Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 3: Problem 46
Give an example of: True or false? Explain your answer: If \(y\) satisfies the equation \(y^{2}+x y-1=0,\) then \(d y / d x\) exists every. where.
Short Answer
Expert verified
False. \( \frac{dy}{dx} \) does not exist where \( 2y + x = 0 \).
Step by step solution
01
Identify the Given Equation
The given equation is a quadratic equation in terms of the variable \( y \): \[ y^2 + xy - 1 = 0. \] It is implicit with respect to \( x \) and \( y \).
02
Analyze Implicit Differentiation
To find \( \frac{dy}{dx} \), we differentiate the given equation implicitly with respect to \( x \). This gives us:\[ \frac{d}{dx}(y^2) + \frac{d}{dx}(xy) - \frac{d}{dx}(1) = 0. \]
03
Apply the Derivatives
Apply the chain rule and product rule to the derivatives:- \( \frac{d}{dx}(y^2) = 2y \frac{dy}{dx} \),- \( \frac{d}{dx}(xy) = y + x \frac{dy}{dx} \),- \( \frac{d}{dx}(1) = 0 \).Substitute these into the equation:\[ 2y \frac{dy}{dx} + y + x \frac{dy}{dx} = 0. \]
04
Solve for the Derivative
Combine the terms involving \( \frac{dy}{dx} \):\[ (2y + x) \frac{dy}{dx} + y = 0. \]Rearrange to solve for \( \frac{dy}{dx} \):\[ (2y + x) \frac{dy}{dx} = -y. \]\[ \frac{dy}{dx} = \frac{-y}{2y + x}. \]
05
Identify the Domain of \( \frac{dy}{dx} \)
The expression for \( \frac{dy}{dx} \) involves division by \( 2y + x \). For \( \frac{dy}{dx} \) to exist, the denominator must not be zero. Thus, \( 2y + x eq 0 \).
06
Conclusion about the Derivative
The derivative \( \frac{dy}{dx} \) does not exist for values of \( x \) and \( y \) that satisfy \( 2y + x = 0 \). Therefore, \( \frac{dy}{dx} \) does not exist everywhere. The statement in the question is false.
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 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equation
A quadratic equation is a polynomial equation of degree two. In standard form, it looks like this: \[ ax^2 + bx + c = 0 \]. Here, the equation involves the variable \( y \), making it a bit different. In the given problem, the equation \( y^2 + xy - 1 = 0 \) is a quadratic equation in terms of \( y \), not \( x \).
To solve or analyze such equations, we often use techniques like factoring, completing the square, or the quadratic formula.
To solve or analyze such equations, we often use techniques like factoring, completing the square, or the quadratic formula.
- Quadratic equations can have two, one, or no real solutions.
- The solutions can be found where the graph of the equation intersects the x-axis when graphed.
Chain Rule
The chain rule is a fundamental technique in calculus used to differentiate compositions of functions. When a variable is nested inside a function, you apply the chain rule to find the derivative.
In the given exercise, the term \( y^2 \) represents a function nested in another, since \( y \) itself is implicitly a function of \( x \). To differentiate \( y^2 \) with respect to \( x \), we need the chain rule, as follows:
In the given exercise, the term \( y^2 \) represents a function nested in another, since \( y \) itself is implicitly a function of \( x \). To differentiate \( y^2 \) with respect to \( x \), we need the chain rule, as follows:
- Differentiating \( y^2 \) directly yields \( 2y \).
- Then multiply by the derivative of \( y \) with respect to \( x \), which is \( \frac{dy}{dx} \).
Product Rule
The product rule is another essential differentiation technique that applies when taking the derivative of a product of two functions.
When we look at the term \( xy \) in the equation, we identify this as a product of \( x \) and \( y \). To differentiate this with respect to \( x \), we utilize the product rule:
When we look at the term \( xy \) in the equation, we identify this as a product of \( x \) and \( y \). To differentiate this with respect to \( x \), we utilize the product rule:
- First, differentiate \( x \) holding \( y \) constant, giving us \( y \).
- Then, differentiate \( y \) holding \( x \) constant, giving \( x \frac{dy}{dx} \).
- Sum these results to find \( \frac{d}{dx}(xy) = y + x \frac{dy}{dx} \).
Existence of Derivative
The existence of a derivative at a point tells us whether the function is differentiable there; it's essentially about the slope's defined nature. In our exercise, this checks whether \( \frac{dy}{dx} \) exists for the given set of \( x \) and \( y \).
Our derived expression for the derivative is \( \frac{dy}{dx} = \frac{-y}{2y + x} \). For this derivative to exist, the denominator \( 2y + x \) must not be zero, otherwise, we would divide by zero, which is undefined.
Our derived expression for the derivative is \( \frac{dy}{dx} = \frac{-y}{2y + x} \). For this derivative to exist, the denominator \( 2y + x \) must not be zero, otherwise, we would divide by zero, which is undefined.
- If \( 2y + x = 0 \), then \( \frac{dy}{dx} \) becomes undefined.
- Thus, the derivative \( \frac{dy}{dx} \) doesn't exist everywhere, particularly when \( 2y + x = 0 \).
