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. $$x^{4}+y^{2}+y=7$$

Short answer

\( \frac{d^2y}{dx^2} = \frac{-24x^2(2y + 1) + 8x^6}{(2y + 1)^2} \)

Step-by-step solution

Differentiate Implicitly with Respect to x

To find the first derivative \( \frac{dy}{dx} \), we differentiate each term of the equation \( x^4 + y^2 + y = 7 \) with respect to \( x \). Applying the chain rule to \( y^2 \) and \( y \), we get:\[ \frac{d}{dx}(x^4) + \frac{d}{dx}(y^2) + \frac{d}{dx}(y) = \frac{d}{dx}(7) \]This results in:\[ 4x^3 + 2y \frac{dy}{dx} + \frac{dy}{dx} = 0 \]Simplify to:\[ 4x^3 + (2y + 1) \frac{dy}{dx} = 0 \]

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

Rearrange the equation from Step 1 to solve for \( \frac{dy}{dx} \):\[ (2y + 1) \frac{dy}{dx} = -4x^3 \]Divide both sides by \( 2y + 1 \):\[ \frac{dy}{dx} = \frac{-4x^3}{2y + 1} \]

Differentiate Again to Find \( \frac{d^2y}{dx^2} \)

Now differentiate \( \frac{dy}{dx} = \frac{-4x^3}{2y + 1} \) implicitly with respect to \( x \). Use the quotient rule:For \( \frac{u}{v} \), where \( u = -4x^3 \) and \( v = 2y + 1 \), the derivative is:\[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \]Calculate \( \frac{du}{dx} = -12x^2 \) and \( \frac{dv}{dx} = 2 \frac{dy}{dx} \).

Substitute and Simplify

Substitute \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) into the quotient rule:\[ \frac{d^2y}{dx^2} = \frac{(2y + 1)(-12x^2) + 4x^3(2 \cdot \frac{dy}{dx})}{(2y + 1)^2} \]Substitute \( \frac{dy}{dx} = \frac{-4x^3}{2y + 1} \):\[ \frac{d^2y}{dx^2} = \frac{(2y + 1)(-12x^2) + 4x^3 \left(2 \cdot \frac{-4x^3}{2y + 1}\right)}{(2y + 1)^2} \]

Simplify Further

Simplify:\[ \frac{d^2y}{dx^2} = \frac{-24x^2(2y + 1) + 8x^6}{(2y + 1)^2} \]Further simplification will result in:\[ \frac{d^2y}{dx^2} = \frac{-24x^2(2y + 1) + 8x^6}{(2y + 1)^2} \]This is the expression for \( \frac{d^2y}{dx^2} \).

Key concepts

Second Derivative

The second derivative gives us insights into the curvature or concavity of a graph at a given point. When dealing with an implicitly defined function, like the one in our problem, finding the second derivative can involve multiple steps. We first found the first derivative using implicit differentiation. This step is crucial because the second derivative is simply the derivative of the first derivative.

Once we have the expression for the first derivative, we differentiate it again to find the second derivative. In this exercise, this involves using more complex rules due to the implicit nature of the function. Finding the second derivative of an implicitly defined function allows us to understand more about how the function behaves, particularly how it changes rate of change (acceleration) at any given point.

Chain Rule

The chain rule is a fundamental tool in calculus for differentiating complex expressions. It is particularly useful when dealing with composite functions. In our implicit differentiation problem, the chain rule helps us differentiate terms like \(y^2\) and \(y\) with respect to \(x\).

Since \(y\) is a function of \(x\), when we differentiate \(y^2\), we actually differentiate two parts: the outer function \(z^2 = (y)^2\) and the inner function \(y\). The chain rule tells us to multiply the derivative of the outside function by the derivative of the inside function. This results in terms like \(2y \cdot \frac{dy}{dx}\), where \(\frac{dy}{dx}\) is the derivative of \(y\) with respect to \(x\).

Using the chain rule effectively simplifies and provides the correct method for differentiating each part of the implicit equation with respect to \(x\).

Quotient Rule

The quotient rule is needed when differentiating expressions where one function is divided by another. In our exercise, it comes into play when finding the second derivative, as the first derivative was a fraction: \(\frac{dy}{dx} = \frac{-4x^3}{2y + 1}\).

The quotient rule formula is \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}\). This is applied by choosing \(u = -4x^3\) and \(v = 2y + 1\).
We first find the derivatives \(\frac{du}{dx} = -12x^2\) and \(\frac{dv}{dx} = 2 \cdot \frac{dy}{dx}\). These are then substituted back into the quotient rule, and simplified to obtain the second derivative. The quotient rule is essential for streamlining the differentiation process strategically, dealing with fractions.

Implicitly Defined Functions

Implicitly defined functions are functions where \(y\) is defined indirectly in terms of \(x\). In these cases, \(y\) might not be expressed directly as \(y = f(x)\), but rather through a relationship like \(x^4 + y^2 + y = 7\).

To handle such cases, implicit differentiation is used. We differentiate each term of the equation separately with respect to \(x\). For terms involving \(y\), we apply the chain rule to capture their dependent relationship with \(x\).
This approach allows us to effectively find the derivatives of complex, indirectly defined relationships by treating \(x\) and \(y\) as mutually interconnected through their shared equation. This method is particularly powerful in scenarios where explicit diversification is cumbersome or impossible.