Problem 9 - The Calculus with Analytic Geometry

Question

Find \(D_{x} y\) by implicit differentiation. $$ x^{2} y^{2}=x^{2}+y^{2} $$

Step-by-Step Solution

Answer

\( \frac{d y}{dx} = \frac{x(1 - y^2)}{x^2 y - y} \)

Step 1 - Differentiate Both Sides with Respect to x

Differentiate the left-hand side and the right-hand side of the equation implicitly with respect to x. Remember to use the product rule and chain rule when necessary.Given: the original equation is \[ x^2 y^2 = x^2 + y^2 \]Differentiating both sides with respect to x:\[ D_{x}(x^2 y^2) = D_{x}(x^2) + D_{x}(y^2) \]

Step 2 - Apply the Product Rule to the Left-Hand Side

On the left-hand side, apply the product rule: \[ D_{x}(x^2 y^2) = D_{x}(x^2) \times y^2 + x^2 \times D_{x}(y^2) \]which becomes:\[ 2x \times y^2 + x^2 \times 2y \frac{d y}{dx} \]

Step 3 - Differentiate the Right-Hand Side

On the right-hand side, differentiate term by term:\[ D_{x}(x^2) + D_{x}(y^2) = 2x + 2y \frac{d y}{dx} \]

Step 4 - Equate and Simplify

Combine the results from Steps 2 and 3 to form an equation:\[ 2x y^2 + 2x^2 y \frac{d y}{dx} = 2x + 2y \frac{d y}{dx} \]Simplify the equation:\[ 2x y^2 + 2x^2 y \frac{d y}{dx} = 2x + 2y \frac{d y}{dx} \]

Step 5 - Group Like Terms and Solve for \( \frac{dy}{dx} \)

Collect all terms involving \( \frac{d y}{dx} \) on one side and factor them out:\[ 2x^2 y \frac{d y}{dx} - 2y \frac{d y}{dx} = 2x - 2x y^2 \]Factor out \( \frac{d y}{dx} \):\[ (2x^2 y - 2y) \frac{d y}{dx} = 2x(1 - y^2) \]Finally, solve for \( \frac{d y}{dx} \):\[ \frac{d y}{dx} = \frac{2x(1 - y^2)}{2(x^2 y - y)} \]Simplify further:\[ \frac{d y}{dx} = \frac{x(1 - y^2)}{x^2 y - y} \]

Key Concepts

Product Rule

When differentiating a product of two functions, we use the product rule. The product rule states that if you have two functions, say \(u(x)\) and \(v(x)\), the derivative of their product, \(u(x)v(x)\), is given by:
\[ D_x[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \]
In our exercise, \(x^2 y^2\), we treat \(x^2\) as \(u\) and \(y^2\) as \(v\).
Using the product rule, we find:
\[ D_x[x^2 y^2] = (2x) y^2 + x^2 (2y \frac{dy}{dx}) \]
This helps us differentiate complex expressions step-by-step by breaking them into simpler parts. Apply the rule whenever you encounter products of functions.

Chain Rule

The chain rule is essential when differentiating compositions of functions. If you have a function \(f(g(x))\), the chain rule states that the derivative of \(f(g(x))\) with respect to \(x\) is:
\[ \frac{d}{dx} f(g(x)) = f'(g(x)) \times g'(x) \]
In our exercise, when differentiating \(y^2\) with respect to \(x\), we treat \(y\) as a function of \(x\). Therefore, the derivative of \(y^2\) using the chain rule becomes:
\[ \frac{d}{dx}[y^2] = 2y \frac{dy}{dx} \]
This is because we are treating \(y\) as \(g(x)\), and \(y^2\) as \(f(g(x))\). The chain rule makes it possible to deal with derivatives inside another function.

Derivative Simplification

After applying the product and chain rules, we aim to simplify the expression for ease of interpretation. Simplification involves combining like terms and solving for \( \frac{dy}{dx} \).
From our example, after differentiating both sides and applying the rules, we have:
\[ 2xy^2 + 2x^2y \frac{dy}{dx} = 2x + 2y \frac{dy}{dx} \]
We then group all \( \frac{dy}{dx} \) terms on one side:
\[ 2x^2y \frac{dy}{dx} - 2y \frac{dy}{dx} = 2x - 2xy^2 \]
Factor out \( \frac{dy}{dx} \):
\[ (2x^2y - 2y) \frac{dy}{dx} = 2x(1 - y^2) \]
Finally, solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{2x(1 - y^2)}{2(x^2y - y)} \]
And simplify to get:
\[ \frac{dy}{dx} = \frac{x(1 - y^2)}{x^2y - y} \]
This process ensures the derivative is in its simplest form, making it easier to understand and use.