Chapter 15: Problem 38
Implicit differentiation Use Theorem 15.9 to evaluate dy/dx. Assume each equation implicitly defines y as a differentiable function of \(x\). $$y e^{x y}-2=0$$
Short Answer
Expert verified
Question: Find the derivative of the implicitly defined function \(y e^{x y}-2=0\) with respect to x.
Answer: The derivative of the implicitly defined function with respect to x is given by \(\frac{dy}{dx} = \frac{- y' - y}{x}\).
Step by step solution
01
Differentiate both sides with respect to x
Differentiating both sides of the given equation with respect to x, we have:
$$\frac{d}{dx}(ye^{xy}-2) = \frac{d}{dx}(0)$$
02
Apply Product Rule and Chain Rule
Recall the product rule: \((uv)' = u'v + uv'\). We apply the product rule to \(ye^{xy}\) and the chain rule for the exponential function.
Differentiating with respect to x, we get:
$$\frac{d}{dx}(y) \cdot e^{xy} + y \cdot \frac{d}{dx}(e^{xy}) - \frac{d}{dx}(2) = 0$$
03
Differentiate e^(xy) and Simplify
Now, we need to differentiate \(e^{xy}\) using the chain rule. Let \(u = xy\), then:
$$\frac{d}{dx}(e^{xy}) = \frac{d}{dx}(e^u) \cdot \frac{d}{dx}(u)$$
Differentiate \(e^u\) with respect to \(u\) and \(u\) with respect to \(x\):
$$\frac{d}{dx}(e^{xy}) = e^u \cdot (y + x \frac{dy}{dx}) = e^{xy}(y + x \frac{dy}{dx})$$
Substitute this result back into our previous equation:
$$y'e^{xy} + y e^{xy}(y + x\frac{dy}{dx}) - 0 = 0$$
04
Solve for dy/dx
Pick out terms containing \(\frac{dy}{dx}\) on the left side and move the other terms to the other side:
$$e^{xy}(y + x\frac{dy}{dx}) = - y'e^{xy} + 0$$
Now isolate \(\frac{dy}{dx}\) on one side:
$$\frac{dy}{dx}(xe^{xy}) = - y'e^{xy} - e^{xy}y$$
Finally, divide both sides by \(xe^{xy}\):
$$\frac{dy}{dx} = \frac{- y'e^{xy} - e^{xy}y}{xe^{xy}}$$
05
Simplify the Answer
Now, simplify the result by canceling the \(e^{xy}\) terms:
$$\frac{dy}{dx} = \frac{- y' - y}{x}$$
So, the derivative of y with respect to x is given by:
$$\frac{dy}{dx} = \frac{- y' - y}{x}$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Product Rule
The Product Rule is a fundamental concept in calculus that is especially useful when differentiating functions that are the product of two other functions. In simple terms, if you have a function that is the product of two functions, say \( u(x) \) and \( v(x) \), the derivative of their product is given by:
In the original exercise, applying the Product Rule to the term \( ye^{xy} \) involves taking the derivative of \( y \) times \( e^{xy} \) plus \( y \) times the derivative of \( e^{xy} \). This helps break down complex expressions into manageable parts for differentiation.
- \((uv)' = u'v + uv'\)
In the original exercise, applying the Product Rule to the term \( ye^{xy} \) involves taking the derivative of \( y \) times \( e^{xy} \) plus \( y \) times the derivative of \( e^{xy} \). This helps break down complex expressions into manageable parts for differentiation.
Chain Rule
The Chain Rule is a powerful tool in calculus used to differentiate composite functions. A composite function is one where a function is applied to the result of another function. When you have such a function in the form \( f(g(x)) \), the Chain Rule states the derivative is:
In the context of implicit differentiation, the Chain Rule is crucial for finding the derivative of \( e^{xy} \). You treat \( xy \) as the inner function and \( e^u \) (where \( u = xy \)) as the outer function. Differentiating accordingly helps in progressing towards finding \( \frac{dy}{dx} \).
- \( f'(g(x)) \, g'(x) \)
In the context of implicit differentiation, the Chain Rule is crucial for finding the derivative of \( e^{xy} \). You treat \( xy \) as the inner function and \( e^u \) (where \( u = xy \)) as the outer function. Differentiating accordingly helps in progressing towards finding \( \frac{dy}{dx} \).
Differentiable Function
A Differentiable Function is a function that can be differentiated everywhere within its domain. This implies that the function has a derivative at every point in its space. For implicit differentiation, it is assumed that the dependent variable is a differentiable function of the independent variable.
In the given exercise, we assume that \( y \) is implicitly defined as a differentiable function of \( x \). This means we can apply differentiation techniques like the Product Rule and Chain Rule to find \( \frac{dy}{dx} \). Keeping in view that the function needs to be smooth and continuous everywhere, ensures that our differentiation methods yield meaningful outputs.
In the given exercise, we assume that \( y \) is implicitly defined as a differentiable function of \( x \). This means we can apply differentiation techniques like the Product Rule and Chain Rule to find \( \frac{dy}{dx} \). Keeping in view that the function needs to be smooth and continuous everywhere, ensures that our differentiation methods yield meaningful outputs.
Exponential Function
An Exponential Function is a mathematical function of the form \( e^x \), where \( e \) is Euler's number, approximately 2.718. Exponential functions have unique properties, especially their characteristic of their derivative being the same as the original function.
In the provided exercise, the exponential function \( e^{xy} \) presents a scenario where you use the Chain Rule to handle differentiation because the exponent itself, \( xy \), is a function needing further differentiation. This highlights the intertwined nature of exponential and chain rules in complex calculus problems.
- The derivative of \( e^u \) is \( e^u \cdot u' \).
In the provided exercise, the exponential function \( e^{xy} \) presents a scenario where you use the Chain Rule to handle differentiation because the exponent itself, \( xy \), is a function needing further differentiation. This highlights the intertwined nature of exponential and chain rules in complex calculus problems.