Problem 24 - University Calculus: Early Transcendentals

Question

Use implicit differentiation to find \(d y / d x\) and then \(d^{2} y / d x^{2}\) $$y^{2}-2 x=1-2 y$$

Step-by-Step Solution

Answer

First derivative: \( \frac{dy}{dx} = \frac{1}{y + 1} \); Second derivative: \( \frac{d^2y}{dx^2} = -\frac{1}{(y + 1)^3} \).

Step 1: Rearrange the equation

We start by rearranging the given equation, bringing all terms to one side:\[ y^2 - 2x - 1 + 2y = 0 \] or equivalently, \[ y^2 + 2y - 2x - 1 = 0 \]. This form will be useful for implicit differentiation.

Step 2: Differentiate implicitly with respect to x

Differentiate both sides of the equation with respect to \(x\). The derivative of \(y^2\) with respect to \(x\) is \(2y \frac{dy}{dx}\); the derivative of \(2y\) is \(2 \frac{dy}{dx}\); for \(-2x\) it is \(-2\); and for \(-1\) it is \(0\). Therefore, the differentiated equation is:\[ 2y \frac{dy}{dx} + 2 \frac{dy}{dx} - 2 = 0 \].

Step 3: Solve for dy/dx

We solve for \( \frac{dy}{dx} \): \[ (2y + 2) \frac{dy}{dx} = 2 \]\[ \frac{dy}{dx} = \frac{2}{2y + 2} \]. This is the first derivative.

Step 4: Simplify first derivative

Further simplify the expression for \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = \frac{2}{2(y + 1)} = \frac{1}{y + 1} \]. This simplification is the expression we'll use to find the second derivative.

Step 5: Differentiate dy/dx implicitly to find d²y/dx²

Differentiate \( \frac{1}{y + 1} \) with respect to \(x\). This requires using the chain rule:\[ \frac{d}{dx} \left( \frac{1}{y + 1} \right) = -\frac{1}{(y + 1)^2} \cdot \frac{dy}{dx} \]. Substitute \( \frac{dy}{dx} = \frac{1}{y + 1} \) into the expression:\[ \frac{d^2y}{dx^2} = -\frac{1}{(y + 1)^2} \cdot \frac{1}{y + 1} = -\frac{1}{(y + 1)^3} \]. This is the second derivative.

Key Concepts

First Derivative

In the context of finding the first derivative using implicit differentiation, we aim to determine how one variable changes with respect to another. In our given exercise, we want to find how the variable \( y \) changes with respect to \( x \).
\( \frac{dy}{dx} \) is the notation used to denote this first derivative. During implicit differentiation, we treat \( y \) as a function of \( x \). So, differentiating \( y^2 \) gives us \( 2y \frac{dy}{dx} \), while differentiating terms with just \( x \), like \(-2x\), gives us their derivative with respect to \( x \) directly, which is \(-2\) in this instance.
In the rearranged equation from the problem, this process leads to the derived formula:

• \( 2y \frac{dy}{dx} + 2 \frac{dy}{dx} - 2 = 0 \)

Solving for \( \frac{dy}{dx} \), we obtain:

• \( \frac{dy}{dx} = \frac{1}{y + 1} \)

This expression signifies the rate at which \( y \) changes with respect to \( x \) in the equation provided.

Second Derivative

Finding the second derivative \( \frac{d^2y}{dx^2} \) involves differentiating the first derivative again. It gives insight into the curvature or concavity of the graph and tells us about the acceleration of the variable change.
Given the first derivative from the previous section, \( \frac{dy}{dx} = \frac{1}{y + 1} \), finding the second derivative requires recognizing it as a composite function, meaning we will need to apply the chain rule.To differentiate \( \frac{1}{y + 1} \) with respect to \( x \), we start with the chain rule. The chain rule states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function.
For our derivative \( \frac{1}{y + 1} \), the outer function is \( \frac{1}{u} \) and its derivative is \(-\frac{1}{u^2}\). Therefore:

• \( \frac{d}{dx} \left( \frac{1}{y + 1} \right) = -\frac{1}{(y + 1)^2} \cdot \frac{dy}{dx} \)

Substituting the first derivative \( \frac{dy}{dx} = \frac{1}{y + 1} \) into this expression:

• \( \frac{d^2y}{dx^2} = -\frac{1}{(y + 1)^3} \)

This result indicates the nature of curvature or the rate of change of the slope for the original equation.

Chain Rule

The chain rule is a fundamental technique in calculus, particularly useful in implicit differentiation. It simplifies finding the derivative of a composite function.
A composite function is a function made up of two or more functions, where the output of one function becomes the input of another.To apply the chain rule, you differentiate the outer function and multiply by the derivative of the inner function.For example, consider differentiating \( f(g(x)) \). Here, \( f \) is the outer function, and \( g(x) \) is the inner function. The derivative is:

• \( f'(g(x)) \times g'(x) \)

In the provided exercise, we apply the chain rule when differentiating terms like \( y^2 \), since \( y \) is not just \( x \) but a function of \( x \). Hence, the derivative of \( y^2 \) involves \( 2y \cdot \frac{dy}{dx} \), linking the processes of implicit differentiation with the chain rule.
This interconnection allows us to effectively differentiate and find higher derivatives, ensuring accurate results in scenarios involving complex relationships between variables.