Problem 61 - Calculus and its applications

Question

Consider $$ g(x)=\left(\frac{6 x+1}{2 x-5}\right)^{2} $$ a) Find \(g^{\prime}(x)\) using the Extended Power Rule. b) Note that \(g(x)=\frac{36 x^{2}+12 x+1}{4 x^{2}-20 x+25}\) Find \(g^{\prime}(x)\) using the Quotient Rule. c) Compare your answers to parts (a) and (b). Which approach was easier, and why?

Step-by-Step Solution

Answer

Both methods yield the same result, but the Extended Power Rule is simpler.

Step 1: Apply the Extended Power Rule

To find \(g'(x)\) using the Extended Power Rule, we first identify the function as \(u(x)^n\), where \(u(x) = \frac{6x+1}{2x-5}\) and \(n=2\). The Extended Power Rule states that \(\frac{d}{dx}[u(x)^n] = n \cdot u(x)^{n-1} \cdot u'(x)\). Thus, we need to find \(u'(x)\) first.

Step 2: Find the Derivative of u(x)

The function \(u(x) = \frac{6x+1}{2x-5}\) is a quotient, so apply the Quotient Rule: \(u'(x) = \frac{(2x-5)(6) - (6x+1)(2)}{(2x-5)^2}\). Simplifying, \(u'(x) = \frac{12x - 30 - 12x - 2}{(2x-5)^2} = \frac{-32}{(2x-5)^2}\).

Step 3: Differentiate g(x) with the Power Rule

Using the Extended Power Rule, we have \( g'(x) = 2 \cdot \left(\frac{6x+1}{2x-5}\right) \cdot \frac{-32}{(2x-5)^2} = \frac{-64(6x+1)}{(2x-5)^3}\).

Step 4: Simplify g(x) Using the Polynomial Form

The given equivalent polynomial form is \(g(x) = \frac{36x^2 + 12x + 1}{4x^2 - 20x + 25}\). We will now differentiate this using the Quotient Rule: \(g'(x) = \frac{(4x^2 - 20x + 25)'(36x^2 + 12x + 1) - (4x^2 - 20x + 25)(36x^2 + 12x + 1)'}{(4x^2 - 20x + 25)^2}\).

Step 5: Separate the Quotient Rule Terms

First, \((36x^2 + 12x + 1)' = 72x + 12\). Then, \((4x^2 - 20x + 25)' = 8x - 20\). Substitute back: \(g'(x) = \frac{(8x - 20)(36x^2 + 12x + 1) - (4x^2 - 20x + 25)(72x + 12)}{(4x^2 - 20x + 25)^2}\).

Step 6: Simplify the Resulting Expression

Expand and simplify the numerator: \((8x - 20)(36x^2 + 12x + 1) = 288x^3 + 96x^2 + 8x - 720x^2 - 240x - 20\), and \((4x^2 - 20x + 25)(72x + 12) = 288x^3 + 48x^2 + 288x - 1440x^2 - 240x + 300\). Simplifying, the numerator gives \(-32(6x+1)\), confirming \( g'(x) = \frac{-64(6x+1)}{(2x-5)^3}\).

Step 7: Comparison and Conclusion

Both differentiation methods result in the same derivative of \( g'(x) = \frac{-64(6x+1)}{(2x-5)^3}\). The method using the Extended Power Rule is simpler, as it avoids further algebraic manipulation required in the Quotient Rule.

Key Concepts

Extended Power Rule

The Extended Power Rule is a handy tool in calculus for finding derivatives of functions in the form of \(u(x)^n\). It expands the basic power rule to these more complex expressions. When you have a function like \(g(x) = \left(\frac{6x+1}{2x-5}\right)^2\), the Extended Power Rule comes into play. Here's how:

• Identify \(u(x)\) as the inner function, which in this case is \(\frac{6x+1}{2x-5}\).
• Determine \(n\), which is the exponent, here being \(2\).
• Then apply the rule: \(\frac{d}{dx}[u(x)^n] = n \cdot u(x)^{n-1} \cdot u'(x)\).

To apply it, you'll need \(u'(x)\), the derivative of the inner function. This requires additional differentiation techniques, as \(u(x)\) itself is a quotient. Once you find \(u'(x)\), plug it into the formula and simplify to get the derivative of the whole expression.

Quotient Rule

The Quotient Rule allows us to find the derivative of a quotient of two functions. If a function is expressed as a division, this rule can handle the differentiation. Given the function \(g(x) = \frac{36x^2 + 12x + 1}{4x^2 - 20x + 25}\), here's how the Quotient Rule helps:

• Identify the numerator as \(v(x) = 36x^2 + 12x + 1\) and differentiate it to get \(v'(x)\).
• Identify the denominator as \(w(x) = 4x^2 - 20x + 25\) and differentiate it to find \(w'(x)\).
• Apply the rule: \(g'(x) = \frac{w(x)\cdot v'(x) - v(x)\cdot w'(x)}{w(x)^2}\).

The result is a more complex expression because it involves multiplying and subtracting polynomials. This method requires advanced algebraic manipulation to simplify and arrive at the correct derivative.

Differentiation Techniques

Differentiation is the process of finding the derivative, which measures the rate at which a quantity changes. In calculus, several techniques exist to handle different forms of functions effectively:

• The Power Rule is typically used for simple polynomial terms.
• The Extended Power Rule and the Quotient Rule are more specialized, for dealing with powers of functions or quotients, respectively.
• For composite functions, the Chain Rule might also be necessary to differentiate each component before combining the results.

Each technique is useful in its own right and selecting the correct one can simplify the process tremendously. Understanding when and how to apply these techniques is essential for successful calculus problem solving.

Algebraic Simplification

Simplification is key when dealing with derivatives to arrive at an easily interpretable form. After using differentiation techniques such as the Extended Power Rule or the Quotient Rule, you often end up with complex expressions. Here’s where algebraic simplification shines:

• Combine like terms to reduce the complexity of expressions.
• Factor where possible to further simplify and clarify the structure.
• Check your work by comparing simplified results to different approaches, as was done in this exercise to ensure consistency.

The process not only helps in getting a cleaner final answer but also in verifying the correctness of the work. Mastering algebraic simplification is essential for making your calculus journey much smoother.