91Ó°ÊÓ

Derivatives allow us to understand how a function changes with respect to its variables. In the context of related rates problems, we're interested in how one quantity changes over time, even as another related quantity also changes over time. Here, we aim to find \( \frac{dy}{dt} \) in the function \( y = \frac{1-x}{1+x^2} \). This involves calculating the derivative \( \frac{dy}{dx} \) of the function with respect to \( x \), which describes the instantaneous rate of change of \( y \) as \( x \) changes.
Once \( \frac{dy}{dx} \) is determined, we use the chain rule to find \( \frac{dy}{dt} \), the rate of change of \( y \) over time. Using the provided values of \( \frac{dx}{dt} = -2 \) and \( x = -1 \) helps to determine this rate, linking the changes in \( x \) over time to changes in \( y \). This calculation requires a clear understanding of how derivatives are used to express these rates mathematically.

The Quotient Rule is an essential tool in calculus, especially when differentiating functions that are given as ratios. To differentiate a function \( y = \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), the Quotient Rule states:

• \( \frac{dy}{dx} = \frac{v(du/dx) - u(dv/dx)}{v^2} \)

In our specific exercise with \( y = \frac{1-x}{1+x^2} \), the functions \( u(x) = 1-x \) and \( v(x) = 1 + x^2 \) are identified, and derivatives of these ( \( du/dx = -1 \) and \( dv/dx = 2x \) respectively) are used within the quotient rule.
Applying these to the formula, you derive \( \frac{dy}{dx} \), assisting in detailing how \( y \) changes as \( x \) changes. This approach allows working through potentially complex derivative problems methodically.

Utilizing the Chain Rule in calculus allows us to find derivatives of composite functions. This rule is vital when a relation involves functions of other functions, which is common in related rates problems. The Chain Rule states that if a variable \( y \) is a function of \( u \), which in turn is a function of \( x \), then the derivative of \( y \) with respect to \( x \) is given by:

• \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \)

In the given exercise, we have already found \( \frac{dy}{dx} \) using the quotient rule. The Chain Rule then connects these steps with the known rate of change \( \frac{dx}{dt} \) to finally evaluate \( \frac{dy}{dt} \).
By multiplying \( \frac{dy}{dx} = -\frac{3}{2} \) by \( \frac{dx}{dt} = -2 \), we arrive at the solution \( \frac{dy}{dt} = 3 \). The Chain Rule elegantly links the rates of change through multiplication, helping to solve real-world problems involving connected changes.

Mathematical Simplification plays a critical role in ensuring solutions to calculus problems are both clear and manageable. Once we compute the initial derivative, simplification is key to making sure the subsequent calculations are more straightforward and help in reducing errors.
For instance, when deriving \( \frac{dy}{dx} = \frac{-1 - x^2 + 2x - 2x^2}{(1+x^2)^2} \), combining like terms we have:

• \( \frac{dy}{dx} = \frac{-1 - 3x^2 + 2x}{(1+x^2)^2} \)

Simplifying expressions not only aids in easier computation but also assists in arriving at precise results. This consolidation of terms avoids unwieldy fractions and lengthy expressions, paving the way for correctly applying subsequent rules, like the Chain Rule, and efficiently using given problem conditions for substitution.
Simplification is not merely an optional step; it's a crucial process in ensuring clarity and accuracy in mathematical solutions.