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For each function, find the partials a. \(f_{x}(x, y)\) and b. \(f_{y}(x, y)\). \(f(x, y)=\frac{x y}{x+y}\)

Short Answer

Expert verified
\( f_{x}(x, y) = \frac{y^2}{(x+y)^2} \), \( f_{y}(x, y) = \frac{x^2}{(x+y)^2} \).

Step by step solution

01

Identify the Function

Given the function is \( f(x, y) = \frac{x y}{x + y} \). We are expected to find the partial derivatives of \( f \) with respect to \( x \) and \( y \).
02

Find Partial Derivative with Respect to x

To find the partial derivative \( f_{x} \), treat \( y \) as a constant. Use the quotient rule for differentiation, which is \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \), where \( u = xy \) and \( v = x + y \).Calculate \( \frac{du}{dx} = y \) and \( \frac{dv}{dx} = 1 \).Apply the quotient rule:\[ f_{x} = \frac{(x+y)(y) - (xy)(1)}{(x+y)^2} = \frac{xy + y^2 - xy}{(x+y)^2} = \frac{y^2}{(x+y)^2}.\]
03

Find Partial Derivative with Respect to y

For the partial derivative \( f_{y} \), treat \( x \) as a constant. Again, use the quotient rule with \( u = xy \) and \( v = x + y \).Calculate \( \frac{du}{dy} = x \) and \( \frac{dv}{dy} = 1 \).Apply the quotient rule:\[f_{y} = \frac{(x+y)(x) - (xy)(1)}{(x+y)^2} = \frac{x^2 + xy - xy}{(x+y)^2} = \frac{x^2}{(x+y)^2}.\]
04

Confirm Results

The partial derivatives have been calculated as follows:For \( f_{x}(x, y) \): \( \frac{y^2}{(x+y)^2} \).For \( f_{y}(x, y) \): \( \frac{x^2}{(x+y)^2} \). These results confirm the application of the quotient rule, while treating the other variable as constant.

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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 powerful tool in calculus for differentiating a function that is the ratio of two other functions. If you have a function formatted as a fraction, such as \(\frac{u}{v}\), where \(u\) and \(v\) are both functions of a variable, the quotient rule becomes your best friend.
\[\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}\]
This rule tells you to first multiply the denominator by the derivative of the numerator, and subtract the numerator multiplied by the derivative of the denominator. All this is divided by the square of the denominator.
Let's break that down further:
  • \(u\): This is the numerator function in our fraction.
  • \(v\): This is the denominator function.
  • \(\frac{du}{dx}\): This is the derivative of \(u\) with respect to \(x\).
  • \(\frac{dv}{dx}\): This is the derivative of \(v\) with respect to \(x\).
Remember, practicing this rule with examples will solidify your understanding and make you quicker at spotting when to use it.
Function of Two Variables
A function of two variables is one where the output depends on two different inputs. These functions are often represented as \(f(x, y)\), where both \(x\) and \(y\) can vary.
This means each point in such a function is a combination of these two inputs.
In various fields, including economics and physics, these types of functions are frequently used to model relationships that depend on more than one factor.
  • Partial Derivatives: In cases with two variables, you often calculate partial derivatives. This means you're interested in the rate of change of the function with respect to one variable, keeping the other constant.
  • Visualizing: You can think of a function of two variables forming a surface in a three-dimensional space. Each pair \((x, y)\) corresponds to a single \(f(x, y)\) value.
Understanding these functions is crucial because they lay the groundwork for multivariable calculus, which is essential in advanced mathematics and its applications.
Calculus Concepts
Calculus is a branch of mathematics focused on change and motion, equipped with tools for analyzing situations where quantities change. It encompasses a variety of concepts, including derivatives, which are central to understanding change.
Derivatives tell us the rate of change of one quantity with respect to another, like how velocity is the derivative of position with respect to time. They are foundational to calculus.
In our problem, partial derivatives are used to analyze functions of more than one variable. These allow us to explore changes in one direction while holding others constant. Differentiation principles include:
  • Chain Rule: Helps find the derivative of a composite function.
  • Product Rule: Useful for differentiating products of two functions.
  • Quotient Rule: Specifically valuable when dealing with ratios, as explained earlier.
Grasping these concepts is essential as calculus extends far beyond simple equations. It's a key component in fields like engineering, physics, economics, and beyond.

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