Problem 1 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

For Problems 1 through 6, differentiate the function given. \(y=3 \tan x-4 \tan ^{-1} x\)

Step-by-Step Solution

Answer

\(y' = 3 \sec^2x - 4 \frac{1}{1+x^2}\)

Step 1: Differentiating the Tangent Function

The derivative of a tangent function is given as \( \frac{d}{dx}(\tan x) = \sec^2x \). Therefore, since our function is \(3 \tan x\), its derivative would be \(3 \sec^2x\).

Step 2: Differentiating the Inverse Tangent Function

The derivative of an inverse tangent function is given as \( \frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}\). Therefore, since our function part is \(4 \tan^{-1} x\), its derivative would be \( \frac{4}{1+x^2}\)

Step 3: Combining the Results

The total derivative of our function would be the sum of the derivatives calculated in Steps 1 and 2. That would give us the expression \(y' = 3 \sec^2x - 4 \frac{1}{1+x^2}\)

Key Concepts

Tangent Function

The tangent function, denoted as \( \tan x \), is a fundamental part of trigonometry. It relates the angle in a right triangle to the ratio of the opposite side over the adjacent side. In calculus, understanding its behavior is essential. One key aspect is the derivative, which tells us how the function changes at any point.

• The derivative of \( \tan x \) is \( \sec^2 x \). This means that the rate of change of the tangent function is dependent on the square of the secant function.
• Remembering this derivative is important for solving various calculus problems, particularly those involving trigonometric expressions.

Breaking down complex functions, like our exercise example, often starts by differentiating basic trigonometric functions like \( \tan x \). This foundational step paves the way for more intricate operations later.

Inverse Trigonometric Functions

Inverse trigonometric functions reverse what their regular counterparts do. For \( \tan^{-1} x \), it finds the angle whose tangent is \( x \). In calculus problems, they pose unique challenges but come with handy derivative rules.

• The derivative of \( \tan^{-1} x \) is \( \frac{1}{1+x^2} \). Knowing this helps in tackling calculus problems where inverse functions appear.
• This rule indicates how the function \( \tan^{-1} x \) changes concerning \( x \), providing insights into curve sketching and function behavior.

Incorporating inverse trigonometric functions in calculus requires diligent application of these rules, ensuring accurate differentiation as shown in our exercise.

Derivative Rules

Derivative rules are the core tools in calculus for finding how functions change. In our exercise, we're applying basic differentiation rules to both tangent and inverse tangent functions.

• Product Rule and Quotient Rule are often used for multiplying or dividing functions, but our example focuses on linear combinations.
• The primary rule used in our exercise is the constant factor rule, which allows us to multiply the derivative of a function by a constant, like \( 3 \) or \( 4 \) in our equation.

By solidifying an understanding of these rules, students can solve complex calculus problems more efficiently and accurately.

Calculus Problems

Calculus problems often involve breaking down a function into manageable parts. Our exercise illustrates this by dealing with a function that combines \( \tan x \) and \( \tan^{-1} x \).

• Step-by-step differentiation is crucial to handle these kinds of problems. Isolate each component and apply the respective rules.
• Combining the derivatives gives a holistic view of the function's behavior, which is seen in the final expression: \( y' = 3 \sec^2 x - \frac{4}{1+x^2} \).

Practicing these problems enhances critical thinking and familiarity with differentiation, leading to adeptness in solving more complex calculus challenges.