Problem 17 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the derivative of each function. \(f(x)=\frac{x-1}{2 x+1}\)

Step-by-Step Solution

Answer

The derivative of the function \(f(x)=\frac{x-1}{2 x+1}\) is \(f'(x) = \frac{3}{(2x+1)^2}\).

Step 1: Identify the form of the function

The given function \(f(x)=\frac{x-1}{2 x+1}\) is a simple division of two smaller functions \(u(x)\) and \(v(x)\). Let's choose the numerator \(x-1\) as function \(u(x)\) and the denominator \(2x+1\) as function \(v(x)\).

Step 2: Apply the quotient rule formula

The quotient rule states that the derivative of the division of two differentiable functions is: \[ \frac{d}{dx}(u/v) = (v\cdot u' - u\cdot v') / v^2 \] where \(u'\) and \(v'\) are the derivatives of \(u\) and \(v\), respectively.

Step 3: Find derivatives of \(u(x)\) and \(v(x)\) ( \(u'\) and \(v'\) )

First, find the derivative of \(u(x) = x-1\). \(u'(x) = 1\) as the derivative of \(x\) is \(1\) and the derivative of a constant is \(0\). Then find the derivative of \(v(x) = 2x+1\). \(v'(x) = 2\) as the derivative of \(2x\) is \(2\) and, again, the derivative of a constant is \(0\).

Step 4: Substitute \(u\), \(v\), \(u'\), and \(v'\) into quotient rule formula

Substitute \(u\), \(v\), \(u'\), and \(v'\) into the quotient rule formula: \[ f'(x) = \frac{v\cdot u' - u\cdot v'}{v^2} \] yielding: \[ f'(x) = \frac{(2x+1)\cdot1 - (x-1)\cdot2}{(2x+1)^2} \]

Step 5: Simplify the result to achieve the final derivative

Simplify the above equation to find the final derivative: \[ f'(x) = \frac{2x+1 - 2x+2}{(2x+1)^2} = \frac{3}{(2x+1)^2} \] In conclusion, the derivative of the function \(f(x)=\frac{x-1}{2 x+1}\) is \(f'(x) = \frac{3}{(2x+1)^2}\).

Key Concepts

Quotient Rule

Understanding how to find the derivative of a quotient of two functions is essential in calculus, and this is where the quotient rule comes into play. The quotient rule is a straightforward method for differentiating problems where one function is divided by another.

Consider a function \(f(x) = \frac{u(x)}{v(x)}\), where both \(u(x)\) and \(v(x)\) are themselves differentiable functions. The quotient rule states that the derivative of this function, \(f'(x)\), is given by \(f'(x) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2}\).

Practically, this means you must first find the derivatives of the numerator, \(u'(x)\), and denominator, \(v'((x)\), separately. Once you have those, multiply the derivative of the numerator by the original denominator and the original numerator by the derivative of the denominator, then subtract one from the other, and finally, divide by the square of the original denominator. This rule is indispensable for complex fractionated functions and understanding it is a stepping stone for mastering calculus.

Differentiable Functions

A function is differentiable at a point if it has a defined derivative at that point. For a function to be differentiable on a range of values, it must be continuous and smooth—without any breaks, corners, or vertical tangents within that range. For the quotient rule to be applicable, both the numerator and denominator functions, \(u(x)\) and \(v(x)\), must be differentiable.

Differentiability implies a function has a distinct slope at each point along its graph. Taking the derivative of a function at any point gives us the slope of the tangent line to the function's graph at that point. The concept of differentiability is fundamental because it allows us to understand the behavior of functions and facilitates the prediction of their behavior around small intervals, which is particularly useful in applied mathematics and physics.

Simplify Derivative

Once you've found the derivative using the quotient rule—or any other rule, for that matter—simplifying your result is the final step to revealing a clean and concise expression that is more suitable for further analysis or application. Simplifying the derivative may involve canceling out like terms, factoring, expanding, and reducing fractions.

For exercise \( f(x)=\frac{x-1}{2x+1} \), after applying the quotient rule, the resulting derivative had terms that could be combined and simplified, transforming an initially complex-looking function into its simplest form \( f'(x) = \frac{3}{(2x+1)^2} \). Simplifying makes the derivative easier to interpret and work with, especially when evaluating the rate of change at particular points or solving equations involving the derivative. It is a skill as crucial as finding the derivative itself and should be practiced regularly to achieve fluency.