Question

An implicitly defined function is given. Find \(\frac{d^{2} y}{d x^{2}} .\) Note: these are the same problems used in Exercises 13 through 16. $$\frac{x}{y}=10$$

Short answer

The second derivative \( \frac{d^2 y}{d x^2} = 0 \).

Step-by-step solution

Differentiate Implicitly

Start by differentiating both sides of the equation \( \frac{x}{y}=10 \) with respect to \( x \). To differentiate the left-hand side, use the quotient rule \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \), where \( u = x \) and \( v = y \). The derivative of the left side is \( \frac{y(1) - x(\frac{dy}{dx})}{y^2} \), and the derivative of the right side is 0.

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

From the differentiated equation \[ \frac{y - x \frac{dy}{dx}}{y^2} = 0 \] solve for \( \frac{dy}{dx} \). Multiply through by \( y^2 \) to get \( y = x \frac{dy}{dx} \). Rearrange to find \( \frac{dy}{dx} = \frac{y}{x} \).

Differentiate Again

Differentiate \( \frac{dy}{dx} = \frac{y}{x} \) implicitly to find \( \frac{d^2y}{dx^2} \). Use the quotient rule again to differentiate: \[ \frac{d}{dx} \left( \frac{y}{x} \right) = \frac{x \frac{dy}{dx} - y}{x^2} \].

Substitute \( \frac{dy}{dx} \) Into Second Derivative

Substitute \( \frac{dy}{dx} = \frac{y}{x} \) into the derivative found in Step 3. This gives us the second derivative: \[ \frac{d^2y}{dx^2} = \frac{x \left( \frac{y}{x} \right) - y}{x^2} = \frac{y - y}{x^2} = 0 \].

Final Result

The second derivative, \( \frac{d^2y}{dx^2} \), is \( 0 \).

Key concepts

Quotient Rule

The Quotient Rule is an essential technique in calculus for finding the derivative of a quotient of two functions. It helps when you have a function divided by another, like in the expression \( \frac{x}{y} \). Using the formula \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \), allows you to differentiate complex fraction-based functions efficiently.

Here’s the breakdown of how it works:

• \( u \) is the numerator function. In the case of \( \frac{x}{y} \), \( u = x \).
• \( v \) is the denominator function. For our example, \( v = y \).
• \( \frac{du}{dx} \) is the derivative of the numerator. For \( x \), this is 1.
• \( \frac{dv}{dx} \) is the derivative of the denominator. Since \( y \) is implicitly a function of \( x \), this becomes \( \frac{dy}{dx} \).

So applying these gives \( \frac{y(1) - x\left(\frac{dy}{dx}\right)}{y^2} \). This result helps us find the first derivative of an implicit function like \( \frac{x}{y} = 10 \).

Using the Quotient Rule is crucial when dealing with implicit differentiation, as it allows us to handle both variables seamlessly.

Second Derivative

The second derivative represents the rate at which the first derivative of a function is changing. In simple terms, it measures the curvature or concavity of a function. For implicit functions, like \( \frac{x}{y} = 10 \), you'll often need to find the second derivative to understand more about the function's behavior.

To find the second derivative \( \frac{d^2y}{dx^2} \), after computing the first derivative \( \frac{dy}{dx} \), you differentiate again. You apply the Quotient Rule, just like before. This is because the derivative you just found, \( \frac{y}{x} \), is in a fraction form.

The process involves:

• Differentiating each component (numerator and denominator separately).
• Applying the rule: \( \frac{x \cdot \frac{dy}{dx} - y}{x^2} \).

For the given example, substituting \( \frac{dy}{dx} = \frac{y}{x} \) into the derivative, simplifies it to \( \frac{y - y}{x^2} \), which equals zero. This tells us that the function is perfectly linear, with no curvature in this region.

Implicitly Defined Functions

Implicitly defined functions are those where the function is not isolated on one side of the equation. Instead of having a clear \( y = f(x) \), you have something like \( \frac{x}{y} = 10 \). Solving such functions involves a method called implicit differentiation, which helps find derivatives without solving the function explicitly for one variable in terms of another.

For example, differentiating \( \frac{x}{y} = 10 \) implicitly involves treating \( y \) as a function of \( x \) even though it isn't clearly spelled out. This means when you derive \( y \), you always accompany it with \( \frac{dy}{dx} \), indicating that \( y \) changes as \( x \) does.

Steps to differentiate implicitly:

• Differentiating both sides of the equation with respect to \( x \).
• Employing rules like the Quotient Rule when necessary.
• Solving the resulting equation to get \( \frac{dy}{dx} \) or \( \frac{d^2y}{dx^2} \).

This technique is powerful as it allows us to work with complex relationships without needing explicit forms, making calculus versatile and robust.