Problem 61 - Essential Calculus Early Transcendentals

Question

$$ y^{\prime} \text { if } e^{x^{2} y}=x+y $$

Step-by-Step Solution

Answer

\( y' = \frac{1 - 2xye^{x^2 y}}{e^{x^2 y} x^2 - 1} \)

Step 1: Implicit Differentiation

Given the equation \( e^{x^2 y} = x + y \), we need to find \( y' \). We use implicit differentiation with respect to \( x \) on both sides of the equation. Start with the left side: differentiate \( e^{x^2 y} \). This requires the chain rule: the derivative of \( e^u \) is \( e^u \frac{du}{dx} \), where \( u = x^2 y \). Differentiate \( x^2 y \) using the product rule: \( x^2 \frac{dy}{dx} + 2xy \). Thus, \( \frac{d}{dx}(e^{x^2 y}) = e^{x^2 y} (x^2 \frac{dy}{dx} + 2xy) \).

Step 2: Differentiate the Right Side

Differentiate the right side, \( x + y \), with respect to \( x \). The derivative is \( 1 + \frac{dy}{dx} \). This is straightforward as \( \frac{d}{dx}(x) = 1 \) and \( \frac{d}{dx}(y) = \frac{dy}{dx} \) since \( y \) is a function of \( x \).

Step 3: Set Both Derivatives Equal

Set the derivatives equal according to the differentiation of the equation: \[e^{x^2 y} (x^2 \frac{dy}{dx} + 2xy) = 1 + \frac{dy}{dx} \]

Step 4: Solve for \( \frac{dy}{dx} \)

Rearrange the equation to solve for \( \frac{dy}{dx} \). Expand:\[ e^{x^2 y} x^2 \frac{dy}{dx} + 2xye^{x^2 y} = 1 + \frac{dy}{dx} \]Move terms involving \( \frac{dy}{dx} \) to one side:\[ e^{x^2 y} x^2 \frac{dy}{dx} - \frac{dy}{dx} = 1 - 2xye^{x^2 y} \]Factor out \( \frac{dy}{dx} \):\[ \frac{dy}{dx}(e^{x^2 y} x^2 - 1) = 1 - 2xye^{x^2 y} \]Solve for \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = \frac{1 - 2xye^{x^2 y}}{e^{x^2 y} x^2 - 1} \]

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus used to find the derivative of composite functions. Suppose we have a function in the form of one function nested within another, such as \(e^{x^2 y}\). To differentiate this, we identify the outer function \(e^u\) and the inner function \(u = x^2 y\). The chain rule states that the derivative of \(e^u\) with respect to \(x\) is \(e^u \cdot \frac{du}{dx}\). This involves:

• Finding the derivative of the outer function: \(e^u\), which is itself \(e^u\).
• Multiplying by the derivative of the inner function \(u\).

In our problem, \(u = x^2 y\). The chain rule helps us manage the derivative of such a product by keeping track of the dependency between the functions.

Product Rule

The product rule is essential when differentiating expressions where two functions are multiplied together. For instance, in the expression \(x^2 y\) in the original problem, both \(x^2\) and \(y\) depend on \(x\). The product rule states that if you have a product \(uv\), then the derivative \( \frac{d}{dx}(uv) \) is given by:

• \(u'v + uv'\)

In our case:

• Let \(u = x^2\), where the derivative \(u' = 2x\).
• Let \(v = y\), where the derivative \(v' = \frac{dy}{dx}\).

So, applying the product rule, we get: \(x^2 \cdot \frac{dy}{dx} + y \cdot 2x\). This step allows the accurate differentiation of the term \(x^2 y\), crucial for applying the chain rule correctly.

Derivatives

Derivatives measure how a function changes as its input changes, essentially expressing the function's rate of change. In implicit differentiation, where a function depends on another unknown function like \(y\), derivatives can be found without explicitly solving the function. Here’s a brief on how derivatives work and apply to this exercise:

• Derivatives of simple terms: For constants and linear terms, such as \(x\) and \(y\), the derivatives are \(1\) and \(\frac{dy}{dx}\) respectively.
• Combining rules: For complex terms, we use combination rules like the chain and product rule.
• Implicit differences: Allow handling equations not solved for \(y\), typically appearing as a relation with \(x\) instead.

Tools like the chain rule make it possible to differentiate intertwined functions, especially when applying implicit differentiation to problems where the dependent variable is not isolated.