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The Chain Rule is a fundamental tool in calculus for differentiating compositions of functions. The basic idea is to take the derivative of the outer function with respect to the inner function, and then multiply it by the derivative of the inner function. This concept is especially important when dealing with complex expressions like \((3xy + 7)^2\) in implicit differentiation.

For this exercise, we apply the Chain Rule to the expression on the left side of the equation. First, we identify \((3xy + 7)^2\) as the outer function and \(3xy + 7\) as the inner function.

• The derivative of the outer function \((u^2)\) is \(2u\).
• The derivative of the inner function \(3xy + 7\) involves applying the product rule for \(3xy\) and then adding the derivative of 7, which is 0.
• Since \(3xy\) is a product of \(3x\) and \(y\), its derivative is \(3y + 3x \frac{dy}{dx}\), where \(\frac{dy}{dx}\) is the derivative of \(y\) with respect to \(x\).

Putting it all together gives \[2(3xy + 7)(3y + 3x \frac{dy}{dx})\]. This shows us how the Chain Rule helps to break down complicated functions into manageable parts.

Partial derivatives come into play when functions have more than one variable, as is often the case in implicit differentiation problems.

In this exercise, we differentiate a function that involves both \(x\) and \(y\), which requires us to find the derivative indirectly with respect to \(x\).

To find \(\frac{dy}{dx}\), we differentiate every term involving \(y\) partially with respect to \(x\), treating \(y\) not as a constant but as a function dependent on \(x\). This requires applying the Chain Rule to terms like \(3xy\):

• The term \(3xy\) is seen as a product of \(3x\) and \(y\). Here, \(x\) is a variable, and \(y\) is a function of \(x\).
• A differentiation approach is used, adding the Chain Rule component to account for both variables.
• This yields terms like \(3y\) and \(3x\frac{dy}{dx}\), clearly showing the influence of both \(x\) and \(y\) in the derivative calculation.

By managing these partial derivatives carefully, we solve the differential equation step-by-step, ultimately finding the implicit derivative.

Solving calculus problems, particularly those involving implicit differentiation, requires a blend of strategic thinking and mathematical technique.

In this exercise, the goal is to find \(\frac{dy}{dx}\), the derivative of \(y\) with respect to \(x\).

Here are some key steps:

• Differentiate Both Sides: Start by differentiating both sides of the equation to express \(\frac{dy}{dx}\) in terms of known quantities.
• Simplify and Rearrange: Collect all terms involving \(\frac{dy}{dx}\) on one side. This step often requires algebraic manipulation like distributing and factoring.
• Factor and Solve: Isolate \(\frac{dy}{dx}\) by factoring it out of the left-hand side of the equation. Then, solve for it by division.

Ensuring each of these steps is done methodically is key. For instance, in our solution, isolating \(\frac{dy}{dx}\) required careful arrangement and factoring, after which we could divide across the equation to solve for the derivative.

Integrating these strategies helps to handle complex calculus problems effectively, providing a clear pathway to the solution.