Problem 105 - Thomas Calculus

Question

The derivative of \(\sin 2 x\) Graph the function \(y=2 \cos 2 x\) for \(-2 \leq x \leq 3.5 .\) Then, on the same screen, graph $$ y=\frac{\sin 2(x+h)-\sin 2 x}{h} $$ for \(h=1.0,0.5,\) and \(0.2 .\) Experiment with other values of \(h,\) including negative values. What do you see happening as \(h \rightarrow 0 ?\) Explain this behavior.

Step-by-Step Solution

Answer

As \(h\to 0\), the difference quotient approaches \(2\cos 2x\), the derivative.

Step 1: Calculate the Derivative

To find the derivative of \(\sin 2x\), we apply the chain rule. The derivative of \(\sin u\) with respect to \(u\) is \(\cos u\), and the derivative of \(2x\) with respect to \(x\) is 2. Thus, the derivative is \(\frac{d}{dx}(\sin 2x) = \cos 2x \cdot 2 = 2\cos 2x\).

Step 2: Graph the Function 2cos 2x

Plot the function \(y = 2\cos 2x\) for \(-2 \leq x \leq 3.5\). This is a cosine wave with twice the frequency of a standard cosine wave, because the factor of 2 inside the cosine function compresses the period. The amplitude is scaled by the factor of 2 outside the cosine function, giving twice the height of a regular cosine wave.

Step 3: Simplify the Difference Quotient

Consider the expression \(\frac{\sin 2(x+h) - \sin 2x}{h}\). Using the trigonometric identity for \(\sin(a + b) = \sin a \cos b + \cos a \sin b\), we can rewrite \(\sin 2(x+h)\) as \(\sin 2x \cos 2h + \cos 2x \sin 2h\). Substituting this into the difference quotient gives:\[\frac{\sin 2x \cos 2h + \cos 2x \sin 2h - \sin 2x}{h} = \frac{\sin 2x (\cos 2h - 1) + \cos 2x \sin 2h}{h}\]

Step 4: Graph the Difference Quotient

Graph the function \(y = \frac{\sin 2(x+h) - \sin 2x}{h}\) using different values of \(h\) such as 1.0, 0.5, and 0.2. You may use a graphing software or calculator. Overlay this graph on the plot of \(y = 2\cos 2x\) to see how the graphs compare as \(h\) gets smaller.

Step 5: Experiment with Different h Values

Try different values of \(h\), including negative values, and observe the graph. As \(h\) approaches zero, you should notice that the difference quotient graph approaches the graph of \(y = 2\cos 2x\). This illustrates the concept of a derivative, where the slopes of secant lines (approximations given by the difference quotient) approach the slope of the tangent line (the derivative) at a point as \(h\) approaches zero.

Step 6: Explain the Behavior as h approaches 0

As \(h \to 0\), the difference quotient \(\frac{\sin 2(x+h) - \sin 2x}{h}\) approaches the derivative \(2\cos 2x\). This is because the limit of a difference quotient as \(h\) approaches zero gives the derivative of the function, demonstrating how instantaneous rates of change are calculated in calculus.

Key Concepts

Understanding the Difference Quotient

The difference quotient is fundamental in calculus, particularly when discussing derivatives. It represents the average rate of change of a function between two points. For a function \( f(x) \), it is defined as:

• \( \frac{f(x+h) - f(x)}{h} \)

Here, \( h \) is the distance between the two points on the function.
In our exercise, the difference quotient \( \frac{\sin 2(x+h) - \sin 2x}{h} \) helps us analyze how \(\sin 2x\) changes as \(x\) changes.
As \( h \to 0 \), the difference quotient gives a better approximation of the derivative at a point, which geometrically translates to the slope of the tangent line at that point.

Mastering the Chain Rule

The chain rule is a powerful tool in differentiation used to find the derivative of composite functions. If you have a function \( y = f(g(x)) \), and both \( f \) and \( g \) are differentiable, the chain rule states:

• \( \frac{dy}{dx} = f'(g(x)) \, g'(x) \)

In simpler terms, you differentiate the outer function with respect to the inner function and multiply it by the derivative of the inner function.
For the derivative of \( \sin 2x \), we recognize \( \sin u \) where \( u = 2x \).
Applying the chain rule, the derivative of \( \sin u \) is \( \cos u \), and the derivative of \( 2x \) is 2. Therefore, the derivative is: \\( \frac{d}{dx}(\sin 2x) = \cos 2x \cdot 2 = 2\cos 2x \).
This step-by-step application of the chain rule is essential for correct differentiation of composite trigonometric functions.

Exploring the Limit Definition of Derivative

The limit definition of a derivative provides the foundation for understanding instantaneous rates of change. The derivative \( f'(x) \) of a function \( f(x) \) at a point \( x \) is given by:

• \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)

This formula tells us that derivatives measure the rate at which a function is changing at any given point.
In the exercise, as \( h \to 0 \), the difference quotient \( \frac{\sin 2(x+h) - \sin 2x}{h} \) approaches the derivative \( 2\cos 2x \).
This behavior illustrates the derivative's concept as the limit of the difference quotient, allowing us to understand how a trigonometric function behaves as \( x \) changes slightly.

Utilizing Trigonometric Identities

Trigonometric identities simplify the calculation and understanding of expressions involving trigonometric functions. These identities are crucial for simplifying expressions in order to find limits or derivatives.
For instance, the identity:

• \( \sin(a+b) = \sin a \cos b + \cos a \sin b \)

was used in our exercise to expand \(\sin 2(x+h) \) as \( \sin 2x \cos 2h + \cos 2x \sin 2h \).
This expansion allows us to simplify the difference quotient to:

• \( \frac{\sin 2x (\cos 2h - 1) + \cos 2x \sin 2h}{h} \)

By understanding and applying such identities, complex trigonometric expressions become more manageable when conducting calculus operations, making the derivative analysis smoother.