Question

Find \(\frac{d y}{d x}\) using implicit differentiation. $$\frac{x}{y}=10$$

Short answer

\(\frac{dy}{dx} = \frac{y}{x}\)

Step-by-step solution

Differentiate Both Sides with Respect to x

To solve the implicit differentiation problem, start by differentiating the equation \(\frac{x}{y}=10\) with respect to \(x\). This requires using the quotient rule on the left-hand side.

Apply the Quotient Rule

The quotient rule is \(\frac{d}{dx} \left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx} }{v^2}\), where \(u = x\) and \(v = y\). Thus, apply it:\[\frac{y \cdot (\frac{d}{dx}(x)) - x \cdot (\frac{d}{dx}(y))}{y^2} = 0\] which leads to\[y \cdot 1 - x \cdot \frac{dy}{dx} = 0\]

Solve for \(\frac{dy}{dx}\)

From \(y - x \frac{dy}{dx} = 0\), isolate \(\frac{dy}{dx}\) by adding \(x \frac{dy}{dx}\) to both sides:\[y = x \frac{dy}{dx}\]Then divide both sides by \(x\):\[\frac{dy}{dx} = \frac{y}{x}\]

Key concepts

Quotient Rule

The Quotient Rule is a vital technique in calculus, especially when differentiating expressions where one function is divided by another. It is essential when dealing with problems like implicit differentiation. The rule can be remembered with the formula: \[\frac{d}{dx} \left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx} }{v^2}\] Think of \(u\) as your numerator and \(v\) as your denominator. By applying the quotient rule, you can differentiate without directly solving for one variable.

• **First, differentiate the numerator (\(u\))**, keeping the denominator (\(v\)) unchanged.
• **Next, differentiate the denominator**, and multiply by the original numerator.
• Finally, combine these results** by subtracting the second from the first, and dividing everything by the square of the original denominator.

This process transforms complex differentiation into manageable steps, useful for problems involving ratios like \(\frac{x}{y}\). The quotient rule is especially powerful when the ratio involves variables promising a simpler approach compared to isolating variables first.

Derivatives

Derivatives represent the rate of change of a function. They tell you how a function changes as its input changes. In the context of implicit differentiation, derivatives help us solve equations where one variable is expressed in terms of another in a "hidden" way. Instead of "extricating" \(y\) and solving \(y\) in terms of \(x\), we differentiate the entire equation with respect to \(x\). This aids in handling equations laid out implicitly. For example, the derivative \(\frac{dy}{dx}\) in the context of \(\frac{x}{y} = 10\) indicates how \(y\) changes concerning \(x\). Differentiating \(x\) with respect to \(x\) is straightforward, but differentiating \(y\) with respect to \(x\) involves recognizing the presence of \(x\) and \(y\) as interwoven components. Using derivatives, we break down complex relationships into bite-sized, analyzable fragments, making sense of how intertwined variables relate through these changes.

Solving Differential Equations

Solving differential equations involves finding a function or functions that satisfy a given equation involving derivatives. These equations display how a quantity changes and help describe various physical and mathematical systems. In our exercise, the differential equation arose from differentiating \(\frac{x}{y}\), resulting in a new equation. With operations like isolating \(\frac{dy}{dx}\), the derivative, we obtain a clearer understanding of relationships between variables. When you "solve" for \(\frac{dy}{dx}\), you're finding a derivative expression; here, it’s \(\frac{y}{x}\). This implies that the rate at which \(y\) changes with respect to \(x\) equals the ratio \(\frac{y}{x}\), highlighting a direct proportional relationship.

• **First, isolate terms with \(\frac{dy}{dx}\)**: This usually means moving terms around to focus on \(dy/dx\).
• **Solve the resulting equation** by performing algebraic manipulations to express the derivative.
• **Interpret the solution** in terms of the original problem, as this derivative will show true change relationships between variables.

Understanding and solving such equations is crucial for modeling real-world phenomena where variables continually change concerning each other.