Problem 15 - Thomas Calculus

Question

In Exercises \(13-16,\) 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

\(y' = 3x^2 + 10x + 2 - x^{-2}\)

Step 1: Part (a) - Product Rule

Let \(u = x^2+1\) and \(v = x+5+x^{-1}\).
\(u' = 2x\), \(v' = 1 - x^{-2}\)
\(y' = u'v + uv' = 2x(x+5+x^{-1}) + (x^2+1)(1-x^{-2})\)
\(= 2x^2 + 10x + 2 + x^2 - 1 + 1 - x^{-2}\)
\(= 3x^2 + 10x + 2 - x^{-2}\)

Step 2: Part (b) - Expand First

\(y = x^3 + 5x^2 + x + x + 5 + x^{-1} = x^3 + 5x^2 + 2x + 5 + x^{-1}\)
Wait, let me redo: \(y = (x^2+1)(x+5+1/x) = x^3 + 5x^2 + x + x + 5 + 1/x = x^3 + 5x^2 + 2x + 5 + x^{-1}\)
\(y' = 3x^2 + 10x + 2 - x^{-2}\)

Step 3: Verify

Both methods give the same result: \(y' = 3x^2 + 10x + 2 - \frac{1}{x^2}\).

Key Concepts

Differentiation

Differentiation is one of the fundamental techniques in calculus, used extensively in finding the rate of change of a function. When you differentiate a function, you're essentially measuring how the function's output value changes with a tiny shift in its input value. This process is essential for understanding various phenomena, such as how fast an object is moving or how a population grows over time. Differentiation leads to the derivative of a function, which can be thought of as the slope of the line tangent to the function's graph at a particular point. This slope indicates the function's "steepness" or rate of change at that point. To perform differentiation, certain rules and formulas are used, like the power rule, chain rule, and importantly for our current focus, the product rule.

Calculus

Calculus is a vast area of mathematics that focuses on change and motion—essentially the mathematics of change. It is divided into two main branches: differential calculus and integral calculus. While differential calculus deals with the rate at which quantities change, integral calculus focuses on accumulation of quantities and the areas under and between curves. Differential calculus, in particular, is used to find the derivative of functions, which helps in solving problems involving rates of change and finding slopes of curves.

• Imagine trying to predict the path of a falling object, or how quickly a car is accelerating. Calculus provides the tools to compute these quantities.
• Calculus is vital across fields such as physics, engineering, and even economics, assisting with problems involving change and motion.

Having a strong grasp of calculus is critical for higher-level mathematics and various real-world applications.

Derivatives

Derivatives are at the heart of differentiation and play a pivotal role in calculus. The derivative of a function gives the rate at which a function is changing at any point, helping to solve problems involving rates and finding instantaneous velocity or slopes. In our exercise, the product rule is used to find the derivative of the function, which is a technique used when dealing with the product of two differentiable functions. So, if you have two functions, say \(u(x)\) and \(v(x)\), their product \(y(x) = u(x) \cdot v(x)\) has a derivative given by the product rule: \[ y'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \]

• This rule tells us how changes in each function individually contribute to changes in their product.
• Finding derivatives is crucial in analyzing functions, optimizing systems, and predicting future behavior in dynamic systems.

Understanding derivatives allows you to analyze and predict how a system behaves under varying conditions—an invaluable skill in both academic and real-world applications.