Problem 6 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the functions

We are given two functions, where the outer function is \(y = \sin u\) and the inner function is \(u = x - \cos x\). We need to find the derivative of \(y\) with respect to \(x\) using the chain rule.

Step 2: Apply the chain rule

By the chain rule, the derivative \( \frac{dy}{dx} \) is calculated as the derivative of \(y\) with respect to \(u\) multiplied by the derivative of \(u\) with respect to \(x\). Thus, \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).

Step 3: Differentiate the outer function

The outer function is \(y = \sin u\). Its derivative with respect to \(u\) is \( \frac{dy}{du} = \cos u \).

Step 4: Differentiate the inner function

The inner function is \(u = x - \cos x\). Differentiating with respect to \(x\), we have:\( \frac{du}{dx} = 1 + \sin x \).

Step 5: Substitute the derivatives back into the chain rule

Substituting the derivatives back into the chain rule formula, we have:\( \frac{dy}{dx} = \cos u \cdot (1 + \sin x) \).

Step 6: Substitute \(u\) back into the expression

Since \(u = x - \cos x\), substitute \(u\) into the expression to get:\( \frac{dy}{dx} = \cos(x - \cos x) \cdot (1 + \sin x) \).

Key Concepts

Derivatives

Derivatives are fundamental in calculus as they express how a function changes at any point. In this exercise, the goal is to compute the derivative of a composite function using the chain rule. While derivatives might initially seem abstract, think of them as a rate of change, like measuring how fast a car travels. Here, you have two functions: the outer function, represented by \( y = \sin u \), and the inner function, \( u = x - \cos x \). To find \( \frac{dy}{dx} \), the rate at which \( y \) changes with respect to \( x \), we utilize their derivatives: \( \frac{dy}{du} = \cos u \) for the outer function and \( \frac{du}{dx} = 1 + \sin x \) for the inner function. The chain rule will help us combine these to find the overall derivative. Remember:

• Derivatives provide the slope of the function at a particular point.
• Think of it as describing the instantaneous rate of change.

Just like a slope tells you how steep a hill is, a derivative tells you how steep a curve is at a particular point.

Trigonometric Functions

Trigonometric functions like \( \sin \), \( \cos \), and \( \tan \) are essential in many mathematical applications, especially involving waves and periodic phenomena. In our exercise, \( \sin u \) and \( \cos x \) play critical roles. The sine function oscillates between -1 and 1, making it periodic and useful in describing cycles. When you differentiate \( \sin u \), it transforms into \( \cos u \). The fact that the derivative of \( \sin \) is \( \cos \) reflects the interconnected nature of these functions:

• Sine and cosine are phase-shifted by 90°, hence their derivatives reflect this offset.
• They model cyclic patterns, seen in fields like physics and engineering.

Understanding their derivatives is crucial in predicting how changes in one part of a cycle affect the whole, much like predicting high and low tides by analyzing the moon's position.

Composite Functions

Composite functions occur when one function is nested inside another, like \( y = \sin(u) \) where \( u = x - \cos x \). This nesting creates layers, and to differentiate them, the chain rule becomes indispensable. Picture composite functions like layered cakes: each layer affects the entire structure. When tackling composite functions:

• Identify both the outer and inner functions.
• Use the chain rule: differentiate the outer function first, followed by the inner one.

As demonstrated, apply \( \frac{dy}{dx} = \cos(u) \cdot (1 + \sin x) \) by substituting \( u = x - \cos x \) into the differentiated expression. This substitution reveals the overall rate of change by considering how changes in \( x \) ripple through both functions, accurately capturing the dynamics at play in this combination.