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The chain rule is a fundamental technique in differentiation that helps us take the derivative of composite functions. A composite function is, simply put, a function made up of other functions. Think of it as a function within a function.

Here's the formula for the chain rule in a simple layout:

• If you have a function written as \(f(g(x))\), the derivative is \(f'(g(x)) \cdot g'(x)\).

Imagine you're peeling an onion — the chain rule helps you differentiate each layer.

In the exercise, we used the chain rule during the differentiation of the term \(\sqrt{xy}\). Since \(\sqrt{xy}\) is really \((xy)^{1/2}\), we treated \((xy)\) as \(u\) and applied the chain rule: first, differentiate \(u^{1/2}\) with respect to \(u\), and then multiply by the derivative of \(xy\) with respect to \(x\). This is crucial when dealing with more complex expressions where one function is nestled inside another.

The product rule comes into play when you're differentiating the product of two or more functions. Essentially, it tells us how to handle the multiplication of two functions when we need to find their derivative.

The simple rule to remember for products is:

• If \(u = f(x) \cdot g(x)\), then \(\frac{d}{dx}(u) = f(x) g'(x) + g(x) f'(x)\).
  

In our exercise, we applied the product rule to the term \(xy\). Here, \(x\) acts as one function and \(y\) as another. By multiplying each by the derivative of the other, we get: \(x \cdot \frac{dy}{dx} + y \cdot 1\).

This step ensures that we accurately capture how the relationship between \(x\) and \(y\) impacts their derivative. It's a vital tool especially when dealing with expressions where variables are multiplied together.

Implicit differentiation is used when dealing with equations where \(y\) is not isolated on one side. Instead of solving for \(y\) and then differentiating, we differentiate both sides of the equation respecting every occurrence of \(y\) as \(f(x)\).

When differentiating \(y\) with respect to \(x\), we treat \(y\) as if it were \(f(x)\). Thus, each time we encounter \(y\), we multiply by \(\frac{dy}{dx}\) to account for its dependency on \(x\).

In this particular problem, the use of implicit differentiation allows us to differentiate the entire equation \(x^2 + \sqrt{xy} = 7\) directly. We invoke implicit differentiation because \(y\) is entwined with \(x\) within the square root function, making it tricky to separate them without this technique. As such, implicit differentiation is indispensable in navigating through situations where \(y\) is embedded within complex equations, ensuring accurate results by taking all variables and their interdependencies into account.