Problem 16 - Calculus Early Transcendentals

Question

Find \(f^{\prime}(x)\) $$ f(x)=\sec ^{2} x-\tan ^{2} x $$

Step-by-Step Solution

Answer

The derivative of the function is 0.

Step 1: Recall the Derivatives

First, let's recall the derivatives of the trigonometric functions involved. The derivative of \(\sec^{2}x\) with respect to \(x\) is \(2\sec^2x \tan x\), and the derivative of \(\tan^{2}x\) is \(2\tan x \sec^2x\).

Step 2: Apply the Derivative Rules

Apply the derivative rules to the function \(f(x) = \sec^2 x - \tan^2 x\). We need to find the derivatives of each term separately. Use the fact that the derivative of a difference \(u(x) - v(x)\) is \(u'(x) - v'(x)\).

Step 3: Derivative of \(\sec^2 x\)

The derivative of \(\sec^2 x\) is \(\frac{d}{dx}(\sec^2 x) = 2\sec^2x \tan x\).

Step 4: Derivative of \(\tan^2 x\)

The derivative of \(\tan^2 x\) is \(\frac{d}{dx}(\tan^2 x) = 2\tan x \sec^2 x\), using the chain rule.

Step 5: Combine the Results

Now subtract the derivative of \(\tan^2 x\) from the derivative of \(\sec^2 x\). We get \[ f^{\prime}(x) = 2\sec^2 x \tan x - 2\tan x \sec^2 x = 0. \]

Step 6: Simplify the Expression

Observe that \(2\sec^2 x \tan x\) and \(2\tan x \sec^2 x\) are equivalent; hence, they cancel each other out, resulting in \[ f^{\prime}(x) = 0. \]

Key Concepts

Secant Function Derivative

The secant function, denoted as \( \sec(x) \), is the reciprocal of the cosine function: \( \sec(x) = \frac{1}{\cos(x)} \). Understanding the derivative of \( \sec(x) \) is crucial in calculus, especially for students encountering trigonometric functions for the first time. To find the derivative of \( \sec(x) \), we need to use the quotient rule and the chain rule. The derivative of \( \sec(x) \) is calculated as follows:

• The quotient rule states that \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2} \), where \( u = 1 \) and \( v = \cos(x) \).
• Applying the rule: \( \sec(x) = \frac{1}{\cos(x)} \), we find that its derivative is \( \sec(x) \tan(x) \).

With this, the derivative of \( \sec^2(x) \) can be found using the power and chain rules:

• The power rule indicates \( \frac{d}{dx} [u^n] = n u^{n-1} u' \), which applies here as \( \frac{d}{dx} (\sec^2(x)) = 2 \sec(x) \cdot \sec(x) \tan(x) \).
• This gives us \( 2\sec^2(x) \tan(x) \) as the derivative.

Tangent Function Derivative

The tangent function, \( \tan(x) \), which is \( \frac{\sin(x)}{\cos(x)} \), also has an important derivative. The derivative of \( \tan(x) \) is derived using the quotient rule, similarly to the secant function.

• Applying the quotient rule, where \( u = \sin(x) \) and \( v = \cos(x) \), gives the derivative \( \sec^2(x) \).

When dealing with \( \tan^2(x) \), you must apply the chain rule:

• Using the chain rule, \( \frac{d}{dx} [u^2] = 2u \cdot \frac{du}{dx} \), and \( u = \tan(x) \), the derivative becomes \( 2 \tan(x) \cdot \sec^2(x) \).
• This provides \( 2\tan(x) \sec^2(x) \).

Knowing these derivatives simplifies working with composite functions involving \( \tan(x) \) and \( \tan^2(x) \).

Calculus Differentiation

Differentiation in calculus is a fundamental concept used to find the rate of change of a function. It's a primary tool to understand how functions behave and change.

• The basic rule for differentiation, often called the power rule, states that \( \frac{d}{dx} [x^n] = nx^{n-1} \).
• When dealing with sums or differences like \( f(x) = u(x) - v(x) \), the derivative is simply \( f'(x) = u'(x) - v'(x) \).

Differentiation can be applied to various functions, including trigonometric ones, using advanced techniques like the chain rule and product rule. In the original problem, these differentiation techniques help compute derivatives for functions involving \( \sec(x) \) and \( \tan(x) \). The ultimate simplification to zero demonstrates how equivalent terms in derivative expressions can cancel each other out, offering insight into the symmetry and canceling effects in calculus.