Problem 300 - Calculus Volume 1

Question

For the following exercises, use implicit differentiation to find \(\frac{d y}{d x}\) $$ x^{2}-y^{2}=4 $$

Step-by-Step Solution

Answer

The derivative \( \frac{dy}{dx} \) is \( \frac{x}{y} \).

Step 1: Differentiate Both Sides with Respect to x

To find \( \frac{dy}{dx} \), we need to differentiate both sides of the equation \( x^2 - y^2 = 4 \) with respect to \( x \). For the left side, differentiate \( x^2 \) to get \( 2x \) and \( -y^2 \) to get \( -2y \frac{dy}{dx} \), due to the chain rule. The right side (4) differentiates to 0. This gives us: \[ 2x - 2y \frac{dy}{dx} = 0 \]

Step 2: Solve for \( \frac{dy}{dx} \)

Now, isolate \( \frac{dy}{dx} \) in the equation. First, add \( 2y \frac{dy}{dx} \) to both sides to get:\[ 2x = 2y \frac{dy}{dx} \]Then, divide both sides by \( 2y \) to solve for \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = \frac{x}{y} \]

Key Concepts

Derivative

Derivatives are a fundamental concept in calculus. They measure how a function changes as its input changes. In simpler terms, the derivative of a function at a point tells us the slope of the tangent line to the function's graph at that point. This slope indicates how steep the function is at that point and the direction in which it is increasing or decreasing.

For example, in the exercise, we start by differentiating the entire equation implicitly. By doing this, we calculate how the equation changes as both variables, \( x \) and \( y \), change. This provides a relationship between their rates of change—exactly what derivatives are about.

This understanding is crucial when solving calculus problems that involve slopes, rates of change, and optimizing functions.

Chain Rule

The chain rule is an essential technique in calculus for finding the derivative of composite functions. When a function is composed of multiple functions, the chain rule helps in differentiating it.

In our problem, we applied the chain rule when differentiating \(-y^2\). Since \(y\) is a function of \(x\), it's not just a simple differentiation. We treat \(y\) as a function of \(x\), so when differentiating, we have to multiply the derivative of \(-y^2\) by \(\frac{dy}{dx}\).

This results in \(-2y \frac{dy}{dx}\). The chain rule is vital for handling more complex equations where different variables depend on each other.

Implicit Functions

Implicit functions are functions where the variables are not separated. Unlike explicit functions, where you have \(y\) expressed directly in terms of \(x\), implicit functions involve a relationship between \(x\) and \(y\) within an equation.

In the given exercise, the equation \(x^2 - y^2 = 4\) is implicit because \(y\) is not isolated—it is mixed within the equation with \(x\).

To differentiate such functions, implicit differentiation is necessary. This allows us to find the derivative \(\frac{dy}{dx}\) without explicitly solving for \(y\). Understanding implicit functions is crucial for working with equations where variables interact in a single expression.

Calculus Problem Solving

Calculus problem solving often involves identifying the type of differentiation needed for a given problem and applying the correct rules. Here, implicit differentiation was required to find \(\frac{dy}{dx}\).

Start by recognizing the implicit nature of the given equation. Use the chain rule effectively when differentiating terms involving \(y\). Simplify your equations step-by-step. For instance, after differentiating, we had \(2x - 2y \frac{dy}{dx} = 0\). Solving this step-by-step, we isolated \(\frac{dy}{dx}\) to get \(\frac{x}{y}\).

Effective calculus problem solving requires practice and understanding of different rules and methods, helping in systematically breaking down and solving intricate mathematical problems.