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Magazine Chapter 3: Problem 22
\(3-32\) Differentiate the function. \(y=\sqrt{x}(x-1)\)
Short Answer
Expert verified
\(\frac{dy}{dx} = \frac{3x - 1}{2\sqrt{x}}\)
Step by step solution
01
Identify the Product Rule
The function \( y = \sqrt{x}(x - 1) \) can be expressed as the product of two functions: \( u = \sqrt{x} \) and \( v = x - 1 \). We will differentiate \( y \) using the product rule, which states: \(\frac{d}{dx}[uv] = u'v + uv'\).
02
Differentiate u = \sqrt{x}
Using the differentiation rule for square roots, differentiate \( u = \sqrt{x} \): \( u' = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}\).
03
Differentiate v = (x - 1)
Since \( v = x - 1 \) is a linear function, its derivative is: \( v' = \frac{d}{dx}(x - 1) = 1\).
04
Apply the Product Rule
Substitute \( u, u', v, \) and \( v' \) into the product rule formula: \( \frac{d}{dx}[uv] = u'v + uv' = \frac{1}{2\sqrt{x}}(x - 1) + \sqrt{x}(1) \).
05
Simplify the Expression
Combine the terms to simplify: \[\frac{1}{2\sqrt{x}}(x - 1) + \sqrt{x} = \frac{x - 1}{2\sqrt{x}} + \sqrt{x}\\]Rewrite \( \sqrt{x} \) in terms of fractions to find common denominators: \( \sqrt{x} = \frac{x}{\sqrt{x}} = \frac{x\sqrt{x}}{\sqrt{x}\sqrt{x}} = \frac{x\sqrt{x}}{x} = \frac{x}{\sqrt{x}}\). Then, combine like terms: \[\frac{x - 1}{2\sqrt{x}} + \frac{2x}{2\sqrt{x}} = \frac{x - 1 + 2x}{2\sqrt{x}} = \frac{3x - 1}{2\sqrt{x}}\\]So, the derivative \( \frac{dy}{dx} = \frac{3x - 1}{2\sqrt{x}}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differentiation
Differentiation is a fundamental concept in calculus, which involves finding the rate at which a function is changing at any given point. More simply, it's about determining how a small change in the input of a function creates a change in the output. This process is primarily accomplished using derivatives.
- Differentiation helps us find the slope of the tangent line to the curve represented by the function.
- The basic idea is to measure how much the function's output value changes as its input value changes infinitesimally.
Derivatives
Derivatives are a core tool in calculus used to calculate the rate of change of a function. A derivative provides vital information about the behavior of functions in mathematics.
- In simple terms, the derivative of a function at a point describes the slope of the tangent line to the graph of the function at that point.
- Finding the derivative involves a set of rules and methods, such as the product rule, the chain rule, and the quotient rule, to name a few.
- Derivatives have wide applications in science and engineering, where they help model instantaneous rates of change.
Square root function
The square root function is a special type of mathematical expression, commonly symbolized as \( \sqrt{x} \) or \( x^{1/2} \). It's an essential aspect of various mathematical concepts and real-world applications.
- The square root function transforms a number into another number, which when squared, returns the original number.
- This function has a derivative that can be found using the power rule of differentiation.
- The power rule states that for any function \( x^n \), its derivative is \( nx^{n-1} \).
