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Understanding multivariable calculus begins with recognizing that we are dealing with functions of more than one variable, like our example function for this exercise: \( z = \frac{x^{2}}{2y} + \frac{4y^{2}}{x} \). In multivariable calculus, partial derivatives play a crucial role in studying how a function changes in response to changes in one variable while keeping the others constant.

• The derivative with respect to \(x\) considers \(y\) as a constant, revealing how \(z\) changes with a slight change in \(x\).
• Conversely, the derivative with respect to \(y\) assumes \(x\) is constant, focusing on \(z\)'s change with variations in \(y\).

This concept is pivotal in scenarios involving multi-dimensional space, such as in physics and engineering, where numerous quantities interact simultaneously. Grasping the function's behavior through partial derivatives provides insights into the way various factors influence the system.

The quotient rule is vital when finding the derivative of a function that is the quotient of two other functions, represented as \( \frac{u(x)}{v(x)} \). For a clearer understanding, let's break down the rule: If \( u(x) \) and \( v(x) \) are differentiable, the derivative is given by:\[\frac{d}{dx}\left( \frac{u(x)}{v(x)} \right) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}\]In our exercise, we applied the quotient rule to differentiate the term \( \frac{4y^{2}}{x} \) with respect to \(x\).

• The function \(u(x) = 4y^{2}\) and its derivative with respect to \(x\) is \(0\) since \(y\) is constant in this scenario.
• The function \(v(x) = x\) has a derivative of \(1\).

Using the quotient rule, we found the derivative for this fraction simply as \( -\frac{4y^{2}}{x^{2}} \), showcasing its power for handling ratios in calculus.

The product rule is applied when differentiating the product of two functions. If you have functions \( u(y) \) and \( v(y) \) dependent on \(y\), and you want to find the derivative of their product, the rule states:\[\frac{d}{dy} [u(y) \cdot v(y)] = u(y) \cdot v'(y) + v(y) \cdot u'(y)\]This rule was useful for our task in finding \( \frac{\partial z}{\partial y} \). We identified the components for differentiation:

• \( u(y) = 2y \) and \( v(y) = \frac{8y}{x} \), with their respective derivatives.
  
• The derivative of \(u(y)\) is \(2\) and for \(v(y) \), it was crucial to correctly differentiate to demonstrate the impact on \( z \).

By using the product rule, we derived the partial derivative \( \frac{\partial z}{\partial y} = \frac{8y}{x} - \frac{x^{2}}{4y^{2}} \), clearly highlighting the change of \(z\) with respect to \(y\). This underscores how interconnected variables in multivariable calculus can be effectively managed with these fundamental differentiation techniques.