Problem 6 - Calculus

Question

Find \(d y\). $$ y=(\tan x+1)^{3} $$

Step-by-Step Solution

Answer

\( dy = 3(\tan x + 1)^2 \sec^2 x \, dx \).

Step 1: Identify the function and its structure

We are given a function in the form of a composition, which makes it suitable to use the chain rule. The function is \( y = (\tan x + 1)^3 \), where the outer function is \( u^3 \) and the inner function is \( u = \tan x + 1 \).

Step 2: Apply the chain rule

To find \( \, dy \, \) or \( dy/dx \), use the chain rule: \( \frac{dy}{dx} = \frac{d}{du}(u^3) \cdot \frac{du}{dx} \).

Step 3: Differentiate the outer function

Differentiate \( u^3 \) with respect to \( u \). The derivative is \( 3u^2 \). Substitute back for \( u \) to get \( 3(\tan x + 1)^2 \).

Step 4: Differentiate the inner function

Differentiate \( \tan x + 1 \) with respect to \( x \). The derivative of \( \tan x \) is \( \sec^2 x \), and the derivative of 1 is 0. Therefore, \( \frac{du}{dx} = \sec^2 x \).

Step 5: Combine the derivatives

Substitute \( \frac{dy}{du} = 3(\tan x + 1)^2 \) and \( \frac{du}{dx} = \sec^2 x \) into the chain rule to get \( \frac{dy}{dx} = 3(\tan x + 1)^2 \cdot \sec^2 x \).

Step 6: Express final derivative

The derivative is \( dy = 3(\tan x + 1)^2 \sec^2 x \, dx \). This is the final expression for \( dy \).

Key Concepts

Understanding Derivatives

Derivatives are a fundamental concept in calculus, representing the rate at which a function changes. When you find the derivative of a function, you're essentially looking for the slope of the tangent line to the curve at any given point.
This is useful because it helps you understand how the function behaves and predict its value at other points.
The process involves taking the limit of the difference quotient of the function. For simple functions, this is straightforward, but complex functions often require rules like the chain rule or product rule.

• The basic idea is expressed as \(\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\).
• For composite functions, the derivative provides insights into how the outer function's rate of change is affected by the inner function's rate of change.

Trigonometric Functions in Calculus

Trigonometric functions, like sine, cosine, and tangent, are periodic functions often used in calculus problems. They model waves and oscillations, making them crucial in fields like physics and engineering.
In calculus, derivatives of trigonometric functions follow specific patterns:

• The derivative of \(\sin x\) is \(\cos x\).
• The derivative of \(\cos x\) is \(-\sin x\).
• The derivative of \(\tan x\) is \(\sec^2 x\).

These derivatives are used to solve problems involving rates of change in cyclical phenomena.
Understanding these relationships is key to solving calculus problems involving trigonometric functions.

Problem Solving with Calculus

Calculus problem solving often involves breaking down complex problems into simpler parts.
Initially, identify the type of problem and the calculus concepts involved. In the case of derivatives, use rules like the chain rule, product rule, or quotient rule as needed.

• The chain rule lets you differentiate composite functions by taking the derivative of the outer function, then multiplying it by the derivative of the inner function.
• This is crucial when dealing with nested functions like \((\tan x + 1)^3\).

Once you have a plan, carefully execute each step, ensuring that you've applied the rules correctly.
Rewriting and simplifying expressions is often necessary to reach the final solution, as shown in exercises like finding \(d y\) for a function like \(y=(\tan x+1)^{3}\).