Problem 24 - APEX Calculus

Question

In Exercises \(23-26,\) find \(\frac{d y}{d x}\) using Implicit Differentiation and Theorem \(12.5 .3 .\) $$ \left(3 x^{2}+2 y^{3}\right)^{4}=2 $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = -\frac{x}{y^2} \)

Step 1: Identify Implicit Function

The equation \( (3x^2 + 2y^3)^4 = 2 \) is given in implicit form. We must find \( \frac{dy}{dx} \) using implicit differentiation, a technique used when \( y \) is given implicitly as a function of \( x \).

Step 2: Apply Implicit Differentiation

Differentiate both sides of the equation with respect to \( x \). On the left side, use the chain rule. Let \( u = 3x^2 + 2y^3 \), so the equation becomes \( u^4 = 2 \). Differentiating with respect to \( x \) gives: \( 4u^3 \frac{du}{dx} = 0 \), since the derivative of a constant is zero.

Step 3: Find \( \frac{du}{dx} \) Using the Chain Rule

We need to find the derivative of \( u = 3x^2 + 2y^3 \) with respect to \( x \). Thus, \( \frac{du}{dx} = \frac{d}{dx}(3x^2) + \frac{d}{dx}(2y^3) \).

Step 4: Differentiate \( 3x^2 \) and \( 2y^3 \) Separately

The derivative of \( 3x^2 \) with respect to \( x \) is \( 6x \). For the term \( 2y^3 \), use the chain rule: \( \frac{d}{dx}(2y^3) = 6y^2 \frac{dy}{dx} \).

Step 5: Substitute \( \frac{du}{dx} \) into Equation

Substitute \( \frac{du}{dx} = 6x + 6y^2 \frac{dy}{dx} \) back into the differentiated equation: \( 4u^3 (6x + 6y^2 \frac{dy}{dx}) = 0 \).

Step 6: Solve for \( \frac{dy}{dx} \)

Simplify and solve for \( \frac{dy}{dx} \): \( 24u^3 x + 24u^3 y^2 \frac{dy}{dx} = 0 \). This rearranges to \( 24u^3 y^2 \frac{dy}{dx} = -24u^3 x \) or \( \frac{dy}{dx} = -\frac{u^3 x}{u^3 y^2} = -\frac{x}{y^2} \).

Step 7: Express in Terms of Original Variables

Recall that \( u = 3x^2 + 2y^3 \). Substitute \( u \) back in terms of its original variables. However, in this case, it cancels since it appears in both terms of the fraction.

Key Concepts

Chain Rule

The chain rule is a powerful differentiation technique used when dealing with composites of functions. It's especially useful in implicit differentiation. In implicit differentiation, you often find functions nested within other functions. The chain rule helps break down these functions into more manageable parts.
For example, consider a function in the format \( (g(x))^n \). To differentiate it, you would use the chain rule by setting it in terms of a simpler function: let \( u = g(x) \), then express the original function as \( u^n \).

• Differentiate \( u^n \) with respect to \( u \) to get \( n u^{n-1} \).
• Then differentiate \( u = g(x) \) with respect to \( x \).
• Combine these using multiplication, i.e., \( n u^{n-1} \cdot g'(x) \).

This method effectively breaks down the derivative into smaller steps, simplifying the differentiation process of complex expressions.

Derivative of a Constant

Understanding the derivative of a constant is fundamental in calculus. A constant is a value that doesn't change, like the number 2 in our exercise's original equation.
The derivative of any constant is zero. This is because differentiation measures how a function changes, and constants, by definition, do not change.
For example, if you have a constant \( c \), then \( \frac{d}{dx}(c) = 0 \). This principle simplifies the differentiation process, as seen in our step-by-step solution.
In the given problem \( (3x^2 + 2y^3)^4 = 2 \), after applying the chain rule, we differentiate 2—the constant on the right side—to get zero. This action effectively removes the constant from influencing further calculations. Recognizing and applying this principle helps streamline your differentiation work.

Implicit Function

An implicit function is one where the variables \( x \) and \( y \) are intertwined in an equation, making it difficult or impossible to solve for \( y \) solely in terms of \( x \). Implicit differentiation allows you to find derivatives without explicitly solving for one variable.
In our exercise, the equation \( (3x^2 + 2y^3)^4 = 2 \) embodies an implicit function. Here \( y \) is not isolated but still depends on \( x \). To differentiate such equations, implicit differentiation considers all terms involving \( y \) and \( x \) simultaneously. You differentiate with respect to \( x \) while treating \( y \) as a function of \( x \).
Employing implicit functions is particularly helpful for complex equations where one variable naturally depends on the other, even if not explicitly shown.

Differentiation Techniques

Differentiation techniques are the backbone of calculus, allowing us to determine rates of change and slopes of curves. Numerous techniques exist, each with a specific purpose.
For implicit differentiation, especially where both \( x \) and \( y \) are present within an equation, these methods are vital. Some key techniques include:

• Product Rule: Used when differentiating products of two functions.
• Quotient Rule: Applied to differentiate the division of one function by another.
• Chain Rule: Effective for nested functions, as we've seen in step-by-step solutions.

In our original problem, applying implicit differentiation alongside the chain rule illustrates how these techniques simplify finding \( \frac{dy}{dx} \). By breaking down each piece of the equation using these rules, we obtain the desired derivative with precision and clarity. Mastery of these techniques equips learners to tackle a wide range of calculus problems.