Problem 32 - Single Variable Calculus

Question

Find the derivative of the function and \(\underline{\text { eval }}\) uate \(f^{\prime}(x)\) at the given value of \(x\). \(f(x)=\frac{2 x+1}{2 x-1} ; \quad x=2\)

Step-by-Step Solution

Answer

The derivative of the function \(f(x)=\frac{2x+1}{2x-1}\) is \(f'(x)=\frac{-4}{(2x-1)^2}\). Evaluating the derivative at \(x=2\), we get \(f'(2)=\frac{-4}{9}\).

Step 1: Apply the quotient rule

The function we are working with is \(f(x)=\frac{2 x+1}{2 x-1}\). To find its derivative, we can apply the quotient rule \[\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2},\] where \(u = 2x+1\) and \(v = 2x-1\).

Step 2: Find the derivatives of u and v

We differentiate u and v with respect to x: \[u'(x) = \frac{d(2x + 1)}{dx} = 2\] and \[v'(x) = \frac{d(2x - 1)}{dx} = 2.\]

Step 3: Use the derivatives in the quotient rule

Put the derivatives of u and v into the quotient rule formula: \[f'(x)=\frac{u'v - uv'}{v^{2}}=\frac{(2)(2x-1)-(2x+1)(2)}{(2x-1)^{2}}.\]

Step 4: Simplify the derivative

Simplify the numerator and denominator in \(f'(x)\): \[f'(x)=\frac{4x-2-4x-2}{(2x-1)^{2}}.\] Thus, the derivative of the function is \[f'(x)=\frac{-4}{(2x-1)^{2}}.\]

Step 5: Evaluate f'(x) at x = 2

Now, we will substitute \(x = 2\) into the derivative function to find the value of \(f'(x)\) at \(x = 2\): \[f'(2) = \frac{-4}{(2(2)-1)^{2}} = \frac{-4}{(4-1)^{2}} = \frac{-4}{(3)^{2}} = \frac{-4}{9}.\] So, the value of \(f^{\prime}(2)\) is \(\frac{-4}{9}\).

Key Concepts

Quotient Rule

In calculus, the quotient rule is essential for differentiating functions that are written as the quotient of two differentiable functions. When you have a function that looks like \( \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), the quotient rule helps you find its derivative.
The rule is given by the formula:

• \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]

This formula tells us to:

• Differentiate \( u \) (the numerator function) to get \( u' \).
• Differentiate \( v \) (the denominator function) to get \( v' \).
• Plug these derivatives into the formula to find \( \frac{u}{v} \)'s derivative.

Understanding the quotient rule is crucial when handling complex fractions, as it shows the change in one function as it divides another function.

Differentiation

Differentiation is a core concept in calculus, utilized to find the derivative of a function. A derivative represents the rate at which a function is changing at any given point. This idea is akin to finding the tangent to a curve at a point, thereby understanding the slope of the function at that point.
The process involves applying specific rules:

• The power rule.
• The product rule.
• The quotient rule (discussed above).
• The chain rule.

By learning these rules, you can effectively find derivatives of virtually any function, revealing essential characteristics like rates of change and critical points, which are indispensable in solving real-world problems.

Calculus

Calculus is the branch of mathematics that studies how things change. It provides tools to deal with change in functions through two main operations: differentiation and integration. While differentiation helps find the slope of a function and understand its rate of change, integration allows us to find the area under curves or accumulate quantities in a given interval.
Calculus is divided into:

• Differential Calculus: Focuses on derivatives and rates of change.
• Integral Calculus: Deals with integrals and accumulation of quantities.

Mastering calculus provides a powerful framework for modeling situations in physics, engineering, economics, statistics, and beyond, turning abstract mathematical notions into practical solutions.

Simplifying Fractions

Simplifying fractions is an essential mathematical skill, especially in calculus when working with derivatives like those derived using the quotient rule. After applying the quotient rule, the resulting derivative expression often needs simplification to become more digestible.
Here are some steps to simplify fractions:

• Factorize the numerator and the denominator, if possible.
• Cancel out any common factors in the numerator and the denominator.
• Combine like terms, if applicable, to simplify the expression further.

In the given solution, simplifying \( \frac{4x-2-4x-2}{(2x-1)^2} \) was crucial in obtaining a clean and understandable derivative expression. It's important to be patient and methodical during simplification to avoid computational errors and ensure you work with the simplest form of a mathematical expression.