Problem 308 - Calculus Volume 1

Question

Use implicit differentiation to find \(\frac{d y}{d x}\). $$ (x y)^{2}+3 x=y^{2} $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{-2xy^2 - 3}{2x^2y - 2y} \)

Step 1: Differentiate Both Sides with Respect to x

Start by differentiating both sides of the equation with respect to \(x\). This involves using the chain rule and the product rule since we have \((xy)^2\).\\[\frac{d}{dx}\left((xy)^2\right) + \frac{d}{dx}(3x) = \frac{d}{dx}(y^2)\]

Step 2: Apply the Product Rule to \((xy)^2\)

Using the product rule, differentiate \((xy)^2\) with respect to \(x\). This gives: \\[ \begin{align*} \frac{d}{dx} \left((xy)^2\right) &= 2(xy) \cdot \frac{d}{dx}(xy) \&= 2(xy) \cdot \left(y + x \frac{dy}{dx}\right) \ \end{align*} \]

Step 3: Differentiate the Remaining Terms

Differentiate the linear term \(3x\) and \(y^2\) with respect to \(x\). \\[\frac{d}{dx}(3x) = 3\] \For \(y^2\), apply the chain rule: \\[ \frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx} \]

Step 4: Combine the Results

Substitute these derivatives back into the equation: \\[ 2(xy)(y + x \frac{dy}{dx}) + 3 = 2y \frac{dy}{dx} \]

Step 5: Solve for \(\frac{dy}{dx}\)

Collect all terms involving \(\frac{dy}{dx}\) on one side and factor it out: \\[2x^2y \frac{dy}{dx} - 2y \frac{dy}{dx} = -2xy^2 - 3\] \Reorganize to express it in terms of \(\frac{dy}{dx}\): \\[ \frac{dy}{dx} (2x^2y - 2y) = -2xy^2 - 3 \]

Step 6: Final Calculation of \(\frac{dy}{dx}\)

Divide both sides by the factor of \(\frac{dy}{dx}\) to isolate it: \\[ \frac{dy}{dx} = \frac{-2xy^2 - 3}{2x^2y - 2y} \]

Key Concepts

The Chain Rule

The chain rule is a powerful tool in calculus used to differentiate compositions of functions. It simplifies the process of handling complex expressions by breaking down differentiation into more manageable steps. In the equation \((xy)^2 + 3x = y^2\), the chain rule is key in differentiating the term \(y^2\). Initially, you identify an "inner function" and an "outer function." Here, consider \(y\) as the inner function and applying power as the outer. You first differentiate the outer function, then multiply by the derivative of the inner function:

• Differentiate the outer function (power): \(2y\)
• Multiply by the derivative of the inner function \(y\) w.r.t. \(x\): \(\frac{dy}{dx}\)

This results in \(\frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}\). Using the chain rule ensures accurate differentiation when variables are functions themselves.

The Product Rule

The product rule is particularly useful when differentiating products of two or more functions. It is cornerstone to handling products like \((xy)^2\) in our original equation. The product rule states that if you have two functions \(u\) and \(v\), then the derivative of their product is:
\[\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}\]
In our case, \(u = x\) and \(v = y\), making \(uv = xy\). We need to differentiate \((xy)^2\):

• First, apply the outer function differentiation: \(2(xy)\)
• Then, use the product rule on the inside part \(\frac{d}{dx}(xy) = y + x \cdot \frac{dy}{dx}\)

Finally, combine these using the product rule result to get \(\frac{d}{dx}((xy)^2) = 2(xy)(y + x \frac{dy}{dx})\).
This allows you to deal with each function sequentially and efficiently.

Differential Equations

Differential equations express relationships involving derivatives of functions, often modeling real-world phenomena. In the problem, we use implicit differentiation to find \(\frac{dy}{dx}\), treating \(y\) as an implicit function of \(x\). Here, we're solving a differential equation derived from an implicit function:

Differential equations have solutions that may describe rates of change, growth processes, or dynamic systems. Finding \(\frac{dy}{dx}\) involves isolating terms containing this derivative and manipulating the equation to express \(\frac{dy}{dx}\):

• First, differentiate the whole equation to express derivatives explicitly.
• Then, gather all terms involving \(\frac{dy}{dx}\) on one side of the equation.
• Solving this gives the expression for \(\frac{dy}{dx}\): \(\frac{-2xy^2 - 3}{2x^2y - 2y}\)

This result allows you to understand how changes in \(x\) affect \(y\), or vice versa, within the constraints of the given implicit relationship.