Problem 11 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Rewrite each of these functions in a form that allows you to differentiate using the tools you have now, but with as little exertion as possible. Then differentiate. Work on strategy; none of these problems require the Quotient Rule. (a) \(f(x)=(x+1)(x-1) x\) (b) \(f(x)=\frac{x+\frac{1}{x}}{x}\)

Step-by-Step Solution

Answer

The derivatives of the given functions are: (a) \(f'(x) = 3x^2 - 1\), (b) \(f'(x) = -\frac{2}{x^3}\)

Step 1: Simplify Function (a)

Simplify the function \(f(x) = (x+1)(x-1)x\) by using the Distributive and the Power Rule. This gives \(f(x) = x^3 - x\)

Step 2: Differentiate Function (a)

Differentiate the simplified function \(f(x) = x^3 - x\) by using the Power Rule. The derivative, \(f'(x)\) would be \(f'(x) = 3x^2 - 1\)

Step 3: Simplify Function (b)

Simplify the function \(f(x) = \frac{x+\frac{1}{x}}{x}\) to \(f(x) = 1 + \frac{1}{x^2}\)

Step 4: Differentiate Function (b)

Differentiate the simplified function \(f(x) = 1 + \frac{1}{x^2}\) using the Power Rule. The derivative, \(f'(x)\), would be \(f'(x) = -\frac{2}{x^3}\)

Key Concepts

Power Rule

The power rule is a cornerstone of differentiation and makes life easier when dealing with polynomial functions. This rule states: if you have a function in the form of \(f(x) = x^n\), where \(n\) is any real number, the derivative is \(f'(x) = nx^{n-1}\). This means you simply multiply by the power and reduce the exponent by one.

For instance, in the function \(f(x) = x^3 - x\), the power rule allows you to quickly find its derivative: the derivative of \(x^3\) is \(3x^2\), and the derivative of \(x\) (which is \(x^1\)) is \(1x^{1-1} = 1\). Thus, the derivative is \(3x^2 - 1\).

Understanding and applying the power rule simplifies differentiation work and is especially useful when combined with other algebraic rules. It's a real time-saver and often the first tool you reach for when taking derivatives.

Function Simplification

Simplifying a function before differentiating it can save you from unnecessary complications. Function simplification involves rewriting a complex expression in a simpler form. This is often done by using algebraic manipulation techniques like expanding, factoring, distributing, or cancelling terms.

Consider the function from part (a) of the exercise: \( f(x) = (x+1)(x-1)x \). By using the distributive property, we first expand \((x+1)(x-1)\) to get \(x^2 - 1\), then multiply by \(x\) to find \(x^3 - x\).

• By doing all this, we make taking the derivative straightforward and quick.
• Function (b) \( f(x) = \frac{x+\frac{1}{x}}{x} \) can be simplified to \(1 + \frac{1}{x^2}\) by separating terms in the numerator and dividing each by \(x\).

Through simplification, we turn a potentially cumbersome problem into something much easier to handle mathematically.

Derivative Calculation

Derivative calculation is the process of finding the rate at which a function is changing at any point. Once you simplify a function, calculating its derivative becomes straightforward, especially when using rules like the power rule.

After simplifying \(f(x) = x^3 - x\) from our example, we use the power rule to calculate its derivative: \(f'(x) = 3x^2 - 1\). Each term is differentiated separately, and then they're combined to give the overall derivative.

Similarly, for the function \(f(x) = 1 + \frac{1}{x^2}\), we rewrite the fractional part as a negative exponent \(x^{-2}\), making it easier to differentiate. The derivative is \(f'(x) = 0 - 2x^{-3}\) which simplifies to \(-\frac{2}{x^3}\).

• Each derivative calculation follows the rules, ensuring you arrive at the correct rate of change.
• The simplification beforehand paves the way for a smoother differentiation process.

Mastering derivative calculations allows you to understand how functions behave and change across their domains.