Problem 71 - Calculus - AP Edition

Question

Recall that $$ \frac{d}{d x}(|x|)=\left\\{\begin{aligned} 1, & x>0 \\ -1, & x<0 \end{aligned}\right. $$ Use this result and the chain rule to find \(\frac{d}{d x}(|\sin x|)\) for nonzero \(x\) in the interval \((-\pi, \pi)\)

Step-by-Step Solution

Answer

\(\frac{d}{dx}(|\sin x|) = \cos x\) for \(x > 0\) and \(-\cos x\) for \(x < 0\).

Step 1: Identify the outer and inner functions

To find the derivative of \(|\sin x|\), we need to identify the functions involved. Here, \(|x|\) is the outer function, and \(\sin x\) is the inner function.

Step 2: Apply the derivative of the outer function

From the given, the derivative of \(|u|\) with respect to \(u\) is \(1\) for \(u > 0\) and \(-1\) for \(u < 0\). Here, \(u = \sin x\).

Step 3: Find the sign of \(\sin x\)

In the interval \((-\pi, \pi)\), the sign of \(\sin x\) depends on the specific value of \(x\). \(\sin x\) is positive in \((0, \pi)\) and negative in \((-\pi, 0)\).

Step 4: Apply the chain rule

The chain rule states \(\frac{d}{dx} \big(f(g(x))\big) = f'(g(x)) \cdot g'(x)\). For \(|\sin x|\), it becomes: \[\frac{d}{dx}(|\sin x|) = \frac{d}{d(\sin x)}(|\sin x|) \cdot \frac{d}{dx}(\sin x)\]

Step 5: Determine the derivative for each sub-interval

For \(x > 0\), \(\frac{d}{d(\sin x)}(|\sin x|) = 1\) and \(\frac{d}{dx}(\sin x) = \cos x\). Thus, \(\frac{d}{dx}(|\sin x|) = \cos x\). For \(x < 0\), \(\frac{d}{d(\sin x)}(|\sin x|) = -1\) and \(\frac{d}{dx}(\sin x) = \cos x\). Thus, \(\frac{d}{dx}(|\sin x|) = -\cos x\).

Step 6: Summarize the results

The derivative \(\frac{d}{dx}(|\sin x|)\) is a piecewise function given by: \[\frac{d}{dx}(|\sin x|) = \begin{cases} \cos x, & x > 0 \ -\cos x, & x < 0\end{cases}\]

Key Concepts

Derivative

The derivative is a fundamental concept in calculus that measures how a function changes as its input changes. It's often thought of as the "slope" of the function at any given point. For any function represented as \( f(x) \), the derivative, denoted as \( f'(x) \) or \( \frac{df}{dx} \), represents the rate at which \( f(x) \) changes with respect to \( x \).

The process of finding the derivative is known as differentiation. When understanding the derivative, it's essential to comprehend the idea of limits, because derivatives are defined as limits of the average rate of change of the function over an interval as that interval approaches zero.

When using techniques like the Chain Rule, the goal is to differentiate complex functions by breaking them into simpler parts. This is especially useful when dealing with composite functions like \(|\sin x|\), where one function is nested inside another. The chain rule allows us to efficiently find a derivative by focusing first on the outer function and then the inner.

Piecewise Function

A piecewise function is a function that is defined by different expressions or formulas over different intervals of the domain. In simpler words, it's like stitching together different functions to construct a singular function.

An example can help clarify:

• For \( x > 0 \), our piecewise function might follow one rule: \( f(x) = 2x \).
• For \( x < 0 \), the rule might change to: \( f(x) = -2x \).

These kinds of functions are extremely useful in modeling real world scenarios where conditions change abruptly. For our exercise, when finding the derivative of \(|\sin x|\), we observed that it changes depending on whether \( x \) is positive or negative, hence making use of piecewise definitions.

Understanding the behavior of a function over various intervals allows us to capture more complex behavior within a single mathematical expression. Piecewise functions need special attention in calculus particularly when finding derivatives or integrals, because each piece may behave differently.

Trigonometric Functions

Trigonometric functions, such as \( \sin x \), \( \cos x \), and \( \tan x \), are the core of many analytical tasks in mathematics, particularly in calculus. These functions are used to describe the relationships of angles and sides in right triangles, and they also have applications in describing periodic phenomena like waves.

For the function \( \sin x \), it represents the vertical component of the point on the unit circle at an angle \( x \). It's important to remember these key properties

• \( \sin x \) repeats its values in a periodic cycle every \( 2\pi \).
• It oscillates between -1 and 1.

When dealing with trigonometric functions in derivatives, recognizing these properties helps streamlining the process and identifying where sign changes occur. In the context of our problem, knowing the intervals where \( \sin x \) is positive or negative directly influences how we apply the chain rule and piecewise functions.

Mastering the rules and properties of trigonometric functions will allow you to efficiently handle problems involving oscillations, waves, and angles.