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Simplify the expression. (This type of expression arises in calculus when using the 鈥渜uotient rule.鈥) $$\frac{\left(1-x^{2}\right)^{1 / 2}+x^{2}\left(1-x^{2}\right)^{-1 / 2}}{1-x^{2}}$$

Short Answer

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

Step by step solution

01

Identify the Expression

The expression given is \( \frac{\left(1-x^{2}\right)^{1 / 2}+x^{2}\left(1-x^{2}\right)^{-1 / 2}}{1-x^{2}} \). We need to simplify this expression.
02

Factor Common Terms

Let's look for common factors in the numerator. We see that \((1-x^2)^{1/2}\) is a common factor. The expression can be rewritten as:\[ \frac{(1-x^2)^{1/2} + x^2(1-x^2)^{-1/2}}{1-x^2} = \frac{(1-x^2)^{-1/2}((1-x^2) + x^2)}{1-x^2} \]
03

Simplify the Inner Expression

Simplify the expression inside the parentheses \((1-x^2) + x^2 = 1\). The expression now becomes:\[ \frac{(1-x^2)^{-1/2}(1)}{1-x^2} = \frac{1}{(1-x^2)^{1/2}(1-x^2)} \]
04

Further Simplify by Division

Since \((1-x^2)^{1/2} = \sqrt{1-x^2}\), the expression can be further simplified by distributing the division across the square root:\[ \frac{1}{(1-x^2)^{3/2}} \]
05

Provide Final Simplified Expression

The final simplified version of the expression is:\[ \frac{1}{(1-x^2)^{3/2}} \]

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Key Concepts

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

Quotient Rule
The quotient rule is a fundamental tool in calculus, especially useful when differentiating expressions that are given as a ratio of two differentiable functions. It helps find the derivative of a quotient of two functions described by the formula:\[ \left( \frac{u}{v} \right)' = \frac{v \cdot u' - u \cdot v'}{v^2} \]where \( u \) and \( v \) are functions of \( x \), and \( u' \) and \( v' \) are their respective derivatives. Applying the quotient rule involves several steps, like:
  • Identifying the numerator \( u \) and the denominator \( v \) of the expression.
  • Calculating the derivatives \( u' \) and \( v' \).
  • Substituting these values into the quotient rule formula.
  • Simplifying the resulting expression after applying the formula.
Although the original exercise on simplifying expressions doesn鈥檛 require differentiation, it sets the foundation for understanding how to handle complex expressions in calculus efficiently. Studying the quotient rule aids in mastering the simplification process. Understanding this rule is crucial as it prevails in various calculus functions and considerably eases the computation of derivatives in more complex expressions.
Algebraic Simplification
Algebraic simplification involves reducing expressions into simpler or more convenient forms. It often includes combining like terms, factoring, or canceling out components, as seen in the provided exercise. Simplifying expressions can significantly ease calculations, especially in complex algebraic and calculus operations. In the original solution, the common factor \((1-x^2)^{1/2}\) was identified and factored out to simplify the given expression. This process involved:
  • Finding and factoring out common components in both parts of the numerator.
  • Rewriting the numerator using these common factors to facilitate its reduction to a more manageable form.
  • Further simplifying the factored expression by canceling or combining terms whenever possible.
The goal is always to transform an expression into its simplest equivalent, making it easier to understand or compute further operations. Such simplifications are a foundational skill in algebra, required for successful execution of more advanced mathematics topics like calculus.
Precalculus
Precalculus serves as the overarching discipline preparing students for calculus by covering essential topics like functions, complex numbers, trigonometry, and algebra. Mastery of these topics is vital as they recur in calculus-related problems and solutions. The original exercise lies well within the precalculus scope, where algebraic simplification is essential, such as:
  • Understanding and manipulating rational expressions.
  • Factoring and expanding expressions into more utilizable forms.
  • Working with exponents and roots to understand their properties in expressions.
Precalculus creates the groundwork necessary for tackling calculus topics like limits, derivatives, and integrals. These subjects can seem daunting without a solid understanding of the foundational principles covered in precalculus. It is crucial to comfortably navigate algebraic manipulations and simplifications at this stage to ensure a seamless transition into more advanced mathematical concepts. Mastery of precalculus not only prepares you for calculus but also enhances problem-solving skills applicable across various mathematical disciplines.

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Most popular questions from this chapter

Test the equation for symmetry. $$x=y^{4}-y^{2}$$

Prove the following Laws of Exponents for the case in which \(m\) and \(n\) are positive integers and \(m>n\) (a) Law \(2: \frac{a^{m}}{a^{n}}=a^{m-n}\) (b) Law \(5:\left(\frac{a}{b}\right)^{n}=\frac{a^{n}}{b^{n}}\)

Simplify the expression. (a) \(\left(\frac{a^{1 / 6} b^{-3}}{x^{-1} y}\right)^{3}\left(\frac{x^{-2} b^{-1}}{a^{3 / 2} y^{1 / 3}}\right)\) (b) \(\frac{(9 s t)^{3 / 2}}{\left(27 s^{3} t^{-4}\right)^{2 / 3}}\left(\frac{3 s^{-2}}{4 t^{1 / 3}}\right)^{-1}\)

Distances in a City \(A\) city has streets that run north and south and avenues that run east and west, all equally spaced. Streets and avenues are numbered sequentially, as shown in the figure. The walking distance between points \(A\) and \(B\) is 7 blocks - that is, 3 blocks east and 4 blocks north. To find the straight-line distance \(d\), we must use the Distance Formula. (a) Find the straight-line distance (in blocks) between \(A\) and \(B\) (b) Find the walking distance and the straight-line distance between the corner of 4 th St. and 2 nd Ave. and the corner of 11 th St. and 26 th Ave. (c) What must be true about the points \(P\) and \(Q\) if the walking distance between \(P\) and \(Q\) equals the straight[line distance between \(P\) and \(Q ?\)

It follows from Kepler's Third Law of planetary motion that the average distance from a planet to the sun (in meters) is $$d=\left(\frac{G M}{4 \pi^{2}}\right)^{1 / 3} T^{2 / 3}$$ where \(M=1.99 \times 10^{30} \mathrm{kg}\) is the mass of the sun, \(G=6.67 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}\) is the gravitational constant, and \(T\) is the period of the planet's orbit (in seconds). Use the fact that the period of the earth's orbit is about 365.25 days to find the distance from the earth to the sun.

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