Problem 55 - Thomas Calculus

Question

In Exercises \(53-58,\) find the value of \((f \circ g)^{\prime}\) at the given value of \(x\) . $$ f(u)=\cot \frac{\pi u}{10}, \quad u=g(x)=5 \sqrt{x}, \quad x=1 $$

Step-by-Step Solution

Answer

\(-\frac{\pi}{5}\)

Step 1: Identify the Composition

First, identify the composition of functions. We have two functions: 1. The outer function is \(f(u) = \cot \left( \frac{\pi u}{10} \right)\).2. The inner function is \(u = g(x) = 5\sqrt{x}\).The composition \((f \circ g)(x)\) becomes \(f(g(x)) = \cot \left( \frac{\pi \cdot 5\sqrt{x}}{10} \right)\).

Step 2: Identify the relevant trigonometric identities

Based on the given expression or equation, identify which trigonometric identities (Pythagorean, double-angle, sum/difference, etc.) are applicable.

Step 3: Apply the identities and simplify

Apply the identified identities to transform the expression. Simplify step by step, combining like terms and reducing fractions where possible.

Step 4: Solve or evaluate

If solving an equation, isolate the trigonometric function and find the angle(s). If evaluating, compute the final numerical value.

Step 5: State the result

Express the final answer, including all solutions in the required domain if solving an equation.

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus. It simplifies the differentiation of composite functions by breaking them into simpler parts. This rule is especially useful when dealing with functions nested within each other, as it allows us to differentiate each part separately and then combine them.
Consider two functions in composition: an outer function \( f(u) \) and an inner function \( g(x) \). The composite, noted as \((f\circ g)(x)\), implies \(f(g(x))\).

• The chain rule states that the derivative of this composition, \((f\circ g)'(x)\), is the product of the derivative of the outer function evaluated at the inner function, \( f'(g(x)) \), and the derivative of the inner function \( g'(x) \).
• This can be formally expressed as: \[(f\circ g)'(x) = f'(g(x)) \cdot g'(x)\]

In practice, once you identify the two functions, you differentiate each separately. First, find \( g'(x) \), then \( f'(u) \) by treating \( u \) as a variable related to \( g(x) \), and finally multiply these derivatives together.

Derivative of Trigonometric Functions

Trigonometric functions like \( \sin(x), \cos(x), \) and \( \cot(x) \) have specific rules for differentiation. It's crucial to memorize these rules, as they are commonly encountered in various calculus problems.
For the exercise involving the function \( f(u) = \cot\left(\frac{\pi u}{10}\right) \):

• The derivative of \( \cot(u) \) with respect to \( u \) is given by: \[\frac{d}{du}\cot(u) = -\csc^2(u)\]
• When dealing with a scaled angle, such as \( \frac{\pi u}{10} \), apply the chain rule again: multiply the derivative \( -\csc^2(\frac{\pi u}{10}) \) by the derivative of the inner angle \( \frac{\pi}{10} \).

Understanding these derivatives is key to effectively applying them. When you encounter a problem that involves finding the derivative of a trigonometric function composed with another, always think of how these basic rules apply.

Function Composition

Function composition combines two functions to form a new function. It is the essence of evaluating one function with the output of another. This concept is pivotal when dealing with nested functions, as it forms the framework for applying the chain rule.
Given two functions, \( f(u) \) and \( g(x) \), the composition \((f \circ g)(x)\) means you first apply \( g \) to \( x \), and then \( f \) to the result.

• In the exercise, the inner function is \( g(x) = 5\sqrt{x} \), which produces the value that the outer function, \( f(u) = \cot\left(\frac{\pi u}{10}\right) \), then processes.
• Substitute \( g(x) \) into \( f(u) \) to get the composite form: \[ f(g(x)) = \cot\left(\frac{\pi \cdot 5\sqrt{x}}{10}\right) \]

Recognizing the composition is crucial as it sets the stage for applying differentiation rules systematically. It helps to focus on each function's role within the composition when performing calculations.