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The first derivative of a function, denoted as \( \frac{dy}{dx} \), represents the rate of change of \( y \) with respect to \( x \). When dealing with implicit differentiation, where your equation involves both \( x \) and \( y \) intertwined, the derivative must be found for each variable even if \( y \) is not isolated. For the equation \( xy - \frac{6}{y} = 4 \), we differentiate each term independently with respect to \( x \).

• The term \( xy \) requires the product rule, as it is a product of two functions of \( x \) (i.e., \( x \) and \( y \)).
• The term \(-\frac{6}{y}\) uses the chain rule since \( y \) depends on \( x \).

After differentiation, we rearrange to extract \( \frac{dy}{dx} \). At the specific point \( P_0 = (2,3) \), substituting gives the slope, which equals \(-\frac{9}{4}\). This value reflects how \( y \) changes with \( x \) at that point.

The second derivative, expressed as \( \frac{d^2y}{dx^2} \), tells us about the concavity of the function—how the rate of change itself is changing. Differentiating \( \frac{dy}{dx} \) gives us \( \frac{d^2y}{dx^2} \). This step involves using implicit differentiation again, applying both the product and chain rules.

• The second derivative allows us to understand the curvature of the graph at \( P_0 \).
• Again, after performing the necessary differentiation and simplifications, substituting points and the found \( \frac{dy}{dx} \) value, we eventually reach \( \frac{d^2y}{dx^2} = -\frac{27}{32} \).

This result not only helps us comprehend how fast \( y \) is moving away from or towards the initial condition but also shows the accelerating/decelerating nature at that point.

The product rule is essential when differentiating expressions where two functions are multiplied. Suppose you're working with a term like \( xy \), which is a product of two dependent variables on \( x \). The product rule states that the derivative \( (xy)' \) is not merely the derivative of \( x \) times \( y \), but requires both parts to be considered, resulting in:
\[(xy)' = x \cdot \frac{dy}{dx} + y \cdot \frac{d}{dx}(x) = x \cdot \frac{dy}{dx} + y\] For the original equation, \( xy \) translates to \( y + x\frac{dy}{dx} \) after differentiation. This shows how each term contributes to the overall rate of change, incorporating the dependency of \( y \) on \( x \). Using the product rule accurately in implicit differentiation is crucial, as it ensures that every part of a multiplied function is correctly accounted for in the derivative.

The chain rule steps in whenever a function is nested within another. In our case, the term \(-\frac{6}{y}\) benefits from the chain rule due to the function \( y \) being in the denominator.

• The function \( y = y(x) \) implies that \( y \) changes with \( x \), and consequently \( -\frac{6}{y} \) becomes differentiated with respect to \( y \), multiplied by the derivative \( \frac{dy}{dx} \).
• This process yields the differentiation of \(-\frac{6}{y}\) as \( \frac{6}{y^2} \cdot \frac{dy}{dx} \).

Utilizing the chain rule ensures that even the most complex part of a differentiated function accounts for the underlying variable dependencies. This correct application is fundamental to solving implicit functions accurately, especially within nested contexts such as division or other operations.