Chapter 3: Problem 26
For each equation, find \(d y / d x\) evaluated at the given values. $$ y^{2}+y+1=x \text { at } x=1, y=-1 $$
Short Answer
Expert verified
\( \frac{dy}{dx} = -1 \) when \( x = 1 \) and \( y = -1 \).
Step by step solution
01
Express the Given Equation
Start with the given equation: \( y^2 + y + 1 = x \) and recognize that it is implicitly defining \( y \) as a function of \( x \). Our job is to find \( \frac{dy}{dx} \).
02
Differentiate Both Sides Implicitly
Use implicit differentiation with respect to \( x \).- Differentiate the left side: \( \frac{d}{dx}(y^2 + y + 1) = 2y \frac{dy}{dx} + \frac{dy}{dx} + 0 \).- Differentiate the right side: \( \frac{d}{dx}(x) = 1 \).Set the derivatives equal: \( 2y \frac{dy}{dx} + \frac{dy}{dx} = 1 \).
03
Factor \( \frac{dy}{dx} \)
Factor \( \frac{dy}{dx} \) out of the left side:\( \frac{dy}{dx} (2y + 1) = 1 \).
04
Solve for \( \frac{dy}{dx} \)
Isolate \( \frac{dy}{dx} \) by dividing both sides by \( 2y + 1 \):\[ \frac{dy}{dx} = \frac{1}{2y + 1} \].
05
Substitute the Given Values
Substitute \( y = -1 \) into the expression for \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = \frac{1}{2(-1) + 1} = \frac{1}{-2 + 1} = \frac{1}{-1} = -1 \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Implicit Differentiation
Implicit differentiation is a method used when dealing with equations where one variable is not isolated on one side. For equations like \( y^2 + y + 1 = x \), instead of \( y \) being expressed solely in terms of \( x \), both variables are tangled together. The goal is to find the derivative of \( y \) with respect to \( x \), written as \( \frac{dy}{dx} \).
To achieve this, we differentiate each part of the equation relative to \( x \). Even though \( y \) appears in the equation, we treat it as a function of \( x \), meaning each instance of \( y \) gets differentiated as \( \frac{dy}{dx} \).
This approach allows us to dissect complex implicit equations, finding the rate at which \( y \) changes concerning \( x \) without ever needing an explicit form of \( y \). This is particularly useful when it's challenging or impossible to solve for \( y \) directly.
To achieve this, we differentiate each part of the equation relative to \( x \). Even though \( y \) appears in the equation, we treat it as a function of \( x \), meaning each instance of \( y \) gets differentiated as \( \frac{dy}{dx} \).
This approach allows us to dissect complex implicit equations, finding the rate at which \( y \) changes concerning \( x \) without ever needing an explicit form of \( y \). This is particularly useful when it's challenging or impossible to solve for \( y \) directly.
Derivative Evaluation
Derivative evaluation is the process through which we calculate the specific value of a derivative, such as \( \frac{dy}{dx} \), at a given point. Once we have isolated the expression, like we did here: \[ \frac{dy}{dx} = \frac{1}{2y + 1} \], we can substitute specific values.
In this exercise, we have the point \( x = 1 \), \( y = -1 \). By substituting \( y = -1 \) into the previously derived expression, \[ \frac{dy}{dx} = \frac{1}{2(-1) + 1} = \frac{1}{-1} \].The derivative evaluation at the point reveals the specific rate at which \( y \) is changing at that point, in this case, \( -1 \).
Evaluating derivatives at particular values allows us to understand the behavior of the function locally, providing insights into how the function behaves in a specific context.
In this exercise, we have the point \( x = 1 \), \( y = -1 \). By substituting \( y = -1 \) into the previously derived expression, \[ \frac{dy}{dx} = \frac{1}{2(-1) + 1} = \frac{1}{-1} \].The derivative evaluation at the point reveals the specific rate at which \( y \) is changing at that point, in this case, \( -1 \).
Evaluating derivatives at particular values allows us to understand the behavior of the function locally, providing insights into how the function behaves in a specific context.
Function of a Variable
Understanding the concept of a function of a variable is key in calculus, especially when dealing with implicit functions. Here, \( y \) is considered a function of \( x \), even though it's not isolated. This means each value of \( x \) potentially corresponds to specific values of \( y \), defining a relationship.
Even if \( y \) doesn’t appear explicitly as \( y = f(x) \), we treat it as such for differentiation purposes. Recognizing this helps bridge the gap between implicit and explicit functions, ensuring that techniques like implicit differentiation still apply.
Functions of a variable highlight the interdependence of these quantities and denote that changes in one affect the other, which is why we seek \( \frac{dy}{dx} \), the rate of change of \( y \) with respect to \( x \). Understanding this interrelationship is central to tackling more advanced calculus problems, as it offers a broader view of dynamic change across a system.
Even if \( y \) doesn’t appear explicitly as \( y = f(x) \), we treat it as such for differentiation purposes. Recognizing this helps bridge the gap between implicit and explicit functions, ensuring that techniques like implicit differentiation still apply.
Functions of a variable highlight the interdependence of these quantities and denote that changes in one affect the other, which is why we seek \( \frac{dy}{dx} \), the rate of change of \( y \) with respect to \( x \). Understanding this interrelationship is central to tackling more advanced calculus problems, as it offers a broader view of dynamic change across a system.