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\(7-30\) Find the first partial derivatives of the function. $$f(x, y)=\frac{x}{y}$$

Short Answer

Expert verified
\(\frac{\partial f}{\partial x} = \frac{1}{y}\); \(\frac{\partial f}{\partial y} = -\frac{x}{y^2}\).

Step by step solution

01

Understanding Partial Derivatives

Partial derivatives involve taking the derivative of a function with respect to one variable while keeping the other variables constant. For the function \(f(x, y)=\frac{x}{y}\), we will find \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\).
02

Finding \(\frac{\partial f}{\partial x}\)

To find the partial derivative of \(f(x, y) = \frac{x}{y}\) with respect to \(x\), treat \(y\) as a constant. The derivative of \(\frac{x}{y}\) with respect to \(x\) is \(\frac{1}{y}\). Thus, \(\frac{\partial f}{\partial x} = \frac{1}{y}\).
03

Finding \(\frac{\partial f}{\partial y}\)

To find the partial derivative with respect to \(y\), treat \(x\) as a constant. Rewrite \(f(x, y) = \frac{x}{y} = x \cdot y^{-1}\). Using the power rule, the derivative of \(x \cdot y^{-1}\) with respect to \(y\) is \(-x \cdot y^{-2}\), or \(-\frac{x}{y^2}\). Thus, \(\frac{\partial f}{\partial y} = -\frac{x}{y^2}\).

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Key Concepts

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

Functions of Two Variables
When we talk about functions of two variables, we're essentially dealing with functions like \(f(x, y)\) that depend on two different inputs, \(x\) and \(y\). A simple way to think of it is as an extension of single variable functions into a two-dimensional space.
Understanding this concept is crucial in fields like calculus and physics where multidimensional analysis is necessary. For example, in our exercise where \(f(x, y) = \frac{x}{y}\), the function describes a relationship between \(x\) and \(y\) that affects the outcome or result. In visual terms, that would appear as a surface in three-dimensional space where each \((x, y)\) pair gives you a particular value of \(f\).
To map out or analyze this function effectively, we often take derivatives in terms of each variable separately, which leads us to the concept of partial derivatives.
Calculus
Calculus is a branch of mathematics that deals with continuous change. It introduces concepts like the derivative, which describes the rate of change of a function, and the integral, which calculates the accumulation of quantities. When dealing with functions of two variables, calculus helps us understand how these changes occur with respect to each variable.
By finding the partial derivative \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) of our function \(f(x, y) = \frac{x}{y}\), we are applying these principles.
  • The derivative \(\frac{\partial f}{\partial x}\) shows how \(f\) changes as \(x\) changes, while keeping \(y\) constant.
  • Similarly, \(\frac{\partial f}{\partial y}\) illustrates how \(f\) changes as \(y\) varies, with \(x\) held constant.
These partial derivatives provide insights into the function's behavior at any given point and can be used in optimization problems, engineering designs, and much more.
Power Rule
The power rule is one of the foundational rules in calculus for finding derivatives, allowing us to differentiate expressions of the form \(x^n\), where \(n\) is any real number. It states that the derivative of \(x^n\) is \(nx^{n-1}\). This makes it easy to handle polynomials or any expression that can be rewritten in this form.
In the context of partial derivatives, we apply the power rule when variables are raised to a power, such as when computing \(\frac{\partial f}{\partial y}\) for \(f(x, y) = \frac{x}{y}\).
Here, the function is rewritten as \(x \cdot y^{-1}\). Using the power rule on \(y^{-1}\), the derivative becomes \(-y^{-2}\) and the partial derivative with respect to \(y\) is \(-\frac{x}{y^2}\). Thus, the power rule provides a straightforward method to tackle derivatives involving exponents, whether they are positive, negative, or fractionary.

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Most popular questions from this chapter

(a) Find the maximum value of $$ f\left(x_{1}, x_{2}, \ldots, x_{n}\right)=\sqrt[n]{x_{1} x_{2} \cdots x_{n}} $$ given that \(x_{1}, x_{2}, \ldots, x_{n}\) are positive numbers and \(x_{1}+x_{2}+\cdots+x_{n}=c,\) where \(c\) is a constant. (b) Deduce from part (a) that if \(x_{1}, x_{2}, \ldots, x_{n}\) are positive numbers, then $$ \sqrt[n]{x_{1} x_{2} \cdots x_{n}} \leqslant \frac{x_{1}+x_{2}+\cdots+x_{n}}{n} $$ This inequality says that the geometric mean of \(n\) numbers is no larger than the arithmetic mean of the numbers. Under what circumstances are these two means equal?

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. $$ f(x, y)=x^{2}+y^{2} ; \quad x y=1 $$

Assume that all the given functions have continuous second-order partial derivatives. Show that any function of the form $$z=f(x+a t)+g(x-a t)$$ is a solution of the wave equation $$\frac{\partial^{2} z}{\partial t^{2}}=a^{2} \frac{\partial^{2} z}{\partial x^{2}}$$ \([\)Hint: Let \(u=x+a t, v=x-a t .]\)

The radius of a right circular cone is increasing at a rate of 1.8 in/s while its height is decreasing at a rate of 2.5 in/s. At what rate is the volume of the cone changing when the radius is 120 in. and the height is 140 in.?

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. $$f(x, y)=e^{x} \cos y$$

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