Problem 34 - Calculus Early Transcendentals

Question

Given the following equations, evaluate \(d y / d x .\) Assume that each equation implicitly defines \(y\) as a differentiable function of \(x\). $$y e^{x y}-2=0$$

Step-by-Step Solution

Answer

Question: Determine the derivative of y with respect to x, dy/dx, for the equation y * exp(xy) - 2 = 0, where y is an implicit function of x. Answer: The derivative of y with respect to x, dy/dx, for the given equation is - (y^2 * exp(xy)) / (exp(xy)(1 + xy)).

Step 1: Implicit differentiation of given equation

To find \(\frac{dy}{dx}\), we will differentiate both sides of the given equation. Differentiate the given equation with respect to x: $$\frac{d}{dx}(y e^{xy} - 2) = \frac{d}{dx}(0)$$

Step 2: Differentiate each term with respect to x

Now, differentiate each term in the equation with respect to x, remembering to apply the Chain Rule and Product Rule as needed. The left side of the equation consists of two terms: \(y e^{xy}\) and \(-2\). Differentiate each term: $$\frac{d}{dx}(y e^{xy}) - \frac{d}{dx}(2) = 0$$ Product Rule tells us that: \((fg)' = f'g + fg'\). So, for our first term, \(f = y\) and \(g = e^{xy}\). Thus, \(f' = \frac{dy}{dx}\) and using the Chain Rule, \(g' = e^{xy}(y + x\frac{dy}{dx})\). Now, applying the product rule and substituting the values in the equation: $$\left(\frac{dy}{dx}e^{xy} + y e^{xy} (y + x\frac{dy}{dx})\right) - 0 = 0$$

Step 3: Solve for dy/dx in the equation

Now we simplify the equation and solve for \(\frac{dy}{dx}\): $$\frac{dy}{dx}e^{xy} + y^2 e^{xy} + x y e^{xy}\frac{dy}{dx} = 0$$ Let's factor \(\frac{dy}{dx}\) out: $$\frac{dy}{dx}(e^{xy} + x y e^{xy}) = -y^2 e^{xy}$$ Now, let's isolate dy/dx by dividing both sides with \((e^{xy} + x y e^{xy})\): $$\frac{dy}{dx} = -\frac{y^2 e^{xy}}{e^{xy}(1 + xy)}$$ This is the derivative of y with respect to x, \(\frac{dy}{dx}\).

Key Concepts

Chain Rule

The Chain Rule is a fundamental concept in calculus used to differentiate composite functions. When a function is composed of two or more functions, the Chain Rule allows us to differentiate it efficiently. For instance, if you have a function inside another function, like in the expression \( e^{xy} \), you must apply the Chain Rule to find the derivative.
The essence of the Chain Rule is: If you have a function \( f(g(x)) \), its derivative is \( f'(g(x)) \cdot g'(x) \). This means you first differentiate the outer function, leaving the inner function intact, and then multiply by the derivative of the inner function.

• Identify the outer and inner functions in the expression.
• Differentiate the outer function.
• Multiply by the derivative of the inner function.

In the problem provided, when differentiating \( e^{xy} \), treat \( x y \) as the inner function and apply the Chain Rule to find its derivative. This process involves finding the derivative of \( e^{xy} \) itself, and then multiplying it by the derivative of \( x y \). It's essential to recognize how the Chain Rule simplifies complex differentiation tasks.

Product Rule

The Product Rule is applied when differentiating expressions involving the product of two functions. In calculus, it allows us to handle derivatives of products like \( y e^{xy} \) in our exercise.
The Product Rule formula is: If you have two functions \( u(x) \) and \( v(x) \), the derivative of the product \( u(x)v(x) \) is given by \( u'(x)v(x) + u(x)v'(x) \).
To use the Product Rule:

• Differentiate \( u(x) \), the first function.
• Multiply by \( v(x) \), the second function.
• Add \( u(x) \) multiplied by the derivative of \( v(x) \).

In the expression \( y e^{xy} \), applying the Product Rule requires treating \( y \) and \( e^{xy} \) as separate functions. First, differentiate \( y \) while keeping \( e^{xy} \) constant, and then differentiate \( e^{xy} \) applying the Chain Rule, as this term is more complex. By coordinating both rules, you're able to fully differentiate the entire product expression.

Differential Equations

Differential equations are equations that include derivatives of a function. These equations describe how a function changes and are fundamental in modeling various natural phenomena.
The key aspect of solving differential equations is recognizing how functions and their derivatives interact.
Implicit differentiation is particularly useful in finding derivatives when the equation isn't solved for one variable explicitly.

• Start by differentiating both sides concerning \( x \). Do this even if \( y \) is not explicitly defined.
• Apply relevant rules, like the Chain Rule and Product Rule, when differentiating each term.
• Isolate \( \frac{dy}{dx} \), which represents the rate of change of \( y \) concerning \( x \).

Such methods are crucial when dealing with implicit functions, where both sides of the equation can interdependently define \( y \) as a function of \( x \).By mastering implicit differentiation and understanding differential equations, you can effectively handle complex dynamic systems expressed through these equations.