91Ó°ÊÓ

The chain rule is a fundamental tool in calculus. It helps us find the derivative of a composite function. For instance, in the equation from the exercise, \(x y^{3}\), we need to find the derivative of \(y^{3}\) with respect to \(x\). Since \(y\) is itself a function of \(x\), we cannot directly differentiate \(y^{3}\). Instead, we use the chain rule, which tells us that the derivative of \(y^{3}\) with respect to \(x\) is \(3y^{2} \frac{dy}{dx}\).

This is why we express the differentiation as \( \frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}\). This represents how the change in \(x\) affects \(y\) and, consequently, \(y^{3}\). The chain rule allows us to handle more complex functions and is essential when dealing with implicit differentiation.

The product rule is crucial when you are differentiating a product of two functions. Our original equation, \(x y^{3} = 2\), involves the product of \(x\) and \(y^{3}\). To differentiate the left-hand side with respect to \(x\), we need to apply the product rule. The product rule tells us that the derivative of \(u \times v\) is \( u \frac{dv}{dx} + v \frac{du}{dx} \).

Applying this to \(x y^{3}\), \(u\) is \(x\) and \(v\) is \(y^{3}\). Differentiating, we get: \( \frac{d}{dx} (x y^{3}) = x \frac{d}{dx}(y^3) + y^3 \frac{d}{dx}(x) \). With \( \frac{d}{dx}(x) \) being \(1\), it simplifies to \( y^{3} + x \frac{d}{dx}(y^3) \). The product rule is essential in splitting the differentiation into manageable parts.

The slope of a graph at any point gives us the gradient or steepness at that specific point. When we use implicit differentiation, we aim to find \( \frac{dy}{dx} \) - the derivative of \(y\) with respect to \(x\). The slope means understanding how \(y\) changes as \(x\) changes. In our exercise, after differentiating and rearranging terms, we derived: \( \frac{dy}{dx} = \frac{-y}{3x} \).

To find the slope at a specific point, indicated in the problem by \(x = -\frac{1}{4}\) and \(y = -2\), we substitute these values into our derived expression. By doing this, we get: \( \frac{dy}{dx} = -\frac{8}{3} \). This value represents the slope of the tangent line to the graph at the given point.

Evaluating the derivative means finding the exact value of the derivative at a specific point. For our exercise, after finding the general derivative formula \( \frac{dy}{dx} = \frac{-y}{3x} \), the next step is to substitute the given values of \(x\) and \(y\). When \(x = -\frac{1}{4}\) and \(y = -2\), plug these into the formula:

\begin{align*} \frac{dy}{dx} = \frac{-(-2)}{3(-\frac{1}{4})} \rightarrow \frac{2}{-\frac{3}{4}} \rightarrow \frac{2 \times 4}{-3} = -\frac{8}{3}. \tag*{} ewline \text {Hence, the slope is } -\frac{8}{3}. \tag*{} ewline \text {This simplified value indicates how steep the graph is at that specific point.} \tag*{} ewline \text {By following these steps, we ensure the derivative’s accuracy and correctly interpret the slope of the graph.} \tag*{} ewline \text {This notion of evaluating the derivative correctly is pivotal in understanding graph behaviors and tangent slopes.} \tag*{} ewline \text {Remember, always double-check values for accurate derivative evaluation.} \tag*{endalign*}.