Chapter 10: Problem 23
In Exercises, find \(d y / d x\) implicitly. $$ x e^{y}-10 x+3 y=0 $$
Short Answer
Expert verified
The derivative \(d y / d x\) is \((10-e^{y}) / (x e^{y}+3)\)
Step by step solution
01
Differentiate each term
Start by differentiating each term. Remember, when differentiating \(x e^{y}\) a product rule is used and for term \(10x\) and \(3y\) simple differentiation rules are used. Differentiation would result in \(e^{y}+x e^{y} d y / d x-10+3 d y / d x = 0\)
02
Group terms involving \(d y / d x\)
Group terms with \(dy/dx\) on one side and move all others on the opposite side. We get, \(x e^{y} d y / d x+3 d y / d x = 10-e^{y}\)
03
Factor out \(d y / d x\) and solve
Now, factor \(d y / d x\) out of the left hand side and isolate \(d y / d x\) to one side of the equation. We get, \(d y / d x = (10-e^{y}) / (x e^{y}+3)\)
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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 powerful technique used when dealing with equations where the variable to be differentiated is not isolated. This approach can be particularly useful when solving problems with variables that are entangled, such as in the equation provided, where we aim to find \(d y / d x\) from \(x e^{y}-10 x+3 y=0\).
Instead of explicitly solving for \(y\) and then differentiating, we differentiate both sides of the equation with respect to \(x\), treating \(y\) as an implicit function of \(x\). This means applying the differentiation rules while bearing in mind that whenever we differentiate terms involving \(y\), we must also include \(dy/dx\) as part of the result, acknowledging that \(y\) is a function of \(x\) and not just a constant. In our exercise, the term \(x e^{y}\) requires the product rule for differentiation which leads to the term \(e^{y}+x e^{y} dy/dx\).
Thus, the process involves a careful balance — recognize when to apply derivative rules such as the product or chain rule, and remember to multiply by \(dy/dx\) appropriately. It enables us to work directly with the given equation, without the need for rearranging it extensively to find an explicit formula for one variable in terms of the other.
Instead of explicitly solving for \(y\) and then differentiating, we differentiate both sides of the equation with respect to \(x\), treating \(y\) as an implicit function of \(x\). This means applying the differentiation rules while bearing in mind that whenever we differentiate terms involving \(y\), we must also include \(dy/dx\) as part of the result, acknowledging that \(y\) is a function of \(x\) and not just a constant. In our exercise, the term \(x e^{y}\) requires the product rule for differentiation which leads to the term \(e^{y}+x e^{y} dy/dx\).
Thus, the process involves a careful balance — recognize when to apply derivative rules such as the product or chain rule, and remember to multiply by \(dy/dx\) appropriately. It enables us to work directly with the given equation, without the need for rearranging it extensively to find an explicit formula for one variable in terms of the other.
Product Rule in Differentiation
The product rule in differentiation is a fundamental rule in differential calculus used when differentiating products of two or more functions. This rule states that for two differentiable functions \(u\) and \(v\), the derivative of their product \(uv\) is given by \(u'v + uv'\), where \(u'\) and \(v'\) are the derivatives of \(u\) and \(v\), respectively.
In the context of our problem, we applied the product rule to differentiate the term \(x e^{y}\), where \(x\) and \(e^{y}\) are the functions we need to differentiate. The derivative of the first function, \(x\), is \(1\), and the derivative of the second function, \(e^{y}\), is \(e^{y} dy/dx\) since \(e^{y}\) is a function of \(y\), which in turn is a function of \(x\). Thus, following the product rule, the derivative of \(x e^{y}\) is \(e^{y} + x e^{y} dy/dx\).
Understanding this rule is crucial because it allows us to handle terms where two functions are multiplied together, which is a common situation in many calculus problems. Since calculus often deals with rates of change and growth, knowing how to differentiate products accurately is essential for modeling and solving real-world scenarios.
In the context of our problem, we applied the product rule to differentiate the term \(x e^{y}\), where \(x\) and \(e^{y}\) are the functions we need to differentiate. The derivative of the first function, \(x\), is \(1\), and the derivative of the second function, \(e^{y}\), is \(e^{y} dy/dx\) since \(e^{y}\) is a function of \(y\), which in turn is a function of \(x\). Thus, following the product rule, the derivative of \(x e^{y}\) is \(e^{y} + x e^{y} dy/dx\).
Understanding this rule is crucial because it allows us to handle terms where two functions are multiplied together, which is a common situation in many calculus problems. Since calculus often deals with rates of change and growth, knowing how to differentiate products accurately is essential for modeling and solving real-world scenarios.
Separating Variables
Separating variables is a method used to simplify the process of solving differential equations. It involves rearranging an equation so that each variable and its corresponding differential are on opposite sides of the equation.
In the context of our implicit differentiation problem, once we have applied the product rule and differentiated all terms, we group the terms involving \(dy/dx\) on one side of the equality and move all other terms to the opposite side. This step enables us to factor \(dy/dx\) from the collected terms, setting the stage to solve for \(dy/dx\) by itself.
To illustrate, consider the way we grouped terms in our problem: \(e^{y}+x e^{y} dy/dx-10+3 dy/dx = 0\) is manipulated into \(x e^{y} dy/dx+3 dy/dx = 10-e^{y}\). After factoring out \(dy/dx\), we have \(dy/dx(e^{y}x+3) = 10 - e^{y}\). The final step in separating variables is to isolate \(dy/dx\) by dividing both sides by \(e^{y}x+3\).
This results in \(dy/dx = (10-e^{y})/(xe^{y}+3)\), separating the variables and providing the derivative of \(y\) with respect to \(x\). Separation of variables transforms a potentially complex differential equation into a simpler form that is easier to solve, making it a crucial technique in calculus.
In the context of our implicit differentiation problem, once we have applied the product rule and differentiated all terms, we group the terms involving \(dy/dx\) on one side of the equality and move all other terms to the opposite side. This step enables us to factor \(dy/dx\) from the collected terms, setting the stage to solve for \(dy/dx\) by itself.
To illustrate, consider the way we grouped terms in our problem: \(e^{y}+x e^{y} dy/dx-10+3 dy/dx = 0\) is manipulated into \(x e^{y} dy/dx+3 dy/dx = 10-e^{y}\). After factoring out \(dy/dx\), we have \(dy/dx(e^{y}x+3) = 10 - e^{y}\). The final step in separating variables is to isolate \(dy/dx\) by dividing both sides by \(e^{y}x+3\).
This results in \(dy/dx = (10-e^{y})/(xe^{y}+3)\), separating the variables and providing the derivative of \(y\) with respect to \(x\). Separation of variables transforms a potentially complex differential equation into a simpler form that is easier to solve, making it a crucial technique in calculus.
