Problem 15 - Calculus - AP Edition

Question

Find \(d^{2} y / d x^{2}\) by implicit differentiation. $$x^{3} y^{3}-4=0$$

Step-by-Step Solution

Answer

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

Step 1: Differentiate implicitly with respect to x

Let's begin by differentiating both sides of the given equation with respect to x. The equation is: \[ x^3 y^3 - 4 = 0 \]Differentiate the left side using the product rule, where we consider \(u = x^3\) and \(v = y^3\). According to the product rule: \[ (uv)' = u'v + uv' \]Thus,\[ d/dx(x^3 y^3) = (x^3)' y^3 + x^3 (y^3)' \]The derivatives are \[ (x^3)' = 3x^2 \] and \[ (y^3)' = 3y^2 (dy/dx) \] (by the chain rule). Substitute into the product rule formula:\[ 3x^2 y^3 + x^3 (3y^2 dy/dx) = 0 \].This simplifies to:\[ 3x^2 y^3 + 3x^3 y^2 dy/dx = 0 \].

Step 2: Solve for dy/dx

Now, we solve the differentiated equation for \(dy/dx\):\[ 3x^2 y^3 + 3x^3 y^2 dy/dx = 0 \]Subtract \(3x^2 y^3\) from both sides:\[ 3x^3 y^2 dy/dx = -3x^2 y^3 \]Divide both sides by \(3x^3 y^2\) to isolate \(dy/dx\):\[ dy/dx = -\frac{x^2 y^3}{x^3 y^2} \]Simplify the expression by canceling \(x^2\) and \(y^2\) where possible:\[ dy/dx = -\frac{y}{x} \]

Step 3: Differentiate dy/dx implicitly to find d²y/dx²

Now, we will differentiate \(dy/dx = -\frac{y}{x}\) with respect to \(x\) to find \(d^2y/dx^2\).Rewrite \(dy/dx\) as: \[ dy/dx = -y/x \]Differentiate both sides:\[ \frac{d}{dx}(-y/x) = d^2y/dx^2 \]Use the quotient rule, \( (v/u)' = (v'u - vu')/u^2 \), where \(v = -y\) and \(u = x\):- \(v' = -dy/dx\)- \(u' = 1\)Substitute these into the formula:\[ d^2y/dx^2 = -\frac{(x)(-dy/dx) - (-y)(1)}{x^2} \]Simplify:\[ = \frac{x(dy/dx) - y}{x^2} \]Substitute \( dy/dx = -y/x \) into the equation:\[ d^2y/dx^2 = \frac{x(-y/x) - y}{x^2} \]This simplifies to:\[ d^2y/dx^2 = \frac{-y - y}{x^2} = \frac{-2y}{x^2} \]

Step 4: Verify and organize the final result

The second derivative is obtained as follows:After all calculations and simplifications, we have:\[ d^2y/dx^2 = \frac{-2y}{x^2} \]This result is simply indicating that the rate of change of the slope with respect to \(x\) is proportional to \(-2y/x^2\).

Key Concepts

Second Derivative

The concept of the second derivative is about finding the rate at which the first derivative changes. Think of it as the "acceleration" of the rate of change. While the first derivative represents the slope of the tangent line to the curve or the rate of change of the function, the second derivative gives us information about how this slope itself is changing.
For example:

• If the second derivative is positive, it means that the slope is increasing, suggesting that the curve is concave up.
• If it is negative, the slope is decreasing, and the curve is concave down.

In solving for the second derivative using implicit differentiation, like we did with the equation \(x^3 y^3 - 4 = 0\), we first find \(dy/dx\). Then, we differentiate \(dy/dx\) again to find \(d^2y/dx^2\). This involves using additional rules like the chain rule and product rule. All these steps allow us to understand how quickly our initial slope \(dy/dx\) itself is changing.

Product Rule

The product rule is a very handy differentiation rule that helps us handle functions that are products of two distinct parts. The rule states that if you have two functions, \(u\) and \(v\), then the derivative of their product \(uv\) is given by \((uv)' = u'v + uv'\).

In our problem, the term \(x^3 y^3\) is such a product where \(u = x^3\) and \(v = y^3\). This requires applying the product rule to find the derivative of \(x^3 y^3\) with respect to \(x\).
Here is a simple breakdown:

• Differentiate \(x^3\) to get \(3x^2\).
• Differentiate \(y^3\) using the chain rule (discussed next) to get \(3y^2 \frac{dy}{dx}\).
• Apply the product rule: substitute these derivatives into \((x^3)' y^3 + x^3 (y^3)'\), which simplifies to \(3x^2 y^3 + 3x^3 y^2 \frac{dy}{dx}\).

Thus, the product rule lets us simplify finding derivatives of products, easing the computation for larger expressions.

Chain Rule

The chain rule is essential when differentiating composite functions, where one function is nested within another. The rule allows you to handle situations where variables are interdependent and can help when dealing with implicit differentiation.

In simple terms, if you have a composition of functions \(f(g(x))\), the chain rule states that \(\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\).

In the context of our example:

• The term \(y^3\) can be seen as a composition where \(y\) is implicitly a function of \(x\).
• To differentiate \(y^3\), you recognize \(f(y) = y^3\) and use the chain rule to find that \(\frac{d}{dx}[y^3] = 3y^2 \cdot \frac{dy}{dx}\).

This was crucial for successfully using the product rule on the term \(x^3 y^3\). The chain rule helps us correctly identify how changes in \(x\) impact \(y\) and its powers when \(y\) is not explicitly given as a function of \(x\). With implicit differentiation, the chain rule becomes an indispensable tool.