Question

Expressions that occur in calculus are given. Write each expression as a single quotient in which only positive exponents and radicals appear. $$\frac{2 x\left(1-x^{2}\right)^{1 / 3}+\frac{2}{3} x^{3}\left(1-x^{2}\right)^{-2 / 3}}{\left(1-x^{2}\right)^{2 / 3}} \quad \neq-1, x \neq 1$$

Short answer

\frac{2x(1 - x^2)^{1 / 3} + \frac{2}{3}x^3(1 - x^2)^{-2 / 3}}{(1 - x^2)^{2 / 3}}

Step-by-step solution

Combine Like Terms

First, identify common terms in the numerator. Notice that both terms in the numerator contain a factor of \((1 - x^2)^{1 / 3}\). The numerator can be written as: \[ 2x (1 - x^2)^{1 / 3} + \frac{2}{3}x^3 (1 - x^2)^{-2 / 3} = \frac{2x (1 - x^2)}{(1 - x^2)^{2 / 3}} + \frac{2}{3}x^3 (1 - x^2)^{-2 / 3} \]

Factor Out Common Term in the Numerator

Next, factor out the common term \( (1 - x^2)^{-2 / 3} \) from the numerator: \[ 2x(1 - x^2)^{1 / 3} \cdot (1 - x^2)^{2 / 3} + \frac{2}{3}x^3 (1 - x^2)^{-2 / 3} (1 - x^2)^{2 / 3} \] Simplify to get: \[ 2x(1 - x^2)^{-1 / 3} + \frac{2}{3}x^3 (1 - x^2)^{-2 / 3} \]

Simplify the Expression

Combine the terms with the common denominator: \[\frac{2x (1 - x^2) + \frac{2}{3} x^3}{(1 - x^2)^{2 / 3}}\ = \frac{2x (1 - x^2)^{1 / 3} + \frac{2}{3} x^3 (1 - x^2)^{-2 / 3}}{(1 - x^2)^{2 / 3}} \]

Simplify the Denominator

Notice that the denominator, \( (1 - x^2)^{2 / 3} \), is unchanged.

Combine the Fraction

Combine the entire expression under a single quotient: \[ \frac{2x (1 - x^2)^{1 / 3} + \frac{2}{3} x^3 (1 - x^2)^{-2 / 3}}{(1 - x^2)^{2 / 3}} \]

Key concepts

Single Quotient

Simplifying complex expressions to a single quotient form is often necessary in calculus to make the solution more manageable and easier to understand. To achieve this, we frequently combine like terms and factor out common elements.
Initially, both terms in the numerator have a common factor: \( \left(1 - x^2\right)^{1 / 3} \). By rearranging and simplifying these common factors, we can write the expression as follows:
\[ 2x \left(1 - x^2\right)^{1 / 3} + \frac{2}{3} x^3 \left(1 - x^2\right)^{-2 / 3} \]
We then focus on getting a common term that can be factored out. This helps in combining the terms into a single quotient.
On simplifying, the expression becomes:
\[ \frac{2x \left(1 - x^2\right) + \frac{2}{3} x^3}{\left(1 - x^2\right)^{2 / 3}} \]
By continuing to factor and combine terms, we ultimately achieve the desired single quotient form. This method makes calculus operations, such as differentiation or integration, easier to handle as the expression is now in a consolidated form.

Positive Exponents

Working with positive exponents simplifies the process and makes the expression easier to interpret or manipulate.
When given an expression like \( \left(1 - x^2\right)^{-2 / 3} \), it is preferable to rewrite terms to involve positive exponents. Negative exponents can indicate a fraction, which might complicate calculations or understanding.
For example, \( \left(1 - x^2\right)^{-2 / 3} \) can be rewritten as \( \frac{1}{\left(1 - x^2\right)^{2 / 3}} \). By doing this, we make it easier to see how terms combine and simplify. Additionally, common denominators become more apparent:
\[ 2x \left(1 - x^2\right)^{1 / 3} + \frac{2}{3} x^3 \left(1 - x^2\right)^{-2 / 3} \]
Factoring in the common terms results in:
\[ \frac{2x \left(1 - x^2\right) + \frac{2}{3} x^3}{\left(1 - x^2\right)^{2 / 3}} \]
This illustrates how maintaining positive exponents simplifies not just writing the expression, but also combining terms effectively.

Calculus

Calculus often involves simplifying expressions to make differentiation and integration straightforward. The process demonstrated in this exercise is crucial for understanding these operations.
Consider the original expression:
\[ \frac{2 x\left(1-x^{2}\right)^{1 / 3}+\frac{2}{3} x^{3}\left(1-x^{2}\right)^{-2 / 3}}{\left(1-x^{2}\right)^{2 / 3}} \]
By simplifying to a single quotient with positive exponents, we make it more manageable. The simplified form:
\[ \frac{2x \left(1 - x^2\right) + \frac{2}{3} x^3}{\left(1 - x^2\right)^{2 / 3}} \]
is easier to differentiate or integrate.
For differentiation, the expression's simplified form can be treated using standard rules such as the product rule, chain rule, or quotient rule. Integration, similarly, benefits from having a clear, consolidated fraction. This method shows just how important simplification is in calculus, enabling efficient problem-solving and clearer results.