Problem 10 - Calculus an Applied Approach

Question

Find \(d y / d x\) \(\frac{x y-y^{2}}{y-x}=1\)

Step-by-Step Solution

Answer

The derivative \(dy/dx\) of the function given by the equation \(\frac{x y-y^2}{y-x}=1\) is \(dy/dx = \frac{2y-x}{2y+x}\)

Step 1: Differentiate both sides of the equation with respect to x

Differentiate both sides of the equation with respect to x using the product rule and quotient rule. The product rule states, when differentiating \(u*v\), that it is equal to \(u'v+uv'\), and the quotient rule states, when differentiating \(\frac{u}{v}\), that it is equal to \(\frac{vu'-uv'}{v^{2}}\). The derivative of x with respect to x is 1, and derivative of y with respect to x is \(d y / d x\), or y'.

Step 2: Set the two differentials equal and rearrange the equation

Apply the differential rules to each term of the function on the left side of the equation and set it equal to the differential of the right side, which is zero because constant's derivative is zero. Collect terms involving y' on one side of the equation, and the remaining ones on other side.

Step 3: Express y' in terms of x and y

Factor out the y', and then isolate y' by dividing both sides of the equation by the remaining factor on the side of y'.

Key Concepts

Quotient Rule

To tackle problems like differentiating fractions or quotients, the Quotient Rule is essential. It helps us differentiate a function that is expressed as one function divided by another, like \( \frac{u}{v} \). Here you apply the rule:

• Differentiate the numerator \( u \) to get \( u' \).
• Differentiate the denominator \( v \) to get \( v' \).
• Then, use the formula: \( \frac{vu' - uv'}{v^2} \).

The process involves combining the derivatives with some arithmetic. Important to note, the denominator squared \( v^2 \) plays a role in ensuring the function remains accurate during differentiation. Always keep track of each piece carefully.

Product Rule

The Product Rule is useful when differentiating the product of two functions, \( u \) and \( v \). If you have a product of functions, you can't just differentiate each part separately; instead, you use this rule:

• Find the derivative of \( u \) giving you \( u' \).
• Find the derivative of \( v \) giving you \( v' \).
• Then, apply the formula: \( u'v + uv' \).

This rule ensures that all interactions between the multiplying parts are considered. The right application of this rule gives a complete derivative of the product, capturing changes due to both functions involved.

Differentiation

Differentiation is the core operation in calculus to find the rate at which a function changes at any point. Using principles like those found in the product and quotient rules, you can differentiate complex functions by considering how each part contributes to the whole. Some general steps for differentiation include:

• Identifying whether to apply rules like the product or quotient rules.
• Computing the derivative of each function identified.
• Simplifying the resulting expression for better insight.

Differentiation unfolds the behavior of functions, giving you the clues to how they rise, fall, or level off at points.

Calculus Problem Solving

Solving calculus problems often requires handling derivatives through a sequence of systematic steps. Appreciating the foundational rules and applying them effectively can drastically reduce complexity. Here's how problem-solving in calculus generally flows:

• First, identify the functions involved and the differentiation rules applicable—product or quotient.
• Next, perform the calculations based on these rules, ensuring all derivations are correct.
• Finally, simplify and interpret the resulting derivative to solve for unknowns or predict behavior.

Being thorough in these steps ensures a correct and insightful result, unlocking the intricacies behind the mathematical relationships at play.