91Ó°ÊÓ

The first derivative is a fundamental concept in calculus, indicating the rate of change of a function. To find the critical points of a function, we first need to calculate its first derivative. This involves differentiating the function with respect to its variable. In the exercise, we started with the function \( f(x) = \frac{1}{x} - \frac{1}{x+3} \).

Using the quotient rule twice, we derived \( f'(x) = -\frac{1}{x^2} + \frac{1}{(x+3)^2} \). This is known as the first derivative.

Setting the first derivative equal to zero helps us find the critical points. This step is crucial because it tells us where the function's rate of change is zero. For our given function, setting \( -\frac{1}{x^2} + \frac{1}{(x+3)^2} = 0 \) led to the equation \( x^2 = (x+3)^2 \). Solving it, we obtained the critical point \( x = -\frac{3}{2} \).

The second derivative test helps determine the nature of critical points, whether they are relative maxima, minima, or neither. After finding the first derivative, we differentiate it again to obtain the second derivative.

For the function \( f(x) = \frac{1}{x} - \frac{1}{x+3} \), we calculated the first derivative as \( f'(x) = -\frac{1}{x^2} + \frac{1}{(x+3)^2} \).

Next, differentiating again, we found the second derivative to be \( f''(x) = \frac{2}{x^3} - \frac{2}{(x+3)^3} \). By evaluating the second derivative at the critical point \( x = -\frac{3}{2} \), we got \( f''\left(-\frac{3}{2}\right) = -\frac{32}{27} \).

Since \( f''\left(-\frac{3}{2}\right) < 0 \), we concluded that the critical point \( x = -\frac{3}{2} \) is a point of relative maximum. When the second derivative at a critical point is less than zero, it indicates that the function is concave down at that point, confirming the presence of a relative maximum.

Relative maxima and minima are points where a function reaches local highs or lows. A relative maximum occurs where the function reaches a peak, and a relative minimum occurs where it reaches a trough within a certain interval.

After identifying the critical points using the first derivative, we use the second derivative to determine their nature. In our exercise, we found a single critical point at \( x = -\frac{3}{2} \).

By evaluating the second derivative at this point, we saw that \( f''\left(-\frac{3}{2}\right) < 0 \). This negative value indicates that the function is concave down at this critical point, confirming it as a relative maximum. Therefore, the function has a local high at \( x = -\frac{3}{2} \).

The quotient rule is a method used in calculus to differentiate functions that are divided by each other. It states that if you have a function \( f(x) = \frac{u(x)}{v(x)} \), then its derivative is given by:

\[ f'(x) = \frac{u'(x) v(x) - u(x) v'(x)}{v(x)^2} \]

In our exercise, we used the quotient rule to differentiate each part of the function \( f(x) = \frac{1}{x} - \frac{1}{x+3} \).

For \( \frac{1}{x} \), we applied the quotient rule to obtain \( -\frac{1}{x^2} \). Likewise, for \( \frac{1}{x+3} \), we derived \( -\frac{1}{(x+3)^2} \). Combining these results, we got the first derivative \( f'(x) = -\frac{1}{x^2} + \frac{1}{(x+3)^2} \).

The quotient rule is indispensable when dealing with complex fractions as it provides a systematic way to find the derivative of divided functions.