Question

In Exercises 61 and \(62,\) use the curve \(x^{2}-x y+y^{2}=1\) Multiple Choice Which of the following is equal to \(d y / d x ?\) (A) \(\frac{y-2 x}{2 y-x} \quad\) (B) \(\frac{y+2 x}{2 y-x}\) (C) \(\frac{2 x}{x-2 y} \quad\) (D) \(\frac{2 x+y}{x-2 y}\) \((\mathbf{E}) \frac{y+2 x}{x}\)

Short answer

(A) \(\frac{y-2x}{2y-x}\)

Step-by-step solution

Differentiate the equation implicitly

Applying the derivative individually to each term on both sides of the equation gives us \[2x - \frac{d}{dx}(xy) + 2yy' = 0\]. The term \(\frac{d}{dx}(xy)\) is differentiated using the product rule, giving us \(x * y' + y \). Therefore, the above equation becomes \[2x - (x * y' + y) + 2yy' = 0\].

Solve for y'

Isolate y' (dy/dx). Rearranging the equation results in \[(2x - y) = (x+2y)y'\]. Dividing both sides by \(x+2y\) to isolate \(y'\) gives us \[y' = \frac{2x-y}{x+2y}\].

Matching the Result with Choices

Compare the result of \(y'\) with the choices provided in the problem. The choice that matches the derived expression for \(y’\) is (A) \(\frac{y-2x}{2y-x}\).

Key concepts

Understanding Calculus

Calculus is a branch of mathematics that focuses on the study of rates of change and the accumulation of quantities. It's split into two main areas: differential and integral calculus. Differential calculus is concerned with the concept of a derivative, which represents an instantaneous rate of change of a function with respect to one of its variables. This is essential in modeling and solving problems involving changing quantities in physics, engineering, economics, and other fields.

For example, in the case of the exercise provided, we use differential calculus to find the rate at which one variable changes relative to another, expressed as \(dy/dx\), which is the derivative of \(y\) with respect to \(x\). In essence, it allows us to find the slope of the curve at any point, which is crucial for understanding how the function behaves locally and for solving real-world problems that depend on this rate of change.

Grasping the Derivative of Implicit Functions

When dealing with implicit functions, where the relationship between the variables \(x\) and \(y\) is expressed in a form that isn't isolated, such as in the equation \(x^2 - xy + y^2 = 1\), we use implicit differentiation to find the derivative \(dy/dx\). Differentiating implicitly means taking the derivative of both sides of the equation with respect to \(x\), treating \(y\) as a function of \(x\) and using the chain rule when necessary.

In our exercise, we differentiate each term with respect to \(x\) and apply the chain rule to the term \(xy\). We handle these mixed terms by remembering that \(y\) is a function of \(x\), hence its derivative involves \(dy/dx\), which is denoted as \(y'\). This leads us to an equation involving \(y'\), which we then solve to isolate \(y'\) and discover the rate of change between \(y\) and \(x\) at any point on the curve described by the function.

Applying the Product Rule

The product rule is a fundamental rule in differentiation when dealing with the product of two functions. In mathematics, it is expressed as \(\frac{d}{dx}(fg) = f'g + fg'\), meaning that to differentiate a product, you take the derivative of the first function \(f\) and multiply it by the second \(g\) as is, and then add the first function \(f\) multiplied by the derivative of the second \(g'\).

In our exercise step, the product rule is applied to the term \(xy\) because it is a product of two functions of \(x\) (note that \(y\) is a function of \(x\) in this context). The product rule helps us find the derivative of this term, which turns out to be \(x*y' + y\), where \(y'\) stands for \(dy/dx\). Understanding and applying the product rule is essential for calculating the derivatives correctly when faced with products of variables that are functions of one another.