Problem 31 - College Algebra and Calculus: An Applied Approach

Question

Find the derivative of the function. State which differentiation rule(s) you used to find the derivative. $$ f(x)=\frac{3 x-2}{2 x-3} $$

Step-by-Step Solution

Answer

The derivative of the function \(f(x) = \frac{3x - 2}{2x - 3}\) is \(f'(x) = \frac{-7}{(2x-3)^2}\). The Quotient Rule was used to find the derivative.

Step 1: Identification of \(u\), \(v\), \(u'\), and \(v'\)

Start with identifying \(u\) and \(v\). Here, \(u=3x-2\) and \(v=2x-3\). Now, find the derivatives of \(u\) and \(v\), denoted as \(u'\) and \(v'\). Hence, \(u'=3\) and \(v'=2\).

Step 2: Application of the Quotient Rule

Next, apply the quotient rule \(\frac{vu' - uv'}{v^2}\). Plugging in \(u\), \(v\), \(u'\), and \(v'\) gives \(\frac{(2x-3)*3 - (3x-2)*2}{(2x-3)^2}\).

Step 3: Simplification of the derivative

Lastly, simplify the derivative. This yields \(\frac{-7}{(2x-3)^2}\).

Key Concepts

Differentiation Rules

Learning how to differentiate functions is a crucial skill in calculus, and it involves the application of several rules. Differentiation rules facilitate finding the derivative, which represents the rate at which a function changes at any given point. Here are the fundamental rules you might encounter:

• Power Rule: If you have a function of the form \(f(x) = x^n\), the derivative, \(f'(x)\), is \(nx^{n-1}\).
• Product Rule: For functions that are products of two functions, \(f(x) = u(x)v(x)\), the derivative is \(f'(x) = u'(x)v(x) + u(x)v'(x)\).
• Quotient Rule: Used when dealing with a quotient of two functions, we'll dive deeper into this next.
• Chain Rule: Helps in differentiating composite functions, involving a function inside another function.

Each rule simplifies calculating derivatives under different circumstances by providing a formulaic approach.

Quotient Rule

The quotient rule is one of the differentiation rules, particularly useful when you have a function that is the division of one function by another. This is applicable when finding the derivative of a function like \( f(x)=\frac{u}{v} \). The formula for the quotient rule is:\[ f'(x) = \frac{v u' - u v'}{v^2} \]Where:- \( u \) and \( v \) are functions of \( x \)- \( u' \) is the derivative of \( u \)- \( v' \) is the derivative of \( v \)In our example, we have \( f(x) = \frac{3x-2}{2x-3} \). We identify \( u = 3x - 2 \) and \( v = 2x - 3 \). The derivatives are \( u' = 3 \) and \( v' = 2 \).Using the quotient rule formula:- Substitute \( u \), \( v \), \( u' \), and \( v' \) into it, resulting in the derivative expression: \[ \frac{(2x-3)*3 - (3x-2)*2}{(2x-3)^2} \]The logic of the quotient rule helps us systematically handle divisions of functions while preserving a clear path to the solution.

Simplification of Expressions

Once you apply the quotient rule, the result is an expression that often requires simplification. Simplifying expressions in calculus is crucial because it makes further computations easier and leads to a more understandable form of your answer.After applying the quotient rule to our original function, we arrived at the derivative expression:\[ \frac{3(2x-3) - 2(3x-2)}{(2x-3)^2} \]The simplification process involves:

• Expanding Terms: Multiply out everything within the numerators, i.e., \(3(2x-3) = 6x - 9\) and \(2(3x-2) = 6x - 4\).
• Combining Like Terms: Subtract the like terms in the numerator: \((6x - 9) - (6x - 4) = -9 + 4 = -5\).
• Final Expression: This simplifies the derivative to \( \frac{-5}{(2x-3)^2} \).

The key here is to perform these steps methodically to avoid errors, resulting in a tidy, simple derivative expression ready for analysis.