Problem 33 - Single Variable Calculus

Question

find \(d^{2} y / d x^{2}\) in terms of \(x\) and \(y\). $$ x y+x^{3}=4 $$

Step-by-Step Solution

Answer

The second derivative \(\frac{d^2y}{dx^2}\) in terms of \(x\) and \(y\) is: \(\frac{d^2y}{dx^2} = -6\)

Step 1: Differentiate the equation implicitly

To find \(\frac{dy}{dx}\), we will differentiate both sides of the given equation with respect to \(x\): \(\frac{d}{dx}(xy+x^3) = \frac{d}{dx}(4)\) Apply the product rule for differentiating the term \(xy\): \((\frac{d}{dx}(x))\cdot y + x(\frac{d}{dx}(y)) + \frac{d}{dx}(x^3) = 0\)

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

Now, we will solve for \(\frac{dy}{dx}\) from the differentiated equation: \(y + x\frac{dy}{dx} + 3x^2 = 0\) \(\frac{dy}{dx} = -\frac{y}{x} - 3x\)

Step 3: Differentiate the first derivative implicitly to find the second derivative \(\frac{d^2y}{dx^2}\)

Differentiate \(\frac{dy}{dx}\) with respect to \(x\): \(\frac{d^2y}{dx^2} = \frac{d}{dx}(-\frac{y}{x} - 3x)\) #echo the first derivative equation for easy reference: The first derivative is \(\frac{dy}{dx} = -\frac{y}{x} - 3x\) Now we'll apply the quotient rule, differentiating \(-\frac{y}{x}\), and the power rule for differentiating \(-3x\): \(\frac{d^2y}{dx^2} = -\frac{x\frac{dy}{dx} - y}{x^2} - 3\)

Step 4: Substitute the first derivative into the second derivative expression and simplify

Substitute the expression for \(\frac{dy}{dx}\) back into the second derivative equation: \(\frac{d^2y}{dx^2} = -\frac{x(-\frac{y}{x} - 3x) - y}{x^2} - 3\) Now, we can simplify to get the final expression for the second derivative: \(\frac{d^2y}{dx^2} = \frac{y - x(3x) - y}{x^2} - 3\) \(\frac{d^2y}{dx^2} = \frac{-3x^2}{x^2} - 3\) \(\frac{d^2y}{dx^2} = -3 - 3\)

Step 5: State the final expression for the second derivative

The final expression for the second derivative \(\frac{d^2y}{dx^2}\) in terms of \(x\) and \(y\) is: \(\frac{d^2y}{dx^2} = -6\)

Key Concepts

Product Rule

Whenever we face a function involving two variables being multiplied, such as \(xy\), we need to apply the product rule for differentiation. This rule tells us how to differentiate products of two functions, say \(u(x)\) and \(v(x)\). According to the product rule, the derivative is:

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

In the exercise, we applied the product rule to differentiate \(xy\). Here, \(x\) is treated as one function and \(y\) as another. So when differentiating \(xy\), it becomes:

• \(\frac{d}{dx}(x) \cdot y + x \cdot \frac{d}{dx}(y)\)

This careful application of the product rule is crucial for solving problems in implicit differentiation correctly.

Quotient Rule

The quotient rule is another essential tool in differentiation, especially when dealing with a function that is a ratio of two expressions. In the problem, we used this rule when handling \(-\frac{y}{x}\). The formula for the quotient rule is:

• \(\frac{d}{dx}\left[ \frac{u}{v} \right] = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}\)

For our function, differentiating \(-\frac{y}{x}\) means \(u = y\) and \(v = x\). By the quotient rule, this gives us:

• \(-\frac{x \cdot \frac{dy}{dx} - y \cdot 1}{x^2}\)

It's highly beneficial in problems with fractions in calculus, ensuring that each term is treated appropriately.

Second Derivative

Finding a second derivative means differentiating the first derivative once more. This requires the application of differentiation rules all over again. In our problem, we started with the first derivative \(\frac{dy}{dx} = -\frac{y}{x} - 3x\). The step involves repeating a similar process with the first derivative as we did with the function originally.

• First, differentiate \(-\frac{y}{x}\) using the quotient rule.
• Then differentiate \(-3x\) normally, which is simple due to it being a linear term.

After these operations, we plug \(\frac{dy}{dx}\) into the expression we derived to find \(\frac{d^2y}{dx^2}\). Successfully computing the second derivative often reveals information about the curvature and concavity of the curve described by the function.

First Derivative

The first derivative of a function with respect to \(x\) gives us the rate at which the function value changes as \(x\) changes. For implicit differentiation, we must consider the relationships between dependent and independent variables as expressed through an equation involving both. In the exercise, we found the first derivative by:

• Applying the product rule to the term \(xy\).
• Adding the derivative of \(x^3\), which is \(3x^2\).

Combining terms properly allowed us to solve for \(\frac{dy}{dx}\), which came out as \(-\frac{y}{x} - 3x\). The first derivative helps analyze slope, growth, and decreasing intervals of functions, providing crucial insights into function behavior.