Chapter 2: Problem 37
Find \(d y / d x\) $$ y=\left(\frac{x-5}{2 x+1}\right)^{3} $$
Short Answer
Expert verified
\( \frac{dy}{dx} = \frac{33(x-5)^2}{(2x+1)^4} \)
Step by step solution
01
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.
02
Apply the chain rule
Given the function \(y = (u)^3\) where \(u = \frac{x-5}{2x+1}\), apply the chain rule: \(\frac{dy}{dx} = 3u^2 \cdot \frac{du}{dx}\).
03
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 \(b' = 2\).
04
Apply the quotient rule
Compute \(\frac{du}{dx}\):\[ \frac{du}{dx} = \frac{(1)(2x+1) - (x-5)(2)}{(2x+1)^2} = \frac{2x + 1 - 2x + 10}{(2x+1)^2} = \frac{11}{(2x+1)^2} \]
05
Substitute back into chain rule formula
Substitute \(u\) and \(\frac{du}{dx}\) back into \(\frac{dy}{dx} = 3u^2 \cdot \frac{du}{dx}\):\[ \frac{dy}{dx} = 3\left(\frac{x-5}{2x+1}\right)^2 \cdot \frac{11}{(2x+1)^2} \]
06
Simplify the expression
Combine the terms:\[ \frac{dy}{dx} = \frac{3 \cdot 11 \cdot (x-5)^2}{(2x+1)^4} = \frac{33(x-5)^2}{(2x+1)^4} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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: \(\frac{dy}{dx} = \text{(derivative of } g \text{ with respect to } u) \times \text{(derivative of } u \text{ with respect to } x)\).
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 \(b' = 2\).
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 \(x eq -\frac{1}{2}\)).