Step 3: Differentiate Each Function

Calculate the derivatives of each component: - For \(x^{-5}\), \(\frac{d}{dx}(x^{-5}) = -5x^{-6}\). - For \(x^{2} + 2x\), \(\frac{d}{dx}(x^{2} + 2x) = 2x + 2\). - For \(4 - 3x\), \(\frac{d}{dx}(4 - 3x) = -3\). - For \(2x^{9} + 1\), \(\frac{d}{dx}(2x^{9} + 1) = 18x^{8}\).

Step 4: Apply the Derivatives to the Product Rule

Apply the derivatives within the product rule:\[-5x^{-6}(x^{2}+2x)(4-3x)(2x^{9}+1) \ {}+ x^{-5}(2x+2)(4-3x)(2x^{9}+1) \ {}+ x^{-5}(x^{2}+2x)(-3)(2x^{9}+1) \ {}+ x^{-5}(x^{2}+2x)(4-3x)(18x^{8}) \]

Step 5: Simplify the Expression

Next, simplify the expression. Each term is expanded individually, and then terms are combined where possible. Factor and simplify where practical for the final structure of the derivative.

Key Concepts

Product Rule

The product rule is a fundamental tool in calculus, especially when dealing with the differentiation of functions that are products of two or more other functions. It provides a method for finding the derivative of such a product efficiently. If you have two functions, say \( u(x) \) and \( v(x) \), the product rule states that their derivative, \( (uv)' \), is given by:

• \( u'v + uv' \)

Now, when dealing with more than two functions, for example \( u, v, w, \text{ and } t \), the rule extends to:

• \( u'vwt + uv'wt + uvw't + uvwt' \)

Each function is differentiated one at a time while keeping the others intact. This allows us to effectively break down complex expressions into manageable pieces.

Differentiation

Differentiation is the process used in calculus to find the rate at which a function is changing at any given point. This is accomplished by finding the derivative of the function. A derivative provides a linear approximation of a function at a certain point, which is essential for understanding the function's behavior.

• For a simple power function like \( x^n \), the derivative is \( nx^{n-1} \).
• For multiple standards functions like \( x^{-5} \) or \( 2x^9 + 1 \), the power rule helps us find \( -5x^{-6} \) and \( 18x^8 \) respectively.

Differentiation lays the groundwork for optimization problems and analyses.
Understanding differentiation rules helps break down complex functions, making them easier to analyze and solve.

Calculus Problem Solving

Solving calculus problems often requires combining rules and techniques to evaluate functions effectively. It's not just about knowing the rules but understanding when and how to apply them. As seen in the problem above:

• Identify each function component within the composite function; for instance, \( x^{-5}, \) contribute distinct behaviors.
• Use known differentiation rules—like the product rule and chain rule—to tackle these parts individually.
• Apply the derivatives in a structured way to understand how the function changes as a whole.

In calculus, breaking down the problem into smaller steps often makes it easier to manage. This approach transforms seemingly complex tasks into a series of simpler, more straightforward operations.
Consistent practice and familiarization with different calculus rules equip you with the skills needed for effective problem-solving.

Chain Rule

The chain rule is a technique in calculus used for differentiating composite functions. When a function is composed of two or more functions, the chain rule helps find its derivative. If you have a composite function \( y = f(g(x)) \), the derivative is:

• \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \)

This means you differentiate the outer function first, leave the inner function untouched initially, and then multiply with the derivative of the inner function.
In our exercise, although primarily solved using the product rule, the chain rule is indispensable when components within composite expressions need differentiation layer by layer. By understanding the application of the chain rule, especially in nested functions, you advance your calculus skills greatly.
The chain rule, along with the product rule, forms a backbone of powerful differentiation techniques necessary for solving advanced calculus problems.