91Ó°ÊÓ

Calculus problems often involve understanding and manipulating equations to find derivatives, which represent how one variable changes with respect to another. This helps us locate slopes and rates of change in different contexts. In the given problem, we are tasked with differentiating the equation implicitly.
Implicit differentiation is a technique used when the relationship between variables is complex, meaning y is not isolated on one side of the equation. Taking the derivative directly with respect to x requires us to assume that y is a function of x and apply chain rules to capture the nuances of their dependency.
It's essential in calculus to grasp how these manipulations connect to finding slopes, rates, and related aspects of curves. This knowledge extends beyond this exercise into more complicated scenarios involving multiple variables and non-linear relationships.

The slope of a curve at a specific point illustrates how steep the curve is at that location. A positive slope means the curve is inclining, whereas a negative slope means it is declining. To find the slope, we need the derivative of the equation at that point. In our problem, after implicit differentiation, we derived the slope formula:

• \( \frac{dy}{dx} = \frac{-y - 1}{x + 2y} \)

For the point (1, 2), substituting these values into the derivative formula gives \( \frac{-3}{5} \). This negative slope confirms that the curve is decreasing at x = 1, y = 2.
Understanding slopes within calculus is crucial. It tells us how a function behaves, how fast y changes as x changes, and provides vital insights into the geometry of curves. It extends beyond simple line slopes to more intricate functional forms.

The product rule is a fundamental calculus tool used when differentiating products of two functions. In this problem, when differentiating \( xy \), we apply the product rule because it is the product of x and y. Following the formula for the product rule:

• \( (uv)' = u'v + uv' \)

For \( xy \), treating x as the first function \( u \) and y as the second function \( v \), we get \( y + x \frac{dy}{dx} \) after differentiating.
The comprehension of the product rule enables us to tackle complex differentiations such as mixed function products and sets a baseline for advanced calculus topics. Combining this rule with implicit differentiation, as shown in the exercise, allows us to solve a broader range of calculus problems effectively.