Problem 32 - Thomas Calculus

Question

find \(d y / d x.\) \begin{equation}\ln y=x e^{y}-2\end{equation}

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{e^{y}}{\frac{1}{y} - x e^{y}} \).

Step 1: Implicit Differentiation Setup

The given equation is \( \ln y = x e^{y} - 2 \). To find \( \frac{dy}{dx} \), we will implicitly differentiate both sides of the equation with respect to \( x \). Remember that \( y \) is a function of \( x \), so we will apply the chain rule wherever necessary.

Step 2: Differentiate Left Side

Differentiate \( \ln y \) with respect to \( x \). Using the chain rule, the derivative of \( \ln y \) with respect to \( y \) is \( \frac{1}{y} \), and multiply by \( \frac{dy}{dx} \) (because of the chain rule):\[\frac{d}{dx}(\ln y) = \frac{1}{y} \frac{dy}{dx}. \]

Step 3: Differentiate Right Side

Differentiate \( x e^{y} - 2 \) with respect to \( x \). Apply the product rule to \( x e^{y} \), which gives:\[\frac{d}{dx}(x e^{y}) = e^{y} + x e^{y} \frac{dy}{dx}. \]Differentiate \( -2 \), which is \( 0 \).

Step 4: Equate Derivatives

The derivatives from both sides give us the equation:\[\frac{1}{y} \frac{dy}{dx} = e^{y} + x e^{y} \frac{dy}{dx}.\]

Step 5: Solve for \( \frac{dy}{dx} \)

Rearrange the equation to solve for \( \frac{dy}{dx} \):\[\frac{1}{y} \frac{dy}{dx} - x e^{y} \frac{dy}{dx} = e^{y}\]Factor out \( \frac{dy}{dx} \):\[\frac{dy}{dx} \left( \frac{1}{y} - x e^{y} \right) = e^{y}.\]Finally, solve for \( \frac{dy}{dx} \):\[\frac{dy}{dx} = \frac{e^{y}}{\frac{1}{y} - x e^{y}}. \]

Key Concepts

Chain Rule

The chain rule is an essential tool when dealing with implicit differentiation, especially when the function involves compositions of functions. It applies when you have a function nested inside another, often seen as \( f(g(x)) \). To differentiate such a function, you multiply the derivative of the outer function, evaluated at the inner function, by the derivative of the inner function. This is written mathematically as: \[\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x). \]In this problem, the chain rule is used on the left side of the equation where we have \( \ln y\). Since \( y \) is a function of \( x \), when differentiating \( \ln y \) with respect to \( x \), you must account for the fact that \( y \) changes with \( x \). This gives us:

• Derivative of \( \ln y \) with respect to \( y \) is \( \frac{1}{y} \)
• Multiply this by the derivative of \( y \) with respect to \( x \) (\( \frac{dy}{dx} \))

Resulting in\[\frac{d}{dx}(\ln y) = \frac{1}{y} \cdot \frac{dy}{dx}.\]This allows the differentiation to bridge from the nested function to the variable of interest, \( x \).

Product Rule

The product rule is vital when differentiating functions that are products of two or more functions. When you have a function \( u(x) \cdot v(x) \), the derivative is not simply the product of their derivatives. Instead, it requires a specific formula:\[\frac{d}{dx}[u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x).\]In our given problem, we use the product rule to tackle the term \( x e^{y} \). Here, we have:

• \( u(x) = x \) and \( v(x) = e^{y} \)
• The derivative of \( u(x) = x \) is \( 1 \)
• The derivative of \( v(x) = e^{y} \) requires the chain rule, giving us \( e^{y} \cdot \frac{dy}{dx} \)

Applying the product rule, the derivative of \( x e^{y} \) becomes:\[\frac{d}{dx}(x e^{y}) = 1 \cdot e^{y} + x \cdot e^{y} \cdot \frac{dy}{dx} = e^{y} + x e^{y} \frac{dy}{dx}.\]This differentiation is crucial in simplifying and manipulating the equation for \( \frac{dy}{dx} \).

Derivatives

Understanding derivatives is fundamental to calculus. Derivatives represent the rate at which a function changes at any given point. More simply, it measures how a function's output changes with respect to variations in its input. In our specific problem, we look at how \( y \) changes with respect to different values of \( x \).Each term from the equation \( \ln y = x e^{y} - 2 \) has a specific derivative:

• \( \ln y \) becomes \( \frac{1}{y} \cdot \frac{dy}{dx} \) using the chain rule
• \( x e^{y} \), a product, requires both the product and chain rule, resulting in \( e^{y} + x e^{y} \frac{dy}{dx} \)
• Constant \( -2 \) has a derivative of \( 0 \), as constants do not change

Ultimately, solving these allows for isolating \( \frac{dy}{dx} \), providing a way to express the derivative explicitly. This calculated derivative is crucial for analyzing how changes in \( x \) directly impact \( y \), revealing the function's behavior.