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The chain rule is a fundamental concept in calculus used when differentiating composite functions. It helps us find the derivative of a function that is nested inside another function.
Let's say you have a function in the form of \( u(x) = f(g(x)) \). The chain rule tells us how to differentiate this by multiplying the derivative of the outer function \( f \) by the derivative of the inner function \( g \). Specifically, the chain rule formula is:

• \( \frac{du}{dx} = f'(g(x)) \cdot g'(x) \)

This rule is particularly useful in implicit differentiation, as seen in our problem, where \( y \) is implicitly defined as a function of \( x \). So, when differentiating terms like \( y^2 \), we treat \( y \) as \( g(x) \), requiring the use of the chain rule. Differentiating \( y^2 \), you apply the rule as follows: \( \frac{d}{dx}[y^2] = 2y \cdot \frac{dy}{dx} \). This highlights how critical chain rule is when dealing with implicit functions.

The product rule is essential when differentiating products of two or more functions. If you have a product of two functions \( f(x) \) and \( g(x) \), the product rule tells us that the derivative is not simply the product of their derivatives. Here's the product rule formula:

• \( \frac{d}{dx}[f(x)g(x)] = f(x) \cdot g'(x) + g(x) \cdot f'(x) \)

In our exercise, the term \( 2xy \) involves two functions of \( x \): the constant \( 2x \) and \( y \). Applying the product rule: \( \frac{d}{dx}[2xy] = 2 \cdot y + 2x \cdot \frac{dy}{dx} \). This application is vital as you can't differentiate each term separately; they need to be treated as one product.

Derivatives represent the rate at which a function is changing at any given point, essentially measuring its slope.
They are the basis of calculus and critical in overestimating and predicting behavior in models.
In the context of our implicit differentiation challenge:

• The derivative of \( x \) with respect to \( x \) is \( 1 \) since \( x \) changes at a constant rate of itself.
• For \( y \), the derivative is \( \frac{dy}{dx} \) because \( y \) is considered a function of \( x \).

Derivatives help transform an equation from its implicit form into a form where one can solve for \( \frac{dy}{dx} \), as seen in our example.

Once you have differentiated both sides of an equation using the appropriate rules, you often need to solve for \( \frac{dy}{dx} \). This is usually the main goal in exercises involving implicit differentiation. The equation that arises can often be complex, requiring steps to isolate terms with \( \frac{dy}{dx} \).
In our case, after differentiating, you should aim to arrange all \( \frac{dy}{dx} \) terms on one side of the equation:

• First, collect like terms: \( 2x \frac{dy}{dx} + 2y \frac{dy}{dx} - \frac{dy}{dx} = 1 - 2y \).
• Then factor out \( \frac{dy}{dx} \): \( \frac{dy}{dx}(2x + 2y - 1) = 1 - 2y \).
• Finally, solve for \( \frac{dy}{dx} \) by dividing: \( \frac{dy}{dx} = \frac{1 - 2y}{2x + 2y - 1} \).

These steps ensure that you isolate \( \frac{dy}{dx} \), resolving the derivative into a usable form.