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In Exercises \(17-28,\) find \(d y\) $$ y=\frac{2 \sqrt{x}}{3(1+\sqrt{x})} $$

Short Answer

Expert verified
\( dy = \frac{dx}{3\sqrt{x}(1+\sqrt{x})^2} \).

Step by step solution

01

Identify the Function and Simplify

The function given is \[ y = \frac{2 \sqrt{x}}{3(1 + \sqrt{x})} \] Here, we need to find the differential \( dy \). Before finding the derivative, it's often helpful to simplify our function if possible to make differentiation easier.
02

Apply Quotient Rule

Our function is in the form of a quotient \( \frac{u}{v} \), where \( u = 2\sqrt{x} \) and \( v = 3(1+\sqrt{x}) \). The quotient rule states that \[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]. We will need to differentiate \( u \) and \( v \) separately.
03

Differentiate the Numerator and Denominator

Differentiate the numerator \( u = 2\sqrt{x} \). Recall that \( \sqrt{x} = x^{0.5} \); thus, using the power rule, \[ \frac{du}{dx} = \frac{d}{dx}(2x^{0.5}) = 2\cdot0.5x^{-0.5} = \frac{1}{\sqrt{x}} \]. Now differentiate the denominator \( v = 3(1+\sqrt{x}) \),\[ \frac{dv}{dx} = 3(0 + \frac{1}{2\sqrt{x}}) = \frac{3}{2\sqrt{x}} \].
04

Substitute and Simplify

Substitute these derivatives back into the quotient rule: \[ \frac{dy}{dx} = \frac{3(1+\sqrt{x})\cdot\frac{1}{\sqrt{x}} - 2\sqrt{x}\cdot\frac{3}{2\sqrt{x}}}{[3(1+\sqrt{x})]^2} \] After simplifying, the expression becomes, \[ \frac{dy}{dx} = \frac{3 + 3x^{-0.5} - 3}{9(1+\sqrt{x})^2} \] Thus, the simplified derivative is \[ \frac{dy}{dx} = \frac{3x^{-0.5}}{9(1+\sqrt{x})^2} = \frac{1}{3\sqrt{x}(1+\sqrt{x})^2} \].
05

Calculate dy

To find \( dy \), multiply \( \frac{dy}{dx} \) by \( dx \): \[ dy = \frac{1}{3\sqrt{x}(1+\sqrt{x})^2} \cdot dx \]. So \( dy \) is expressed as a differential: \[ dy = \frac{dx}{3\sqrt{x}(1+\sqrt{x})^2} \].

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Key Concepts

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

Quotient Rule
When you have to find the derivative of a function that is the ratio of two other functions, the quotient rule is your go-to tool. It's essential to break down complex fraction-related problems into simpler parts. The rule is all about differentiating a function expressed as the quotient \( \frac{u}{v} \). Here, \( u \) and \( v \) are both differentiable functions. The quotient rule formula is:
  • \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \)
This rule tells us that the derivative of the quotient is the denominator \( v \) times the derivative of the numerator \( du/dx \) minus the numerator \( u \) times the derivative of the denominator \( dv/dx \), all divided by the square of the denominator \( v^2 \).
This might seem complicated initially, but it's just a consistent process of applying the formula.
With practice, it becomes second nature.Why use the quotient rule? Because it simplifies the differentiation process for complex fractional expressions, making calculus problems much more straightforward.
Differentiation
Differentiation is one of the two main operations in calculus, the other being integration. At its core, differentiation refers to the process of finding a derivative.
A derivative represents how a function changes as its input changes. In a sense, it measures the function's rate of change or slope at any given point.Here's a quick breakdown of key points:
  • The derivative of a function tells us the slope or steepness at any point along the curve of that function.
  • When finding \( dy/dx \), you're determining the rate of change of \( y \) with respect to \( x \).
  • This is useful for understanding motion, growth patterns, and more.
Differentiation isn't just a tool for simple variables; it's adaptable and can handle composite functions through techniques like the chain rule, and, of course, the quotient rule.
It's a powerful skill for analyzing behaviors within mathematical models and real-world phenomena.
Power Rule
The power rule is an easy-to-use shortcut that greatly simplifies the differentiation of functions that are polynomials or have terms of the form \( x^n \). It allows quick computation of the derivative without requiring more complex differentiation techniques.Here's how the power rule works:
  • If \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \).
This means that you bring down the exponent as a coefficient and reduce the exponent by one to find the derivative. This rule even applies to roots and fractional exponents, making it versatile in the context of calculus.For example, when differentiating \( 2x^{0.5} \):
  • Using the power rule, we first bring down the exponent 0.5, multiply it by 2, and then decrement the exponent by 1, leading to \( 1x^{-0.5} \), which is \( \frac{1}{\sqrt{x}} \).
This consistency across different forms of functions makes the power rule a fundamental component of differentiation, helping to solve problems quickly and efficiently.

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