Problem 31 - Calculus Early Transcendentals

Question

Given the following equations, evaluate \(d y / d x .\) Assume that each equation implicitly defines \(y\) as a differentiable function of \(x\). $$x^{2}-2 y^{2}-1=0$$

Step-by-Step Solution

Answer

Answer: The derivative, \(\frac{dy}{dx}\), is given by the expression \(\frac{x}{2y}\).

Step 1: Differentiate the equation with respect to x

Differentiate both sides of the given equation with respect to x using the implicit differentiation technique. Apply the chain rule when differentiating terms involving y. $$\frac{d}{dx}(x^2) -2 \frac{d}{dx}(y^2) -\frac{d}{dx}(1)=\frac{d}{dx}(0)$$

Step 2: Calculate the derivatives of both sides

On the left-hand side of the equation, we'll calculate the derivative of each term, and on the right-hand side, we'll have zero since the derivative of any constant is zero. $$2x - 2(2y)\frac{dy}{dx} - 0 = 0$$

Step 3: Solve for dy/dx

Now, in order to find \(\frac{dy}{dx}\), we need to isolate it in the equation and solve for the variable. $$2x - 4y\frac{dy}{dx} = 0$$ Now, let's isolate \(\frac{dy}{dx}\) from the equation: $$4y\frac{dy}{dx} = 2x$$ Now, divide both sides by \(4y\): $$\frac{dy}{dx} = \frac{2x}{4y}$$

Step 4: Simplify the expression

Lastly, let's simplify the expression by dividing both numerator and denominator by 2: $$\frac{dy}{dx} =\frac{x}{2y}$$ Now we have found the derivative of y with respect to x, so the solution is: $$\frac{dy}{dx} =\frac{x}{2y}$$

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus, especially when it comes to implicit differentiation. It’s used to find the derivative of composite functions. In our problem, we differentiate

• Since the original equation involves both \(x\) and \(y\), when we differentiate terms that include \(y\), we apply the chain rule. For instance, differentiating \(y^2\) with respect to \(x\) does not simply become \(2y\), but instead \(2y \frac{dy}{dx}\) because \(y\) itself is a function of \(x\).

The application of the chain rule in implicit differentiation often requires assigning a placeholder for \(\frac{dy}{dx}\). This ensures we account for how \(y\) changes with \(x\). Whenever you see terms like \(y^2\) in these equations, remember to apply the chain rule for accurate results. It’s what allows you to maintain the relationship between the variables correctly.

Differentiable Function

A differentiable function is essential in calculus because it signifies that a function can have a derivative wherever it is defined. When we say that \(y\) is a differentiable function of \(x\), this implies that everywhere the function \(y\) is defined, we can compute a meaningful slope or rate of change.

• This is crucial for implicit differentiation. It ensures that when we differentiate with respect to \(x\), we have a clear and defined transformation from \(y\) to \(\frac{dy}{dx}\).
• In this type of problem, implicit equations like \(x^2 - 2y^2 -1=0\) indicate that there is some derived function \(y(x)\) that satisfies the equation.

Ensuring that \(y\) is a differentiable function lets you smoothly progress through solutions, like the one above, without encountering undefined terms.

Derivative of a Function

The derivative of a function represents the rate at which the function changes as its input changes. It is a core concept in calculus that describes the steepness or slope of the graph at any point.

• In the context of our problem, we were tasked with finding \(\frac{dy}{dx}\), which is the derivative of \(y\) with respect to \(x\).
• This derivative tells us how \(y\) changes for a tiny change in \(x\). Our solution shows that the derivative is \(\frac{x}{2y}\), derived by differentiating the implicit equation separately and rearranging.
• To derive it: we differentiate with respect to \(x\), rearrange to isolate \(\frac{dy}{dx}\), and simplify, which creates a functional relationship of change between \(x\) and \(y\).
• This derivative doesn't just give us a number, but a function that helps us understand the dynamics between \(x\) and \(y\) across their domain.

Understand this connection and skill at deriving derivatives will be pivotal as you handle different calculus problems in your studies.