91Ó°ÊÓ

The quotient rule is an essential tool in calculus used to find the derivative of a function expressed as a quotient of two other functions. If we have a function of the form \( f(x) = \frac{u(x)}{v(x)} \), the quotient rule states that its derivative is given by:

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

This rule is particularly useful when dealing with complex fractions, allowing us to differentiate the numerator and denominator separately.
In the given problem, where \( f(x) = \frac{\sin x}{x} \), we apply the quotient rule by setting \( u(x) = \sin x \) and \( v(x) = x \). Here, the derivatives \( u'(x) = \cos x \) and \( v'(x) = 1 \) are used to find the derivative \( f'(x) \). This gives us:

• \( f'(x) = \frac{(\cos x)x - \sin x \cdot 1}{x^2} \)

Understanding how to utilize the quotient rule simplifies the process of finding derivatives for rational functions.

Once the derivative of a function is known, analyzing its sign over a particular interval provides insights into the function's behavior. The sign of the derivative indicates whether the function is increasing or decreasing:

• A positive derivative means the function is increasing.
• A negative derivative signifies the function is decreasing.

In the given context, determining the sign of \( f'(x) \) helps to show that \( \frac{\sin x}{x} \) is decreasing.
We observe the interval \( (0, \frac{\pi}{2}) \), where \( \cos x \), and \( x \) are both positive, making \( x \cdot \cos x \) positive. Since \( \sin x < x \) over this range, the term \( \cos x \cdot x - \sin x \) remains positive. Thus:

• For \( 0 < x < \frac{\pi}{2} \), \( f'(x) > 0 \).

This result indicates that the function is increasing over the specified interval, a foundation for the given inequality.

Trigonometric inequalities involve inequalities containing trigonometric functions such as sine, cosine, or tangent. In this exercise, the core inequality is \( 2x/\pi < \sin x < x \) for \( 0 < x < \frac{\pi}{2} \).
Here, understanding the behavior of \( \sin x \) is crucial. It's known that \( \sin x \) grows from 0 to 1 as \( x \) increases from 0 to \( \frac{\pi}{2} \). This means:

• \( \sin x < x \) because the curve of \( \sin x \) lies below the line \( y = x \) in this interval.
• \( 2x/\pi < \sin x \) arises as a result of comparing the rate at which \( \sin x \) approaches 1 compared to the linear function \( 2x/\pi \).

Applying derivative sign analysis to show that the function \( \frac{\sin x}{x} \) is increasing establishes that \( \sin x \) stays between these bounding lines over \( 0 < x < \frac{\pi}{2} \).

An increasing function is one where the function value increases as the input increases. To mathematically confirm whether a function is increasing, its derivative must be evaluated:

• If \( f'(x) > 0 \) for all \( x \) in a given interval, the function is increasing over that interval.

In this problem, \( f(x) = \frac{\sin x}{x} \) was shown to have a positive derivative over \( (0, \frac{\pi}{2}) \). Thus, \( f(x) \) is increasing.
As a consequence, as \( x \) approaches \( \frac{\pi}{2} \), \( \frac{\sin x}{x} \) approaches its maximum within this interval. Establishing that the function is increasing was key in proving the given trigonometric inequality, bridging the connection between increasing functions and trigonometric bounds. This reinforces the understanding of how functions behave over specific intervals and how they relate to inequalities.