Problem 11 - University Calculus: Early Transcendentals

Question

Find \(d y / d x\). $$y=\frac{\cot x}{1+\cot x}$$

Step-by-Step Solution

Answer

The derivative is \( \frac{-\csc^2 x}{(1 + \cot x)^2} \).

Step 1: Set the Original Function

We begin with the given function:\[ y = \frac{\cot x}{1 + \cot x} \]

Step 2: Identify Derivative Components

The function is a quotient of two expressions, so we'll use the quotient rule for derivatives. If \( f(x) = \frac{g(x)}{h(x)} \), then \( \frac{d}{dx}[f(x)] = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \). Let's identify \(g(x)\) and \(h(x)\):- \( g(x) = \cot x \)- \( h(x) = 1 + \cot x \)

Step 3: Find Derivatives of \(g(x)\) and \(h(x)\)

Calculate the derivatives:- The derivative of \(g(x) = \cot x\) is \( g'(x) = -\csc^2 x \).- The derivative of \(h(x) = 1 + \cot x\) is \( h'(x) = -\csc^2 x \).

Step 4: Apply the Quotient Rule

Using the quotient rule, substitute the components and their derivatives:\[\frac{d}{dx}\left(\frac{\cot x}{1 + \cot x}\right) = \frac{(-\csc^2 x)(1 + \cot x) - (\cot x)(-\csc^2 x)}{(1 + \cot x)^2}\]

Step 5: Simplify the Expression

Expand and simplify the terms in the numerator:\[= \frac{-\csc^2 x - \cot x \csc^2 x + \csc^2 x \cot x}{(1 + \cot x)^2}\]Notice the \(\cot x \csc^2 x\) terms cancel each other:\[= \frac{-\csc^2 x}{(1 + \cot x)^2}\]

Step 6: Finalize the Derivative

The derivative \( \frac{dy}{dx} \) is:\[\boxed{\frac{-\csc^2 x}{(1 + \cot x)^2}}\]

Key Concepts

Quotient Rule for Differentiation

The quotient rule is a technique used in calculus to find the derivative of a function that is expressed as the ratio of two differentiable functions. Imagine you have a function represented as \( f(x) = \frac{g(x)}{h(x)} \). The quotient rule formula is then given by:\[\frac{d}{dx}[f(x)] = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}\]Here's a gentle breakdown to make it easier:

• Start by determining the two functions: \( g(x) \) is the top function, and \( h(x) \) is the bottom function.
• Compute the derivatives: \( g'(x) \) for the numerator and \( h'(x) \) for the denominator.
• Plug them into the quotient rule formula to find the derivative of the function.

Use the quotient rule when differentiating ratios, which is key to functions involving divisions of two terms. Catching on to this can simplify many calculus problems.

Trigonometric Derivatives

Trigonometric functions often appear in calculus, making it important to understand how their derivatives behave. Here's a quick reference for some common derivatives:

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

In the exercise, the function involved \( \cot x \), which has a derivative of \( -\csc^2 x \). Knowing these derivatives by heart helps in efficiently solving calculus problems involving trigonometric expressions.

Simplifying Expressions

Simplifying expressions is the art of making a complex equation easier to understand or work with. After applying calculus operations like differentiation, the result often needs simplification. A few strategies include:

• Combine like terms: Separate and then group similar terms in the numerator or denominator.
• Cancel terms: Identify and cancel terms that appear in both the numerator and denominator.
• Use identities: Apply trigonometric identities to transform expressions into simpler forms.

In the given example, after using the quotient rule, we simplified the derivative:\\[\frac{-\csc^2 x - \cot x \csc^2 x + \csc^2 x \cot x}{(1 + \cot x)^2} \]Observe how terms were canceled and combined to achieve the final, simplified form:\[\boxed{\frac{-\csc^2 x}{(1 + \cot x)^2}}\]Simplifying expressions can make them more manageable and reveal insights into the behaviors of functions.