Problem 52 - Thomas Calculus

Question

Find \(y^{\prime \prime}\) in Exercises \(49-52\) $$ y=9 \tan \left(\frac{x}{3}\right) $$

Step-by-Step Solution

Answer

The second derivative is \( y'' = 2 \sec^2\left( \frac{x}{3} \right) \tan\left( \frac{x}{3} \right) \).

Step 1: Find the First Derivative

To find the second derivative, we first need the first derivative of the function. If \( y = 9 \tan\left( \frac{x}{3} \right) \), let's find \( y' \). The derivative of \( \tan(u) \) with respect to \( u \) is \( \sec^2(u) \). Thus, \( y' = 9 \cdot \sec^2\left( \frac{x}{3} \right) \cdot \left( \frac{1}{3} \right) \) because of the chain rule. So, \( y' = 3 \sec^2\left( \frac{x}{3} \right) \).

Step 2: Apply the Chain Rule Again

Next, we need to find the second derivative \( y'' \). Start by differentiating \( y' = 3 \sec^2\left( \frac{x}{3} \right) \). The derivative of \( \sec^2(u) \) is \( 2\sec^2(u)\tan(u) \) \( \cdot \frac{du}{dx} \). Here, \( u = \frac{x}{3} \), so \( \frac{du}{dx} = \frac{1}{3} \).

Step 3: Calculation of Second Derivative

Compute the total derivative: \( y'' = 3 \cdot 2 \cdot \sec^2\left(\frac{x}{3}\right) \cdot \tan\left(\frac{x}{3}\right) \cdot \frac{1}{3} \). Simplifying this, we get \( y'' = 2 \sec^2\left( \frac{x}{3} \right) \tan\left( \frac{x}{3} \right) \).

Key Concepts

Trigonometric Functions

Understanding trigonometric functions is crucial for calculus. They often describe oscillating motions, such as waves or circular rotations. In our exercise, the core trigonometric function is the tangent function, written as \( \tan(x) \).
This function is one of the fundamental ratios in trigonometry. It is defined in terms of sine and cosine functions, specifically: \( \tan(x) = \frac{\sin(x)}{\cos(x)} \). Because it is a ratio of two periodic functions, tangent has its own periodic behavior, repeating every \( \pi \) radians.
One important property of \( \tan(x) \) is that it becomes undefined where \( \cos(x) = 0 \), leading to vertical asymptotes on a graph. These specifics make understanding tangent essential for solving problems involving periodic behaviors. Related derivatives involve the secant function, \( \sec(x) \), which is \( \frac{1}{\cos(x)} \), and significantly affects the calculation of first and second derivatives.

Chain Rule

The chain rule is a fundamental technique in calculus used when differentiating composite functions. Composite functions are those where one function is inside another, like \( \tan\left(\frac{x}{3}\right) \) in our exercise.
Understanding the chain rule requires identifying the inner and outer functions. The outer function is \( \tan(u) \), and the inner function is \( u = \frac{x}{3} \). To apply the chain rule, take the derivative of the outer function with respect to the inner function. Then multiply by the derivative of the inner function with respect to \( x \).
This method sounds complex but simplifies the process of finding derivatives. In this exercise, using the chain rule helps in handling the differentiation of functions like \( \sec^2(u) \). Applying it correctly results in correctly calculating the derivative terms crucial for finding the second derivative.

Second Derivative

The second derivative of a function provides insights into the curvature and concavity of the graph of the function. It tells us how the rate of change of the slope itself is changing.
For our function, we start by getting the first derivative. This gives information about the rate of change of \( y \). The second derivative, \( y'' \), takes this a step further by detailing how this rate changes over time or within a space.
In practical terms, calculating the second derivative involves differentiating our first derivative again. For functions involving trigonometric ratios, like \( \tan(x) \), this simplifies considerably with rules like the chain rule and basic differentiation rules for trigonometric identities.
Once calculated, the second derivative informs us about the nature of the graph. Positive second derivatives signify concave upward portions, while negative signify concave downward parts. This information is essential in graphing and understanding the behavior of functions over intervals.