Problem 28 - University Calculus: Early Transcendentals

Question

Use implicit differentiation to find \(d y / d x\) and then \(d^{2} y / d x^{2}\) If \(x y+y^{2}=1,\) find the value of \(d^{2} y / d x^{2}\) at the point (0,-1)

Step-by-Step Solution

Answer

The value of \(\frac{d^2 y}{d x^2}\) at \((0, -1)\) is \(-\frac{1}{4}\).

Step 1: Differentiate Implicitly for First Derivative

Starting with the equation \( x y + y^2 = 1 \), apply implicit differentiation with respect to \( x \). Use the product rule on the term \( xy \). The differentiation yields:\[ \frac{d}{dx}(xy) = x \frac{dy}{dx} + y \] Differentiating \( y^2 \) with respect to \( x \) yields:\[ 2y \frac{dy}{dx} \]So the differentiated equation is:\[ x \frac{dy}{dx} + y + 2y \frac{dy}{dx} = 0 \]

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

Rearrange the differentiated equation to solve for \( \frac{dy}{dx} \):\[ x \frac{dy}{dx} + 2y \frac{dy}{dx} = -y \]Factor out \( \frac{dy}{dx} \):\[ (x + 2y) \frac{dy}{dx} = -y \]Solve for \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = \frac{-y}{x + 2y} \]

Step 3: Differentiate Again for Second Derivative

With \( \frac{dy}{dx} = \frac{-y}{x + 2y} \), differentiate again implicitly with respect to \( x \) to find \( \frac{d^2y}{dx^2} \). Apply the quotient rule:For \( \frac{d}{dx} \left( \frac{-y}{x + 2y} \right) \), let \( u = -y \) and \( v = x + 2y \), so:\[\frac{d}{dx}(v) = 1 + 2 \frac{dy}{dx} \]By quotient rule:\[\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}\]Substitute \( u \) and \( v \):\[\frac{d^2y}{dx^2} = \frac{(x + 2y)(-\frac{dy}{dx}) - (-y)(1 + 2\frac{dy}{dx})}{(x + 2y)^2}\]

Step 4: Evaluate at the Given Point (0, -1)

Substitute \( x = 0 \) and \( y = -1 \) into the first derivative \( \frac{dy}{dx} = \frac{-y}{x + 2y} \):\[ \frac{dy}{dx} \bigg|_{(0, -1)} = \frac{-(-1)}{0 + 2(-1)} = \frac{1}{-2} = -\frac{1}{2} \]Now substitute these values into the second derivative formula:\[\frac{d^2 y}{dx^2} = \frac{(0 + 2(-1))(-(-\frac{1}{2})) - (-1)(1 + 2(-\frac{1}{2}))}{(0+2(-1))^2}\]Simplify the expression:\[\frac{d^2 y}{dx^2} = \frac{(-2)(\frac{1}{2}) - (-1)(1 - 1)}{4} = \frac{-1 + 0}{4} = -\frac{1}{4} \]

Step 5: Conclusion

The second derivative at the point \((0, -1)\) is \( -\frac{1}{4} \). This represents the concavity of the function at this point, indicating that the curve is concave down.

Key Concepts

Second Derivative

The second derivative, denoted as \( \frac{d^2y}{dx^2} \), provides insights into the curvature or concavity of a function at a certain point. When we find the second derivative, we are essentially evaluating the rate of change of the rate of change of the function. Knowing the second derivative allows us to determine whether the function is concave up or concave down in a given section of its graph.

Finding the second derivative often involves differentiating once to find the first derivative and then again to find the second. This requires careful application of differentiation rules, especially when dealing with implicitly defined functions. In our problem, the second derivative \( \frac{d^2y}{dx^2} \) helps in understanding the behavior of the equation \(xy + y^2 = 1\) at the specific point (0, -1). After calculating, the second derivative at this point is \(-\frac{1}{4}\), which indicates a particular concavity.

Product Rule

The product rule is a fundamental tool in calculus used to differentiate functions that are products of two simpler functions. If you have a function \( f(x) = u(x)v(x) \), then the derivative of \( f(x) \) with respect to \( x \) is:

• \( \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} \)

This rule is essential when dealing with the term \( xy \) from the original equation \( xy + y^2 = 1 \).

Using the product rule requires us to differentiate each part of the product while keeping the product's other half constant, then summing these results. For instance, in \( xy \), you treat \( x \) as one function and \( y \) as another, applying the product rule to calculate its derivative properly. In our problem, the product rule helped to transform the left side of the equation to \( x \frac{dy}{dx} + y \). This newly formed expression is crucial for setting up the path to finding the first and second derivatives.

Quotient Rule

The quotient rule comes in handy when you need to differentiate a function that is a ratio of two other functions. If a function \( f(x) \) can be expressed as \( \frac{u(x)}{v(x)} \), the derivative \( \frac{d}{dx} \left(\frac{u}{v}\right) \) is:

• \( \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \)

This rule is pivotal in dealing with the implicit form of the second derivative we found, \( \frac{dy}{dx} = \frac{-y}{x + 2y} \).

In the context of our exercise, applying the quotient rule to differentiate the expression \( \frac{-y}{x + 2y} \) allows us to further explore \( \frac{d^2y}{dx^2} \). Given that the first derivative involves a ratio, the quotient rule provides the algebraic framework to find the second derivative accurately. It's a systematic approach to unravel how changes in \( y \) with respect to \( x \) affect the curvature or concavity of the function at a given point.

Concavity

Understanding concavity is important in calculus as it describes how a curve opens. Concavity indicates the direction of the curve's bend. If the second derivative of a function at a point is positive, the function is concave up at that point, resembling a "cup" shape. Conversely, if the second derivative is negative, the function is concave down, resembling a "cap" shape.

In our problem, the second derivative at point (0, -1) is \(-\frac{1}{4}\), which suggests that the curve is concave down at this point. This information is critical when analyzing the behavior of functions because it gives insights into the nature of turning points. An understanding of concave and convex regions helps in interpreting the overall shape of the graph and predicting future values.