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Chapter 3: Problem 21

Find \(d y / d x\). \(y=\sqrt[3]{x}+\frac{1}{\sqrt[3]{x}}\)

Short Answer

Expert verified
\( \frac{dy}{dx} = \frac{1}{3}x^{-2/3} - \frac{1}{3}x^{-4/3} \).

Step by step solution

01

Rewrite the function using exponent notation

The given function is \( y = \sqrt[3]{x} + \frac{1}{\sqrt[3]{x}} \). First, rewrite the roots in terms of exponents. We have \( \sqrt[3]{x} = x^{1/3} \) and \( \frac{1}{\sqrt[3]{x}} = x^{-1/3} \). Thus, the function can be expressed as: \[ y = x^{1/3} + x^{-1/3}. \]
02

Use the power rule to differentiate each term

Differentiate each term in the function individually using the power rule. The power rule states that \( \frac{d}{dx}[x^n] = n \cdot x^{n-1} \). For the term \( x^{1/3} \), apply the power rule:\[ \frac{d}{dx}[x^{1/3}] = \frac{1}{3}x^{1/3 - 1} = \frac{1}{3}x^{-2/3}. \]For the term \( x^{-1/3} \), apply the power rule:\[ \frac{d}{dx}[x^{-1/3}] = -\frac{1}{3}x^{-1/3 - 1} = -\frac{1}{3}x^{-4/3}. \]
03

Combine the derivatives of each term

Add the derivatives of each term to find the overall derivative. Combine the results from Step 2:\[ \frac{dy}{dx} = \frac{1}{3}x^{-2/3} - \frac{1}{3}x^{-4/3}. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Rule
The power rule is a fundamental concept in calculus, especially useful for differentiation. Its primary use is to find the derivative of any function expressed as a power of a variable. The rule is elegantly simple: if you have a function of the form \( f(x) = x^n \), its derivative is \( f'(x) = n \cdot x^{n-1} \). This means you multiply the power by the coefficient of the term, and then reduce the exponent by one.
  • Start with identifying terms in power form.
  • Apply the rule to each term separately, which works because differentiation is a linear operation.
  • Remember, it's valid for all real numbers \( n \), including fractions and negatives.
For example, in the exercise above, we used it to derive \( \frac{d}{dx}[x^{1/3}] = \frac{1}{3}x^{-2/3} \) and \( \frac{d}{dx}[x^{-1/3}] = -\frac{1}{3}x^{-4/3} \). This showcases the power rule's applicability to terms with fractional and negative exponents.
Exponent Notation
Exponent notation is crucial for working with roots and powers. Instead of writing \( \sqrt[3]{x} \), you can express it as \( x^{1/3} \). This format makes it easier to manipulate mathematical expressions, especially during differentiation.
  • Convert roots to fractional exponents, such as \( \sqrt{a} = a^{1/2} \) or \( \sqrt[3]{a} = a^{1/3} \).
  • Handle inverse operations with negative exponents, like \( \frac{1}{x^b} = x^{-b} \).
  • Both fractional and negative exponents follow the same rules as integer exponents.
In the original exercise, rewriting \( \sqrt[3]{x} \) as \( x^{1/3} \) and \( \frac{1}{\sqrt[3]{x}} \) as \( x^{-1/3} \) allows easy application of the power rule. The expressions become more manageable and differentiation follows naturally.
Differentiation
Differentiation is the process that calculates the rate at which a function is changing. In essence, it finds the derivative \( \frac{dy}{dx} \) of a given function. This reveals slopes of tangent lines, rates of change, and is foundational in finding critical points in calculus.
  • The process involves applying rules like the power rule, chain rule, and product rule to express rates of change.
  • It's a linear operation, allowing derivatives of sums to be handled term by term.
  • Provides insights into functions' behavior, such as increasing or decreasing trends.
In our provided solution, differentiation of the function \( y = x^{1/3} + x^{-1/3} \) used the power rule to derive expressions \( \frac{1}{3}x^{-2/3} \) and \( -\frac{1}{3}x^{-4/3} \). Combining these derivatives gives the overall derivative of the function, showcasing how differentiation directly applies mathematical rules to real-world problems.

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