91Ó°ÊÓ

Differentiation is a mathematical process that helps us find the rate at which a function is changing at any given point. In our case, we have an equation that involves both variables, \(x\) and \(y\). When executing differentiation, it's vital to respect each term's structure. In our original equation, \(2y^{3/2} + xy - x = 0\), we used several differentiation techniques to differentiate each term separately with respect to \(x\).

• The first term, \(2y^{3/2}\), requires the use of the chain rule, which is essential when differentiating functions of another function, like \(y^{3/2}\).
• The second term, \(xy\), uses the product rule since it's a product of two functions: \(x\) and \(y\).
• The final term \(-x\) involves a basic differentiation of a single-variable polynomial, which is simple but vital.

By applying these techniques methodically, we can handle more complex functions found in implicit equations. Understanding and using these techniques ensures accurate differentiation and shows how each part of a complex function contributes to its overall rate of change.

An implicit function is where the relationship between \(x\) and \(y\) is given by an equation rather than expressed explicitly as \(y=f(x)\). In our exercise, the equation \(2y^{3/2} + xy - x = 0\) represents an implicit function, as \(y\) is intertwined with \(x\) and not isolated on one side of the equation.

Implicit differentiation is particularly useful here. Unlike explicit functions, where \(y\) is by itself, implicit functions require differentiating all terms with respect to \(x\) while treating \(y\) as a function of \(x\). Thus, whenever you differentiate a term involving \(y\), you must multiply by \(\frac{dy}{dx}\) to account for \(y\) being a function of \(x\). This process involves applying various differentiation rules, such as the chain rule and product rule, to successfully differentiate implicit equations.

Implicit differentiation allows us to find \(\frac{dy}{dx}\) even if we can't easily express \(y\) explicitly in terms of \(x\). This approach is necessary for handling equations where \(x\) and \(y\) are inseparably linked.

The product rule is essential for differentiating functions that are multiplied together. In our problem, one term, \(xy\), is an expression that requires the product rule. In this rule, if we have two functions, \(u\) and \(v\), the derivative of their product, \(uv\), with respect to \(x\) is given by \(u'v + uv'\).

For the term \(xy\):

• We let \(u = x\), so \(u' = 1\).
• We let \(v = y\), so \(v' = \frac{dy}{dx}\), since \(y\) is dependent on \(x\).

Applying the product rule gives us: \((1)y + x\frac{dy}{dx}\). Here, we first multiply the derivative of \(x\) by \(y\), then add \(x\) times the derivative of \(y\).

Understanding when and how to apply the product rule is crucial for correctly differentiating terms where multiple functions are multiplied together. It ensures that all changes in each component function are accounted for in the derived expression. This is especially important in problems involving implicit functions, where intertwined variables must be differentiated simultaneously.