Problem 15 - University Calculus: Early Transcendentals

Question

Find \(y^{\prime}\) (a) by applying the Product Rule and (b) by multiplying the factors to produce a sum of simpler terms to differentiate. $$y=\left(x^{2}+1\right)\left(x+5+\frac{1}{x}\right)$$

Step-by-Step Solution

Answer

The derivative is \( y' = 3x^2 + 10x + 2 - \frac{1}{x^2} \).

Step 1: Identify the Two Factors of the Product

The expression given is \( y = (x^2 + 1)(x + 5 + \frac{1}{x}) \). Here, the two factors are \( u = x^2 + 1 \) and \( v = x + 5 + \frac{1}{x} \).

Step 2: Differentiate Each Factor

First, find the derivative of \( u \):\[ u = x^2 + 1 \ u' = \frac{d}{dx}(x^2 + 1) = 2x\] Next, find the derivative of \( v \):\[ v = x + 5 + \frac{1}{x} \ v' = \frac{d}{dx} \left( x + 5 + \frac{1}{x} \right) = 1 - \frac{1}{x^2}\]

Step 3: Apply the Product Rule

The Product Rule states: \((uv)' = u'v + uv'\). Apply this to find the derivative of \( y \):\[ y' = (u'v + uv') = (2x)(x + 5 + \frac{1}{x}) + (x^2 + 1)(1 - \frac{1}{x^2})\]

Step 4: Expand and Simplify the Expression

Expand and simplify:The first part:\[2x(x + 5 + \frac{1}{x}) = 2x^2 + 10x + 2\]The second part:\[(x^2+1)(1 - \frac{1}{x^2}) = x^2 + 1 - 1 - \frac{1}{x^2} \ = x^2 - \frac{1}{x^2}\]Combine both results:\[y' = 2x^2 + 10x + 2 + x^2 - \frac{1}{x^2} = 3x^2 + 10x + 2 - \frac{1}{x^2}\]

Step 5: Multiply the Factors Before Differentiating

Expand the original expression first: \((x^2 + 1)(x + 5 + \frac{1}{x})\). Multiply each term:- \(x^2(x) = x^3\)- \(x^2(5) = 5x^2\)- \(x^2(\frac{1}{x}) = x\)- \(1(x) = x\)- \(1(5) = 5\)- \(1(\frac{1}{x}) = \frac{1}{x}\)Combine these:\[y = x^3 + 5x^2 + x + x + 5 + \frac{1}{x} = x^3 + 5x^2 + 2x + 5 + \frac{1}{x}\]

Step 6: Differentiate Each Term After Expansion

Differentiate each term in the expanded form:\[y' = \frac{d}{dx}(x^3) + \frac{d}{dx}(5x^2) + \frac{d}{dx}(2x) + \frac{d}{dx}(5) + \frac{d}{dx}(\frac{1}{x})\]The derivatives:- \(\frac{d}{dx}(x^3) = 3x^2\)- \(\frac{d}{dx}(5x^2) = 10x\)- \(\frac{d}{dx}(2x) = 2\)- \(\frac{d}{dx}(5) = 0\)- \(\frac{d}{dx}(\frac{1}{x}) = -\frac{1}{x^2}\)Combine these:\[y' = 3x^2 + 10x + 2 - \frac{1}{x^2}\]

Step 7: Compare Results from Both Methods

In both methods—applying the Product Rule and expanding before differentiating—the derivative obtained is the same: \[ y' = 3x^2 + 10x + 2 - \frac{1}{x^2} \]

Key Concepts

Differentiation

Differentiation is the process by which we calculate the derivative of a function. This is a fundamental operation in calculus. The basic idea is to find the rate at which a function changes at any given point. Think of it as finding out how steep a hill is at a particular spot.
To perform differentiation:

• We use the rules of differentiation which are straightforward once you get familiar with them.
• One of the key rules is the Product Rule used when differentiating a product of two functions.

In this exercise, we differentiate a product of functions using the Product Rule, which requires differentiating each factor separately before combining them using the rule. That’s what makes the process structured and manageable. Understanding differentiation is key to mastering problems in calculus.

Derivatives

The concept of a derivative is foundational in calculus. It represents the rate of change of one variable with respect to another. In simpler terms, it tells us how a function behaves as its inputs change. Derivatives can also reflect the slope of a function at a specific point.
For example:

• The derivative of a constant is zero, because a constant function does not change.
• If we have a function like \(y = x^2\), its derivative \(y' = 2x\) shows us how the function value will alter with small changes in \(x\).

In applied calculus problems, derivatives can represent things like speed or acceleration. That is why they are useful beyond theoretical exercises. They have real-world applications, helping to solve many practical problems.

Calculus problems

Solving calculus problems typically involves several steps. You need a strategy to work through these problems effectively. Understanding the principles and rules is crucial.
Key strategies include:

• Yeah always follow a systematic approach. Breaking down a problem into manageable parts simplifies the process.
• Applying rules like the Product Rule, Quotient Rule, or Chain Rule can help tackle more complex expressions.

In this exercise, we applied two different methods to solve the same problem. First, using the Product Rule directly and then by altering the form of the equation before differentiation. Seeing the process from both perspectives deepens understanding. This is especially valuable since in calculus there is often more than one way to find a solution. Practice and problem-solving confidence are vital for mastering calculus.