91Ó°ÊÓ

The quotient rule is an essential tool in differentiation, specifically when dealing with functions expressed as a fraction of two other functions. When you have a function in the form of \( y = \frac{u}{v} \), the quotient rule allows you to find its derivative efficiently. The formula is as follows:

• \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2} \)

Here, \( u \) and \( v \) are the top and bottom functions of the fraction, respectively, and \( u' \) and \( v' \) are their derivatives. This formula specifically helps manage the differentiation process when both the numerator and denominator are polynomials or other standard functions.

In the given problem, \( u = x + 3 \) and \( v = 1 - x \), with their derivatives, \( u' = 1 \) and \( v' = -1 \). Applying the quotient rule allowed us to find the derivative \( y' \) of \( y = \frac{x+3}{1-x} \) as \( \frac{4}{(1-x)^2} \), showing the technique's power in simplifying complex problems.

The slope of the tangent line is a key concept in calculus as it describes how a curve behaves at a particular point. To find this slope, we need the function's derivative at the specific point of interest.

• The function's derivative represents the instantaneous rate of change, akin to the slope of the tangent line.
• For the function \( y = \frac{x+3}{1-x} \), the derivative was calculated to be \( y' = \frac{4}{(1-x)^2} \).

Once you have the derivative, you simply substitute the given value of \( x \) to find the slope at that point. For \( x = -2 \), substituting into \( y' \) results in the slope \( \frac{4}{9} \).

This slope not only provides the gradient of the tangent at a point but also reflects the direction and rate at which the function changes. A positive slope like \( \frac{4}{9} \) indicates an increasing function at that point.

Derivative evaluation ties together differentiation techniques and application, focusing on solving for the derivative at a specific value of the variable. This process often consists of using rules like the quotient rule and then substituting the given variable value to find particular information like the slope.

• In practice, it involves taking the general form of the derivative you've calculated and plugging in the specific \( x \) value.
• From our example, with \( y' = \frac{4}{(1-x)^2} \), substituting \( x = -2 \) allowed us to derive \( y'(-2) = \frac{4}{3^2} = \frac{4}{9} \).

Evaluation yields a concrete number representing something tangible about the function, such as the slope of a tangent. It culminates the differentiation process by concretely linking derivatives to real-world problems as well as graphically interpreting how the function curves or flattens at specific points.