Question

\(2 y=x^{2}-x y, \quad x=2, \quad d x=-0.05\)

Short answer

The value of \(dy\) can be found using the equation \(dy = \frac{y - 4}{4} * -0.05\) after necessary substitutions have been made. \(dy\) is the change in y for a small change \(dx = -0.05\) in x.

Step-by-step solution

Differentiate both sides with respect to x

To find the derivative, each term in the equation must be differentiated. For the left side, differentiating \(2y\) gives \(2 \frac{dy}{dx}\). For the right side, using the power rule, \(\frac{d}{dx}(x^{2}) = 2x\). Then, differentiate \(- xy\), yielding \(-x \frac{dy}{dx} - y\) after using the product rule.

Apply the derivatives and form the equation

By applying these derivatives, we get \(2 \frac{dy}{dx} = 2x - x\frac{dy}{dx} - y.\) This allows us to solve for \(\frac{dy}{dx}\), the derivative of y with respect to x.

Simplify the equation

To isolate \(\frac{dy}{dx}\), we add \(x\frac{dy}{dx}\) on both sides, subtract \(2x\) on both sides and divide by 2. This gives us \(\frac{dy}{dx} = \frac{y - 2x}{2 + x}\).

Substitute given values

Now, we substitute \(x = 2\) as provided. This gives \(\frac{dy}{dx} = \frac{y - 4}{4}\).

Find value of dy

The value of \(dx\) is given as -0.05. To find \(dy\), we multiply \(\frac{dy}{dx}\) by \(dx\). That is \(dy = \frac{dy}{dx} * dx = \frac{y - 4}{4} * -0.05\)

Key concepts

Differentiating Equations

Differentiating equations is akin to unraveling a code. When presented with an equation like \(2 y=x^{2}-x y\), the challenge is to find how one variable changes in response to the other. This process begins with applying derivative rules to each term in the equation, meticulously considering the interplay between the variables.

The act of differentiation turns the focus from mere numbers and variables to their rates of change. For example, when differentiating \(2y\) with respect to \(x\), it becomes \(2 \frac{dy}{dx}\), a reflection of how \(y\) changes as \(x\) nudges forward. The transformation of the right side involves the power rule turning \(x^{2}\) into \(2x\) and the product rule dissecting \(-xy\) into \(-x \frac{dy}{dx} - y\).

By carrying out these steps, we transition from a static algebraic statement to a dynamic description of the relationship between \(x\) and \(y\). This concept is fundamental to understanding how to interpret and tackle calculus problems.

Applying Derivatives

Applying derivatives is like putting the pieces of a puzzle in place. Once each term of an equation has been differentiated, integrating these derivatives shapes the overall image—a formula for the derivative of one variable with respect to another.

Consider our initial outcome \(2 \frac{dy}{dx} = 2x - x \frac{dy}{dx} - y\). This equation is the culmination of applying the derivative to each term individually. The next step invokes algebra: maneuvering the terms to isolate \(\frac{dy}{dx}\), the rate of change of \(y\) with respect to \(x\), unveils the pattern we're searching for.

Through the application of derivatives, we're essentially constructing the slope of the tangent line to the curve defined by our original equation at any point \((x, y)\). Grasping this application is crucial in calculus, as it underpins much of the analysis techniques that allow us to understand rates of change and motion.

Solving for y'

Solving for \(y'\), or \(\frac{dy}{dx}\), is unearthing the heartbeat of a function—its instantaneous rate of change at any point. This search involves isolation of \(\frac{dy}{dx}\) within our derivative equation, transforming it from a conjoined status to the sole variable of interest.

To elucidate \(\frac{dy}{dx}\), we perform delicate algebraic surgery: adding alike terms to one side, subtracting others, and finally dividing the equation to reveal \(\frac{dy}{dx}\) in its purest form, like \(\frac{dy}{dx} = \frac{y - 2x}{2 + x}\).

It's a vital practice that strengthens one's ability to move from general expressions to specific insights regarding how a variable is expected to change instantaneously, paving the path to deeper analysis in calculus—whether it be optimizing a function or predicting the nature of its graph.

Product Rule in Differentiation

The product rule in differentiation is a critical ally when confronting terms where two variables are multiplied together. It states that to differentiate a product, one must take the derivative of the first function multiplied by the second function and then add the product of the first function and the derivative of the second function.

For instance, differentiating \(-xy\) involves computing \(-x \frac{dy}{dx} - y\). This results by first keeping \(x\) constant and differentiating \(y\) to get \(-x \frac{dy}{dx}\), and then by keeping \(y\) constant and differentiating \(-x\) to simply get \(-y\).

This rule is not only a cornerstone in calculus but a testament to the subject's elegance, revealing how compound relationships can be systematically deconstructed into their rate components. Understanding and applying the product rule is foundational in tackling calculus problems involving products of functions.

Calculus Education

Calculus education represents the journey through a landscape of functions, limits, derivatives, integrals, and infinite series. Breaking down complex problems into easily digestible concepts, like the methodical steps required for implicit differentiation, equips students with the analytical tools necessary for a comprehensive understanding of the mathematical world.

Every concept, from differentiating equations to applying the product rule, must be communicated with clarity and connected to its applications in real-world phenomena. Instructors aim to scaffold students' learning experiences to foster problem-solving skills and develop the ability to convey mathematical ideas effectively.

Quality calculus education transcends the memorization of formulas; it nurtures a mindset attuned to patterns and changes, empowering students not only to excel in mathematics but to interpret and analyze the ever-dynamic world around them.