Problem 181 - Calculus Volume 1

Question

For the following exercises, find \(\frac{d y}{d x}\) for the given functions. $$y=(x+\cos x)(1-\sin x)$$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = 1 - 2\sin x + \sin^2 x - x\cos x - \cos^2 x \)

Step 1: Identify the Rule

We must find the derivative of the product of two functions. Here, our functions are \(u = x + \cos x\) and \(v = 1 - \sin x\). To find the derivative \(\frac{dy}{dx}\), use the product rule: \(\frac{d}{dx}[uv] = u'v + uv'\).

Step 2: Differentiate First Function

Differentiate \(u = x + \cos x\). The derivative is \(u' = \frac{d}{dx}(x) + \frac{d}{dx}(\cos x) = 1 - \sin x\).

Step 3: Differentiate Second Function

Differentiate \(v = 1 - \sin x\). The derivative is \(v' = \frac{d}{dx}(1) - \frac{d}{dx}(\sin x) = -\cos x\).

Step 4: Apply the Product Rule

Substitute \(u\), \(v\), \(u'\), and \(v'\) into the product rule: \(\frac{d}{dx}[(x+\cos x)(1-\sin x)] = (1-\sin x)(1-\sin x) + (x+\cos x)(-\cos x)\).

Step 5: Simplify the Expression

Simplify the expression: \((1-\sin x)^2 - (x+\cos x)\cos x = 1 - 2\sin x + \sin^2 x - x\cos x - \cos^2 x\).

Key Concepts

Product Rule

The product rule is a fundamental technique in calculus used to differentiate expressions where two functions are multiplied together. It states that if you have two functions, say \( u(x) \) and \( v(x) \), the derivative of their product is given by:

• \( \frac{d}{dx}[u \cdot v] = u'v + uv' \)

This means you need to differentiate each function separately and then apply the rule by multiplying the derivative of the first function with the second function, and vice versa, before adding them together.

In the case of the original problem, the functions to differentiate are \( u = x + \cos x \) and \( v = 1 - \sin x \). By using the product rule, one can efficiently find the overall derivative \( \frac{dy}{dx} \). This serves as a powerful tool when working with more complex expressions composed of multiple types of functions.

Trigonometric Differentiation

Trigonometric differentiation involves finding the derivatives of functions that include trigonometric terms like \( \sin x \) and \( \cos x \). These derivatives are standard and should be memorized for quick use:

• The derivative of \( \sin x \) is \( \cos x \).
• The derivative of \( \cos x \) is \(-\sin x \).

Trigonometric differentiation is used extensively when solving calculus problems that feature trigonometric functions in their expressions.

In our problem, to differentiate \( u = x + \cos x \), we use the trigonometric rule to discover that its derivative is \( u' = 1 - \sin x \). Similarly, differentiating \( v = 1 - \sin x \) gives \( v' = -\cos x \) using \( \sin x \)'s derivative.
This knowledge simplifies handling derivatives of expressions with trigonometric functions, allowing us to apply the product rule effectively.

Derivative of Trigonometric Functions

Understanding the derivative of trigonometric functions is crucial when you are working on problems that involve sine and cosine terms. These functions are periodic in nature and have derivatives that are also trigonometric, creating a cyclical pattern:

• The derivative of \( \sin x \) is \( \cos x \), indicating the rate of change of the sine function relative to x.
• The derivative of \( \cos x \) is \(-\sin x \), showing how it changes direction as \( x \) changes.

In calculus, these derivatives are foundational, especially when functions like \( u(x) \) and \( v(x) \) include both polynomial and trigonometric components as seen in composite functions.

When we apply these derivatives in combination with the product rule, we can accurately deduce the derivative of a product of two functions containing trigonometric elements. For instance, when dealing with \( y = (x+\cos x)(1-\sin x) \), knowing these derivatives allows us to confidently compute the overall derivative using the steps outlined with the product rule.