Problem 37 - Calculus Early Transcendentals

Question

Find \(d y / d x\) $$ y=\left(\frac{x-5}{2 x+1}\right)^{3} $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{33(x-5)^2}{(2x+1)^4} \)

Step 1: Identify the type of function

The function given is in the form of a rational function raised to a power. This indicates that the chain rule for derivatives is needed.

Step 2: Apply the chain rule

Given the function \(y = (u)^3\) where \(u = \frac{x-5}{2x+1}\), apply the chain rule: \(\frac{dy}{dx} = 3u^2 \cdot \frac{du}{dx}\).

Step 3: Differentiate the inside function (u)

Differentiate \(u = \frac{x-5}{2x+1}\) with respect to \(x\) using the quotient rule, \(\frac{d}{dx}\left(\frac{a}{b}\right) = \frac{a'b - ab'}{b^2}\). Here, \(a = x-5\), \(b = 2x+1\), \(a' = 1\), and \(b' = 2\).

Step 4: Apply the quotient rule

Compute \(\frac{du}{dx}\):\[ \frac{du}{dx} = \frac{(1)(2x+1) - (x-5)(2)}{(2x+1)^2} = \frac{2x + 1 - 2x + 10}{(2x+1)^2} = \frac{11}{(2x+1)^2} \]

Step 5: Substitute back into chain rule formula

Substitute \(u\) and \(\frac{du}{dx}\) back into \(\frac{dy}{dx} = 3u^2 \cdot \frac{du}{dx}\):\[ \frac{dy}{dx} = 3\left(\frac{x-5}{2x+1}\right)^2 \cdot \frac{11}{(2x+1)^2} \]

Step 6: Simplify the expression

Combine the terms:\[ \frac{dy}{dx} = \frac{3 \cdot 11 \cdot (x-5)^2}{(2x+1)^4} = \frac{33(x-5)^2}{(2x+1)^4} \]

Key Concepts

Chain Rule

The Chain Rule is a fundamental concept in calculus used to differentiate composite functions. It is particularly helpful when you have functions nested within other functions. In the given exercise, the function \(y = \left(\frac{x-5}{2x+1}\right)^{3}\) is a composite function. It consists of the outer function \(g(u) = u^3\) and the inner function \(u = \frac{x-5}{2x+1}\). Here's how the Chain Rule works:

• The derivative of the outer function is found first. For \(g(u) = u^3\), it yields \(3u^2\).
• The inner function \(u = \frac{x-5}{2x+1}\) is then differentiated separately using other rules, like the Quotient Rule.
• Finally, you multiply the results: \(\frac{dy}{dx} = \text{(derivative of } g \text{ with respect to } u) \times \text{(derivative of } u \text{ with respect to } x)\).

By applying the Chain Rule, you're essentially breaking down complex differentiation into manageable steps, making it easier to handle functions layered within one another.

Quotient Rule

The Quotient Rule is essential for differentiating functions that involve division, which is exactly what we deal with in the inner function \(u = \frac{x-5}{2x+1}\). When you have a function that is a ratio \(\frac{a}{b}\), the Quotient Rule provides a method for finding its derivative:The formula is:\[ \frac{d}{dx}\left(\frac{a}{b}\right) = \frac{a'b - ab'}{b^2} \]where \(a'\) and \(b'\) are the derivatives of \(a\) and \(b\) respectively.For the function \(u\):

• \(a = x-5\) and its derivative \(a' = 1\).
• \(b = 2x+1\) and its derivative \(b' = 2\).

Apply the formula:- The numerator becomes \((1)(2x+1) - (x-5)(2) = 2x + 1 - 2x + 10\).- The denominator is \((2x+1)^2\).This yields \(\frac{du}{dx} = \frac{11}{(2x+1)^2}\). Using the Quotient Rule allows you to accurately find derivatives of functions expressed as ratios.

Rational Functions

Rational functions are fractions that involve polynomials in the numerator and the denominator. In our example, \(y = \left(\frac{x-5}{2x+1}\right)^3\), the expression \(\frac{x-5}{2x+1}\) is a rational function. Understanding these kinds of functions is important because:

• They're common in many areas of math and science.
• They can be simplified or transformed using rules like the Quotient Rule.
• Rational functions have certain properties such as asymptotes and domain restrictions (denominator cannot be zero, here \(2x+1 eq 0\) implies \(x eq -\frac{1}{2}\)).

With rational functions, you're often combining or manipulating them with algebraic methods to prepare them for differentiation or integration. Therefore, knowing how they behave both graphically and algebraically is essential for deeper mathematical understanding and applications.