Problem 6 - University Calculus: Early Transcendentals

Question

In Exercises \(1-8,\) given \(y=f(u)\) and \(u=g(x),\) find \(d y / d x=\) \(f^{\prime}(g(x)) g^{\prime}(x)\). $$y=\sin u, \quad u=x-\cos x$$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \cos(x - \cos(x)) (1 + \sin(x)) \)."

Step 1: Differentiate the Outer Function

First, we need to find the derivative of the outer function. Here, the outer function is given as \( y = \sin(u) \). The derivative of \( \sin(u) \) with respect to \( u \) is \( \cos(u) \). This is noted as \( f'(u) = \cos(u) \).

Step 2: Differentiate the Inner Function

Next, differentiate the inner function \( u = x - \cos(x) \) with respect to \( x \). We have:1. The derivative of \( x \) is \( 1 \).2. The derivative of \( \cos(x) \) is \( -\sin(x) \).Thus, the derivative of the entire function \( g'(x) = 1 + \sin(x) \).

Step 3: Apply the Chain Rule

We apply the chain rule to find \( \frac{dy}{dx} \), which is \( f'(g(x)) \cdot g'(x) \). Substituting what we found in Steps 1 and 2:1. \( f'(g(x)) = \cos(x - \cos(x)) \)2. \( g'(x) = 1 + \sin(x) \)Therefore, \( \frac{dy}{dx} = \cos(x - \cos(x)) (1 + \sin(x)) \).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus focusing on finding the rate at which a function changes at any given point. This rate of change is known as the derivative. Understanding differentiation helps in analyzing how changes in one variable affect another in functions of a single variable.
For example, if you were driving a car, differentiation allows you to find the speed at any given moment by looking at how your distance changes over time.
In the exercise provided, we differentiate two parts of a composite function: the outer and the inner functions. By working out the derivative of each, we can find the derivative of the composite function as a whole.

Composite Functions

Composite functions occur when one function is inside another, such as in the problem where we have the composition of functions: \( y = f(u) \) and \( u = g(x) \). Think of it as a machine where the output of one function becomes the input for another.
To understand composite functions better, consider replacing \( u \) in \( f(u) \) with \( g(x) \), forming \( f(g(x)) \).
In differentiation, we often encounter this scenario and thus, have to unravel these layers correctly. Doing so involves using a specific technique called the Chain Rule, which is tailor-made for differentiating composite functions, ensuring we correctly capture how each layer of the function affects the overall derivative.
By breaking down complex functions into manageable parts, composite functions illustrate how interconnected functions are in the world of calculus.

Calculus

Calculus is the mathematical study of continuous change and deals with quantities changing over time. Major branches of calculus include differentiation and integration.
The focus of this exercise is on differentiation, which is a central piece of calculus. It helps us find instantaneous rates of change and slopes of curves.
In this particular exercise, through using the Chain Rule, we see how calculus provides a systematic approach to handling multiple layers in functions, simplifying the process of finding derivatives.
Through calculus, mathematicians and scientists can describe the precise nature of motion, growth, decay, and many other dynamic processes, making it an essential tool in a variety of disciplines.