Chapter 2: Problem 22
Differentiate the function. \(y=\frac{\sqrt{x}+x}{x^{2}}\)
Short Answer
Expert verified
The derivative is \( \frac{dy}{dx} = -\frac{3}{2}x^{-5/2} - x^{-2} \).
Step by step solution
01
Simplify the Function
Rewrite the function making sure all terms are more straightforward to differentiate. We have: \[ y = \frac{\sqrt{x}+x}{x^{2}} = \frac{x^{1/2}}{x^{2}} + \frac{x}{x^{2}} \]which simplifies to:\[ y = x^{-3/2} + x^{-1} \]
02
Differentiate the First Term
Differentiate the term \( x^{-3/2} \) using the power rule, which states \( \frac{d}{dx} [x^n] = nx^{n-1} \). Thus:\[ \frac{d}{dx}[x^{-3/2}] = -\frac{3}{2}x^{-5/2} \]
03
Differentiate the Second Term
Differentiate the term \( x^{-1} \) using the power rule. We have:\[ \frac{d}{dx}[x^{-1}] = -x^{-2} \]
04
Combine the Derivatives
Combine the results from Steps 2 and 3 to find the derivative of the function:\[ \frac{dy}{dx} = -\frac{3}{2}x^{-5/2} - x^{-2} \]
05
Simplify the Derivative
Optionally, convert the derivative back to a form that can be easier to interpret:\[ \frac{dy}{dx} = -\frac{3}{2} \frac{1}{x^{5/2}} - \frac{1}{x^{2}} \]
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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 technique in calculus used to differentiate functions of the form \( x^n \), where \( n \) is any real number. The power rule states that the derivative of \( x^n \) is \( nx^{n-1} \). Here's how it works: - Take the power \( n \) in the exponent: e.g., if \( n = -\frac{3}{2} \).- Multiply \( n \) by the coefficient (if any).- Subtract one from the exponent \( n \). For example, to differentiate \( x^{-3/2} \), apply the power rule: - Multiply the exponent by the coefficient: \(-\frac{3}{2}\times1 = -\frac{3}{2}\).- Subtract 1 from the exponent: \(-\frac{3}{2} - 1 = -\frac{5}{2}\).The derivative becomes \(-\frac{3}{2}x^{-5/2}\). This simple rule speeds up differentiation tremendously, especially in the context of polynomial-like expressions.
Simplifying Rational Expressions
Simplifying rational expressions involves reducing fractions to their simplest form, which makes them easier to work with, especially in calculus. Let's consider the function \( y = \frac{\sqrt{x} + x}{x^2} \).Here are the steps:- Rewrite \( \sqrt{x} \) as \( x^{1/2} \). This helps when you apply other algebraic operations.- Break down the fraction: \( \frac{x^{1/2}}{x^2} + \frac{x}{x^2} \).- Simplify each term: \( x^{1/2}x^{-2} = x^{-3/2} \) and \( x^1x^{-2} = x^{-1} \).These steps reduce the complexity, transforming the original function into \( y = x^{-3/2} + x^{-1} \). Such simplification is crucial as it paves the way for integrating or differentiating the expression neatly.
Fractional Exponents
Fractional exponents are a way to express roots in terms of powers. Instead of writing \( \sqrt{x} \), it is often presented as \( x^{1/2} \). This is incredibly helpful in calculus as it allows one to leverage the power rule directly.Consider the conversion:- \( \sqrt{x^n} = x^{n/2} \), meaning the square root is an exponent of \( \frac{1}{2} \).- Other roots can follow similarly: \( \sqrt[3]{x^n} = x^{n/3} \).Transforming roots into exponents simplifies differentiation using algebraic rules. It provides a consistent way to apply the power rule and simplify calculations. This technique was used to convert \( \sqrt{x} \) to \( x^{1/2} \), and then further manipulation led to \( x^{-3/2} \). Fractional exponents not only make calculations straightforward but also reveal deeper connections between polynomials and roots.
Calculus
Calculus is a branch of mathematics focused on change and motion. It primarily deals with concepts such as derivatives and integrals. Differentiation, as seen in this exercise, is a central part of calculus, used to find the rate at which something changes.Key aspects of differentiation:- Provides a way to find the slope of a function at any point.- Utilizes rules, such as the power rule, to make finding derivatives easier.Take the function being differentiated, \( y = \frac{\sqrt{x} + x}{x^2} \):- After simplifying the function, derivatives of each term can be determined, showcasing the function's behavior.- Combine these derivatives to find \( \frac{dy}{dx} \) which tells us how \( y \) changes with respect to \( x \).Thus, calculus isn't just about finding derivatives but understanding the real-world problems and patterns they represent. It provides tools to analyze complex systems, predict future behavior, and optimize solutions.