Problem 209 - Calculus Volume 1

Question

Find the requested higher-order derivative for the given functions. $$ \frac{d^{3} y}{d x^{3}} \text { of } y=3 \cos x $$

Step-by-Step Solution

Answer

The third derivative is \( y''' = 3 \sin x \).

Step 1: Identify the function and the requirement

We begin with the function \( y = 3 \cos x \), and we need to find the third derivative \( \frac{d^{3} y}{d x^{3}} \).

Step 2: Find the First Derivative

To find the first derivative, take the derivative of \( y = 3 \cos x \). The derivative of \( \cos x \) is \( -\sin x \). Hence, \( y' = \frac{d}{dx}(3 \cos x) = -3 \sin x \).

Step 3: Find the Second Derivative

We now find the second derivative by differentiating \( y' = -3 \sin x \). The derivative of \( \sin x \) is \( \cos x \), so \( y'' = \frac{d}{dx}(-3 \sin x) = -3 \cos x \).

Step 4: Find the Third Derivative

Finally, we find the third derivative by differentiating \( y'' = -3 \cos x \). The derivative of \( \cos x \) is \( -\sin x \), thus \( y''' = \frac{d}{dx}(-3 \cos x) = 3 \sin x \).

Key Concepts

Higher-Order Derivatives

Higher-order derivatives involve taking the derivative of a function multiple times. If the first derivative of a function provides information about its rate of change, higher-order derivatives delve deeper into how this rate of change itself evolves. In many calculus problems, especially those involving motion, acceleration is expressed through the second derivative. The third derivative can relate to "jerk," or the rate of change of acceleration, although less often encountered in basic problems. For the function given in the exercise, the original function is differentiated three times to determine the third derivative, providing insights into the intricate changes of the function over an interval.

Trigonometric Functions

Trigonometric functions such as sine (\( \sin x \)) and cosine (\( \cos x \)) are fundamental in calculus. These functions model periodic phenomena like waves or circular motion, which are prevalent in natural sciences and engineering.Derivative patterns of these functions are important:

• The derivative of \( \sin x \) is \( \cos x \).
• The derivative of \( \cos x \) is \( -\sin x \).
• Both \( \sin x \) and \( \cos x \) are periodic, with a cycle every \( 2\pi \) radians.

When differentiating trigonometric functions, recognizing these patterns allows for quick and accurate computation of derivatives. This understanding is crucial when solving calculus problems that involve wave patterns or circular functions.

Calculus Problems

Calculus problems can range from simple differentiation exercises to complex applications in physics and engineering. Differentiation, particularly, offers a method to find how functions change and to solve real-world problems involving rates. In this exercise, using differentiation repeatedly on a trigonometric function demonstrates how calculus systematically breaks down complex expressions. Successfully solving a calculus problem, like finding the third derivative in this example, typically involves:

• Accurate identification of the function.
• Step-by-step calculation of each subsequent derivative.
• Ability to handle mathematical notation carefully.

By organizing the steps clearly, we can understand each phase of the differentiation process more easily.

Mathematics Education

Mathematics education plays a vital role in fostering analytical skills. It empowers students to solve problems across numerous disciplines. Learning strategies like the step-by-step process illustrated in the differential calculus problem helps students approach complex concepts methodically. In the case of higher-order derivatives,

• Students learn to build upon foundational derivatives.
• They enhance their understanding of mathematical operations.
• It encourages recognizing patterns, especially in functions with recurring behaviors like trigonometric functions.

Good mathematics education involves exposing students to diverse problems, practicing problems from different angles, and continuously building on existing mathematical knowledge. This helps form a comprehensive understanding of calculus and its applications.