Question

In Exercises, find \(d y / d x\) $$ \frac{x y-y^{2}}{y-x}=1 $$

Short answer

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

Step-by-step solution

Rearrange the equation

The given equation can be rearranged as \(x y - y^2 = y - x\). This form is easier to handle as we can differentiate each side of the equation separately.

Differentiate both sides

Applying product rule \((u.v)' = u'.v + u.v'\) on the left side where \(u = x\) and \(v = y\) we get \((x y)' = 1.y + x.\frac{dy}{dx}\). For \((-y^2)'\), apply power rule to get \(2y.\frac{dy}{dx}\). Differentiating the right side, using chain rule for \(y\) and the fact that \(x' = 1\), we get \(\frac{dy}{dx} - 1\). So, the differential equation becomes: \( y + x.\frac{dy}{dx} - 2y.\frac{dy}{dx} = \frac{dy}{dx} - 1 \).

Solve for \(dy/dx\)

Rearrange to solve for \(dy/dx\). Group the dy/dx terms on one side and the non-dy/dx terms on the other. Subtracting \(\frac{dy}{dx}\) from both sides and then factoring, we get \((1 + x - 2y).\frac{dy}{dx} = -y + 1 \). Dividing by \((1+ x -2y)\), final expression is \(\frac{dy}{dx} = \frac{- y + 1}{1 + x - 2y}\)

Key concepts

Derivative

The concept of a derivative is central to calculus. Derivatives represent the rate at which a function is changing at any given point. In simple terms, if you have a function that describes a line drawn on a graph, the derivative tells you how steep that line is at any point.
For any function \(f(x)\), the derivative, written as \(f'(x)\) or \(\frac{df}{dx}\), provides a way to predict how \(f\) will behave near any point \(x\).
It is calculated as the limit of the difference quotient, essentially a fancy way of saying it's the slope of the function as the interval becomes infinitesimally small. Derivatives are very useful when working with directions of curves, optimizations, or any scenario where you need to understand how change can impact your function.

Product rule

When you need to take the derivative of a function that is the product of two other functions, you'll use the product rule. The product rule states that if you have two functions \(u(x)\) and \(v(x)\), the derivative of their product \(u(x)v(x)\) is:\[(u.v)' = u'v + uv'\]
This rule is particularly important because it allows you to differentiate products of functions without expanding them beforehand, which can be cumbersome and complex.
In the example exercise, the function \(u(x) = x\) and \(v(x) = y\) are multiplied, so apply the rule to find \((xy)'\) by calculating \(1 \cdot y + x \cdot \frac{dy}{dx}\). This is foundational in managing and simplifying the differentiation process.

Chain rule

The chain rule is vital when dealing with composite functions, where one function is nested inside another. It helps find the derivative of such functions efficiently. The chain rule states that if a function \(y = f(g(x))\), then its derivative is given by:\(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\).
Essentially, you differentiate the outer function and multiply it by the derivative of the inner function.
In our problem, although the direct application is clearer on the right side with \(y\), the nature of the left side requires observing how variables interconnect as you differentiate expressions involving multiple variables adjusting themselves as some functions of another variable.

Algebraic manipulation

Algebraic manipulation refers to rearranging and simplifying expressions to facilitate solving equations. This is often needed in calculus when equations grow complex after differentiation.
Manipulating an equation correctly sets up a clearer pathway to solving for derivatives like \(\frac{dy}{dx}\).
In our exercise, we begin by rearranging the equation to more manageable terms, transforming \(xy - y^2 = y - x\). This allows clearer differentiation of each side, leading to solving the equation more straightforwardly.

• Move terms involving \(dy/dx\) to one side of the equation.
• Combine like terms or factor common elements, literally simplifying the expression.
• Solve for the unknown derivative by isolating its terms.

Mastery of algebraic manipulation is crucial as it directly affects your ability to solve equations efficiently.