Problem 74 - Calculus Volume 1

Question

For the following exercises, find the differential and evaluate for the given \(x\) and \(d x .\) $$ y=\tan x, \quad x=0, \quad d x=\frac{\pi}{10} $$

Step-by-Step Solution

Answer

The differential \( dy \) evaluates to \( \frac{\pi}{10} \).

Step 1: Differentiate the Function

The first step is to find the derivative of the function. We have \( y = \tan x \). The derivative of \( \tan x \) with respect to \( x \) is \( \sec^2 x \). So, \( \frac{dy}{dx} = \sec^2 x \).

Step 2: Find the Differential

The differential \( dy \) can be described as the product of the derivative \( \frac{dy}{dx} \) and \( dx \). Thus, we have \( dy = \sec^2 x \cdot dx \).

Step 3: Substitute Given Values

Substitute \( x = 0 \) and \( dx = \frac{\pi}{10} \) into the equation from Step 2. We get \( dy = \sec^2(0) \cdot \frac{\pi}{10} \).

Step 4: Evaluate \( \sec^2(0) \)

Evaluate \( \sec^2(0) \). Since \( \sec(0) = 1 \), we have \( \sec^2(0) = 1^2 = 1 \).

Step 5: Calculate the Differential

Using the result from Step 4, substitute \( \sec^2(0) = 1 \) into the equation for \( dy \): \( dy = 1 \cdot \frac{\pi}{10} \). Thus, \( dy = \frac{\pi}{10} \).

Key Concepts

Derivative of Tangent

Differential calculus involves the study of how functions change, and one of the key concepts is taking derivatives. A derivative essentially measures the rate at which a quantity changes. For the function \( y = \tan x \), we are interested in finding how the tangent function changes as \( x \) changes. When you differentiate \( \tan x \), you get \( \sec^2 x \). This result is crucial because it tells us the slope of the tangent line to the function \( y = \tan x \) at any point \( x \).

The process of differentiation for \( \tan x \) involves applying the rules of derivatives. The specific rule for derivatives of trigonometric functions tells us that the derivative of tangent is secant squared.

• Derivative of \( \tan x \) is \( \sec^2 x \)

This shows the instantaneous rate of change of the tangent function, helping us understand how steep or flat the curve is at any point along its path.

Differential of a Function

The differential of a function allows us to approximate how much the function's value changes when its input changes by a small amount. For a function \( y = f(x) \), the differential \( dy \) is given by \( dy = f'(x) \cdot dx \), where \( f'(x) \) is the derivative of the function, and \( dx \) is a small change in \( x \).

In our exercise where \( y = \tan x \), the differential \( dy \) is determined by multiplying the derivative \( \sec^2 x \) by \( dx \), which is a small increment in \( x \).

• Formula: \( dy = \sec^2 x \cdot dx \)

This expression gives an approximation of how much \( y \) will change for a small change in \( x \), which is foundational for understanding small changes in functions.

Evaluating Differential at a Point

Once you have the expression for the differential of a function, you can evaluate it at a specific point. This means you plug in particular values for \( x \) and \( dx \) to find out how much the function value will change at that point.

In the problem, we've been given \( x = 0 \) and \( dx = \frac{\pi}{10} \). By substituting \( x = 0 \) into the expression \( dy = \sec^2 x \cdot dx \), you first evaluate \( \sec^2(0) \), which equals 1.

• \( \sec(0) = 1 \)
• \( \sec^2(0) = 1^2 = 1 \)

After evaluating \( \sec^2(0) \), you multiply it by \( dx \) to find \( dy = 1 \cdot \frac{\pi}{10} \). Thus, the change in \( y \) when \( x \) changes by \( \frac{\pi}{10} \) is \( \frac{\pi}{10} \). This process helps us find the differential for specific values and understand the immediate effects on the function.