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Chapter 7: Problem 12

Find \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) for each of the following functions. $$ f(x, y)=\sqrt{x^{2}+y^{2}} $$

Short Answer

Expert verified
\frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}}; \frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}}

Step by step solution

01

Understand the function

The given function is: \[ f(x, y) = \sqrt{x^2 + y^2} \]We need to find the partial derivatives with respect to both variables, x and y.
02

Find the partial derivative with respect to x

To find \( \frac{\partial f}{\partial x} \), treat y as a constant. The general rule for the derivative of a square root function \( f(u) = \sqrt{u} \) is \( f'(u) = \frac{1}{2\sqrt{u}} \).Here, let \( u = x^2 + y^2 \). The derivative of \( \sqrt{u} \) with respect to \( u \) is:\[ \frac{d}{du} \sqrt{u} = \frac{1}{2\sqrt{u}} \]Now apply the chain rule. Since \( u = x^2 + y^2 \):\[ \frac{\partial f}{\partial x} = \frac{d}{du} \sqrt{x^2 + y^2} \cdot \frac{\partial (x^2 + y^2)}{\partial x} \]Compute \( \frac{\partial (x^2 + y^2)}{\partial x} \):\[ \frac{d}{dx} (x^2 + y^2) = 2x \]Putting it all together:\[ \frac{\partial f}{\partial x} = \frac{1}{2\sqrt{x^2 + y^2}} \cdot 2x = \frac{x}{\sqrt{x^2 + y^2}} \]
03

Find the partial derivative with respect to y

To find \( \frac{\partial f}{\partial y} \), treat x as a constant. Using the same steps as above, we start with:\[ \frac{d}{du} \sqrt{u} = \frac{1}{2\sqrt{u}} \]Now apply the chain rule. Since \( u = x^2 + y^2 \):\[ \frac{\partial f}{\partial y} = \frac{d}{du} \sqrt{x^2 + y^2} \cdot \frac{\partial (x^2 + y^2)}{\partial y} \]Compute \( \frac{\partial (x^2 + y^2)}{\partial y} \):\[ \frac{d}{dy} (x^2 + y^2) = 2y \]Putting it all together:\[ \frac{\partial f}{\partial y} = \frac{1}{2\sqrt{x^2 + y^2}} \cdot 2y = \frac{y}{\sqrt{x^2 + y^2}} \]

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Key Concepts

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

Chain Rule
The chain rule is a foundational concept in calculus. It is used to find the derivative of a composite function. In simple terms, if you have a function inside another function, you use the chain rule to differentiate it.
Suppose you have a function \ f(g(x)). The chain rule states that the derivative of that composite function with respect to x is: \ \ \( \frac{d}{dx} f(g(x)) = f'(g(x)) \times g'(x) \).
This rule helps break down complex functions into simpler parts for differentiation. In our exercise, we use the chain rule to differentiate \ \ \( f(x, y) = \sqrt{x^{2}+y^{2}} \) with respect to both x and y. First, we treat the inside function, \( u = x^2 + y^2 \), then apply the chain rule to differentiate with respect to x and y separately. This allows us to manage the complexity step-by-step.
Square Root Function
The square root function is a mathematical function that returns the non-negative root of a number. Its general form is \( f(u) = \sqrt{u} \).
When differentiating the square root function, we use the general derivative rule for square roots, which is: \ \ \( \frac{d}{du} \sqrt{u} = \frac{1}{2\sqrt{u}} \).
In our exercise, we handle the square root of the sum of squares function \ \( f(x, y) = \sqrt{x^2 + y^2} \). By recognizing \( x^2 + y^2 \) as the inside function, we simplify the derivative process. This step is crucial when using the chain rule to obtain the partial derivatives with respect to x and y.
Understanding how to differentiate square root functions helps in a variety of calculus problems where roots appear within more complex expressions.
Calculus
Calculus is a branch of mathematics that studies change. There are two main areas: differential calculus and integral calculus. Differential calculus focuses on finding the rate of change (the derivative) of functions. Integral calculus, on the other hand, deals with finding the area under a curve (the integral).
In this exercise, we are specifically dealing with differential calculus. We find the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) of the function \( f(x, y) = \sqrt{x^2 + y^2} \).
Partial derivatives measure how a function changes as only one of its inputs changes while the other inputs remain constant. These concepts are useful for studying functions of multiple variables, which are common in physics, engineering, and economics.
By mastering partial derivatives and the chain rule, you gain powerful tools to analyze and solve complex, real-world problems.

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Most popular questions from this chapter

Find the two positive numbers whose product is 25 and whose sum is as small as possible.

Both first partial derivatives of the function \(f(x, y)\) are zero at the given points. Use the second-derivative test to determine the nature of \(f(x, y)\) at each of these points. If the second-derivative test is inconclusive, so state. $$ f(x, y)=\frac{1}{x}+\frac{1}{y}+x y ;(1,1) $$

In a certain suburban community, commuters have the choice of getting into the city by bus or train. The demand for these modes of transportation varies with their cost. Let \(f\left(p_{1}, p_{2}\right)\) be the number of people who will take the bus when \(p_{1}\) is the price of the bus ride and \(p_{2}\) is the price of the train ride. For example, if \(f(4.50,6)=7000\), then 7000 commuters will take the bus when the price of a bus ticket is $$\$ 4.50$$ and the price of a train ticket is $$\$ 6.00$$. Explain why \(\frac{\partial f}{\partial p_{1}}<0\) and \(\frac{\partial f}{\partial p_{2}}>0\).

Let \(p_{1}\) be the average price of MP3 players, \(p_{2}\) the average price of audio files, \(f\left(p_{1}, p_{2}\right)\) the demand for MP3 players, and \(g\left(p_{1}, p_{2}\right)\) the demand for audio files. Explain why \(\frac{\partial f}{\partial p_{2}}<0\) and \(\frac{\partial g}{\partial p_{1}}<0 .\)

The function \(f(x, y)=\frac{1}{2} x^{2}+2 x y+3 y^{2}-x+2 y\) has a minimum at some point \((x, y)\). Find the values of \(x\) and \(y\) where this minimum occurs.
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