/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 Assuming that the equations in E... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Assuming that the equations in Exercises \(25-30\) define \(y\) as a differentiable function of \(x,\) use Theorem 8 to find the value of \(d y / d x\) at the given point. $$ x y+y^{2}-3 x-3=0, \quad(-1,1) $$

Short Answer

Expert verified
At the point (-1,1), \( \frac{dy}{dx} = 2 \).

Step by step solution

01

Differentiate Implicitly

Given the equation \( x y + y^2 - 3x - 3 = 0 \), we have to find \( \frac{dy}{dx} \) using implicit differentiation. Differentiating both sides with respect to \( x \), we use the product rule for \( xy \):\[ \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) - \frac{d}{dx}(3x) - \frac{d}{dx}(3) = 0 \] This gives: \[ y + x \frac{dy}{dx} + 2y \frac{dy}{dx} - 3 = 0 \]
02

Group Terms with Dy/Dx

Combine terms involving \( \frac{dy}{dx} \) on one side of the equation:\[ x \frac{dy}{dx} + 2y \frac{dy}{dx} = 3 - y \]
03

Factor Out Dy/Dx

Factor \( \frac{dy}{dx} \) from the left side of the equation:\[ (x + 2y) \frac{dy}{dx} = 3 - y \]
04

Solve for Dy/Dx

Solve the equation for \( \frac{dy}{dx} \) by dividing both sides by \( x + 2y \):\[ \frac{dy}{dx} = \frac{3 - y}{x + 2y} \]
05

Substitute the Point (-1, 1)

Substitute the point \((-1, 1)\) into the derived formula for \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = \frac{3 - 1}{-1 + 2(1)} = \frac{2}{1} = 2 \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus that allows us to find the rate at which a function changes at any given point.
In simpler terms, it helps us determine how a variable changes in relation to another. This is particularly useful when dealing with complex equations that cannot be solved algebraically.
In the context of implicit differentiation, which is the method used in the original exercise, we handle equations where the dependent variable isn't isolated. Implicit differentiation involves differentiating both sides of an equation with respect to an independent variable, while treating the other variable as a function of the independent variable. This technique often requires applying the product rule and the chain rule, as seen in the differentiation of the equation \( x y + y^2 - 3x - 3 = 0 \).
The goal is to express derivative values even when they don't appear explicitly in an equation.
Derivatives
In calculus, derivatives are essentially the output of differentiation.
They provide us with the slope of the tangent line to a curve at any given point, signifying how the function's value changes as its input changes. Derivatives are represented by the notation \( \frac{dy}{dx} \), which can be read as "the derivative of \( y \) with respect to \( x \)."Understanding the calculation of derivatives is crucial. It gives insight into the behavior of functions, such as identifying maxima, minima, or points of inflection. In the original exercise, the derived expression \( \frac{dy}{dx} = \frac{3 - y}{x + 2y} \) is crucial as it captures the change of \( y \) in relation to \( x \), allowing for the calculation of the specific derivative at the point \((-1, 1)\).
Finding the derivative at specific points like this helps in analyzing the behavior of the curve at that exact location.
Calculus
Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series.
It forms the foundation for advanced mathematics and applied sciences because it provides powerful tools to model and solve problems involving change.
Calculus allows scientists and engineers to understand and describe the world in dynamic ways, ranging from the motion of planets to the changes in a stock market. The use of implicit differentiation in calculus exemplifies how versatile and important these concepts are.
When students learn calculus, they're not just memorizing formulas—they're developing a toolkit that will empower them to analyze real-world scenarios. Whether you're examining the trajectory of a projectile or the rate of a chemical reaction, calculus helps break down these complex processes into manageable parts.
In the exercise provided, calculus techniques like differentiation offer a pathway to understanding non-linear relationships between variables, showcasing the depth and utility of these mathematical concepts.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Can you conclude anything about \(f(a, b)\) if \(f\) and its first and second partial derivatives are continuous throughout a disk centered at the critical point \((a, b)\) and \(f_{x x}(a, b)\) and \(f_{y y}(a, b)\) differ in sign? Give reasons for your answer.

Suppose that \(f(x, y, z, w)=0\) and \(g(x, y, z, w)=0\) determine \(z\) and \(w\) as differentiable functions of the independent variables \(x\) and \(y,\) and suppose that $$ \frac{\partial f}{\partial z} \frac{\partial g}{\partial w}-\frac{\partial f}{\partial w} \frac{\partial g}{\partial z} \neq 0 $$ Show that $$ \left(\frac{\partial z}{\partial x}\right)_{y}=-\frac{\frac{\partial f}{\partial x} \frac{\partial g}{\partial w}-\frac{\partial f}{\partial w} \frac{\partial g}{\partial x}}{\frac{\partial f}{\partial z} \frac{\partial g}{\partial w}-\frac{\partial f}{\partial w} \frac{\partial g}{\partial z}} $$ and $$ \left(\frac{\partial w}{\partial y}\right)_{x}=-\frac{\frac{\partial f}{\partial z} \frac{\partial g}{\partial y}-\frac{\partial f}{\partial y} \frac{\partial g}{\partial z}}{\frac{\partial f}{\partial z} \frac{\partial g}{\partial w}-\frac{\partial f}{\partial w} \frac{\partial g}{\partial z}} $$

If \(z=x+f(u),\) where \(u=x y,\) show that $$ x \frac{\partial z}{\partial x}-y \frac{\partial z}{\partial y}=x $$

Least squares and regression lines When we try to fit a line \(y=m x+b\) to a set of numerical data points \(\left(x_{1}, y_{1}\right)\) \(\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right),\) we usually choose the line that minimizes the sum of the squares of the vertical distances from the points to the line. In theory, this means finding the values of \(m\) and \(b\) that minimize the value of the function $$w=\left(m x_{1}+b-y_{1}\right)^{2}+\cdots+\left(m x_{n}+b-y_{n}\right)^{2}$$ (See the accompanying figure.) Show that the values of \(m\) and \(b\) that do this are $$ \begin{array}{l}{m=\frac{\left(\sum x_{k}\right)\left(\sum y_{k}\right)-n \sum x_{k} y_{k}}{\left(\sum x_{k}\right)^{2}-n \sum x_{k}^{2}}} \\\ {b=\frac{1}{n}\left(\sum y_{k}-m \sum x_{k}\right)}\end{array} $$ with all sums running from \(k=1\) to \(k=n .\) Many scientific calculators have these formulas built in, enabling you to find \(m\) and \(b\) with only a few keystrokes after you have entered the data. The line \(y=m x+b\) determined by these values of \(m\) and \(b\) is called the least squares line, regression line, or trend line for the data under study. Finding a least squares line lets you 1\. summarize data with a simple expression, 2\. predict values of \(y\) for other, experimentally untried values of \(x\) 3\. handle data analytically.

In Exercises \(63-66,\) use the limit definition of partial derivative to compute the partial derivatives of the functions at the specified points. $$f(x, y)=1-x+y-3 x^{2} y, \quad \frac{\partial f}{\partial x} \quad \text { and } \quad \frac{\partial f}{\partial y} \quad \text { at }(1,2)$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.