Problem 8 - Calculus Early Transcendentals

Question

Find \(d y / d x\) by implicit differentiation. \(x^{2}=\frac{x+y}{x-y}\)

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{(x-y)^2}{2x} \)

Step 1: Differentiate Both Sides with Respect to x

To find \( \frac{dy}{dx} \) using implicit differentiation, first differentiate the equation \( x^2 = \frac{x+y}{x-y} \) with respect to \( x \). Start with the left side: the derivative of \( x^2 \) is \( 2x \).

Step 2: Differentiate the Right Side Using the Quotient Rule

The right side \( \frac{x+y}{x-y} \) needs the quotient rule: if \( u = x+y \) and \( v = x-y \), then \( \frac{d}{dx}\left( \frac{u}{v} \right)= \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \). Compute \( \frac{du}{dx} = 1 + \frac{dy}{dx} \) and \( \frac{dv}{dx} = 1 - \frac{dy}{dx} \).

Step 3: Apply the Quotient Rule

Substitute the derivatives \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) into the quotient rule formula: \[ \frac{d}{dx} \left( \frac{x+y}{x-y} \right) = \frac{(x-y)(1 + \frac{dy}{dx}) - (x+y)(1 - \frac{dy}{dx})}{(x-y)^2} \].

Step 4: Simplify the Resulting Equation

Simplify the expression from Step 3: \((x-y)(1+\frac{dy}{dx}) - (x+y)(1-\frac{dy}{dx}) = (x-y) + (x-y)\frac{dy}{dx} - (x+y) + (x+y)\frac{dy}{dx}\). This simplifies to \((x-y)\frac{dy}{dx} + (x+y)\frac{dy}{dx} \), which is equivalent to \(2x\frac{dy}{dx}\) since the other terms cancel.

Step 5: Equate and Solve for dy/dx

Equate the derivatives from Steps 1 and 4: \(2x = \frac{2x\frac{dy}{dx}}{(x-y)^2}\). Solve this equation for \( \frac{dy}{dx} \) by multiplying by \((x-y)^2\) and then dividing by \( 2x \). This yields \( \frac{dy}{dx} = \frac{(x-y)^2}{2x} \).

Key Concepts

Quotient Rule

The Quotient Rule is essential when differentiating expressions where one function divides another. In the formula, if you have a function expressed as a quotient of two functions, like \( \frac{u}{v} \), the derivative with respect to \( x \) is given by:\[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \]Here's a simple breakdown to help understand this rule:

• Identify your parts: Assign the top function \( u \) and the bottom function \( v \).
• Differentiate separately: Find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \).
• Apply the formula: Plug these into the Quotient Rule formula.

These steps create a structured approach to handle challenges like our problem where the right-side expression is a quotient. Understanding and applying the Quotient Rule sharpens your differentiation skills!

Derivative

A derivative represents how a function changes as its input changes. When we say we're finding the derivative of a function, we're generally discovering its rate of change or slope at any given point.For basic functions, like polynomial functions, derivatives are relatively straightforward. For instance, the derivative of \( x^2 \) is just \( 2x \), expressing how quickly \( x^2 \) increases as \( x \) itself increases. In implicit differentiation, derivatives are more nuanced. Here, derivative calculation involves both \( x \) and \( y \), often when \( y \) isn't isolated. This technique requires:

• Chain Rule application: Assume \( y \) is a function of \( x \) when differentiating.
• Solving for \( \frac{dy}{dx} \): After differentiation, rearrange to find \( \frac{dy}{dx} \).

Implicit differentiation enables us to find derivatives where traditional methods may pose challenges.

Solving Equations

Solving equations in calculus, specifically with derivatives, involves algebraic manipulation to find an unknown variable. When dealing with implicit differentiation and derivative rules, you eventually arrive at an equation needing to be solved for \( \frac{dy}{dx} \).Here's a step-by-step approach:

• Isolate \( \frac{dy}{dx} \): Manipulate the equation so that all instances of \( \frac{dy}{dx} \) appear on one side.
• Use algebraic routines: Apply basic algebra—like factoring or distributing—to simplify.
• Final solution: Find \( \frac{dy}{dx} \) explicitly.

This process helps transition from the derivative form to a usable expression for \( \frac{dy}{dx} \), revealing the underlying dynamics between \( x \) and \( y \) in the function.