91Ó°ÊÓ

To find the derivative of a function that is the quotient of two functions, like \( F(x) = \frac{f(x) + 3}{x - g(x)} \), we utilize the quotient rule. This rule is essential in differential calculus and helps us differentiate functions that are "divided" by each other. The quotient rule states that given two differentiable functions, \( u(x) \) and \( v(x) \), the derivative \( \left( \frac{u}{v} \right)' \) is\[\frac{u'v - uv'}{v^2}\].
Let's break this down:

• \( u(x) = f(x) + 3 \) and \( v(x) = x - g(x) \)
• Compute \( u'(x) = f'(x) \)
• Compute \( v'(x) = 1 - g'(x) \)

Substituting these into the formula helps us find how the function \( F(x) \) changes with respect to \( x \), by considering both the functions \( u \) and \( v \) and their changes (derivatives). This method is crucial for finding the slope of the tangent line to the curve at a specific point.

The derivative is a fundamental concept in calculus that measures how a function changes as its input changes. In simple terms, it represents the slope of the tangent line to a curve at a specific point.
In the given exercise, finding the derivative of \( F(x) = \frac{f(x) + 3}{x - g(x)} \) at \( x = 2 \) involves applying the quotient rule. Once we have the formula for \( F'(x) \), we substitute the values of \( f(2) = 3 \), \( f'(2) = -1 \), \( g(2) = -4 \), and \( g'(2) = 1 \) into it.
This gives us\[F'(2) = \frac{-1 \cdot (2 + 4) - (3 + 3) \cdot (1 - 1)}{(2 + 4)^2} = \frac{-6}{36} = -\frac{1}{6}\].
This result tells us the slope of the tangent to the curve at \( x = 2 \), which is a key step in determining the equation of a line related to this curve.

A perpendicular line to a curve at a given point means that the line forms a right angle (90 degrees) with the tangent line at that point. To find such a line's equation, one must first determine the slope of the tangent line, as the slope of the perpendicular line is the negative reciprocal of the tangent slope.
In our exercise, the tangent slope is given by the derivative \( F'(2) = -\frac{1}{6} \). The slope of the line perpendicular to this is \( 6 \) because the negative reciprocal of \(-\frac{1}{6}\) is \( 6 \).
Now, with a point on the curve \((2, F(2))\), where \( F(2) = 1 \), we use the point-slope form of the line equation:\[y - y_1 = m(x - x_1)\].

• The point is \((2, 1)\) and the slope \( m = 6 \).
• This gives: \( y - 1 = 6(x - 2) \).

Simplifying this yields the equation \( y = 6x - 11 \), representing the line perpendicular to the curve at that specific point.