Problem 12 - Thomas Calculus

Question

In Exercises \(1-12,\) find \(d y / d x\) $$ y=x^{2} \cos x-2 x \sin x-2 \cos x $$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = -x^2 \sin x\).

Step 1: Identify the Functions

Recognize that the function consists of four terms: \(x^2\), \(\cos x\), \(-2x\sin x\), and \(-2\cos x\). The terms will need to be differentiated, some using product rule and others using basic differentiation techniques.

Step 2: Apply the Product Rule to the First Term

For the term \(x^2 \cos x\), use the product rule: if \(u = x^2\) and \(v = \cos x\), then \(d(uv)/dx = u'v + uv'\). Thus, \(\frac{d}{dx}(x^2\cos x) = 2x \cos x - x^2 \sin x\).

Step 3: Differentiate the Second Term

The second term is \(-2x\sin x\). Using the product rule here as well, if \(u = -2x\) and \(v = \sin x\), then \(\frac{d}{dx}(-2x\sin x) = -2\sin x - 2x\cos x\).

Step 4: Differentiate the Third Term

The third term is \(-2\cos x\). The derivative of \(-2\cos x\) with respect to \(x\) is \(2\sin x\).

Step 5: Combine All Derivatives

Combine all the derivatives from Steps 2, 3, and 4 together: \(2x \cos x - x^2 \sin x - 2\sin x - 2x\cos x + 2\sin x\).

Step 6: Simplify the Expression

Combine like terms in the expression: \(2x \cos x - x^2 \sin x - 2x\cos x + 2\sin x - 2\sin x = -x^2 \sin x\). Thus, the derivative is: \(\frac{dy}{dx} = -x^2 \sin x\).

Key Concepts

Product Rule

The product rule is a fundamental differentiation technique used in calculus when dealing with functions that are products of two differentiable functions. It provides a systematic way to differentiate these products easily. When we have a function expressed as the product of two functions, say \( u(x) \) and \( v(x) \), the product rule formula is given by: \[ \frac{d}{dx}[u(x) \, v(x)] = u'(x) \, v(x) + u(x) \, v'(x) \]This means that to differentiate \( u(x) \, v(x) \), we take the derivative of the first function and multiply by the second function, then add it to the first function multiplied by the derivative of the second function. Let's consider an example: if \( u(x) = x^2 \) and \( v(x) = \cos x \), then applying the product rule would look like this:

• \( u'(x) = 2x \)
• \( v'(x) = -\sin x \)

Therefore, the derivative of \( x^2 \cos x \) is \( 2x \cos x - x^2 \sin x \). Understanding the product rule can simplify the process of differentiation and is essential for tackling complex problems involving products of functions.

Differentiation Techniques

Differentiation techniques encompass various methods used to find the derivative of different types of functions. Each method allows us to approach differentiation efficiently, depending on the structure of the function involved. Some of the main techniques include:

• **Basic Differentiation:** This technique involves using standard derivative formulas such as \( \frac{d}{dx}[x^n] = nx^{n-1} \), \( \frac{d}{dx}[\sin x] = \cos x \), and \( \frac{d}{dx}[\cos x] = -\sin x \).
• **Product Rule:** Used when differentiating products of two functions, as explained in detail above.
• **Chain Rule:** Commonly applied in composite functions, where the derivative of an outer function is multiplied by the derivative of the inner function.

To practice these techniques, you should identify the type of function you have and choose the appropriate technique. For instance, in our original exercise, the product rule and basic differentiation were used. Mastery of these techniques is crucial for solving a wide range of calculus problems.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent are vital elements in calculus, often appearing in various mathematical models and equations. Understanding how to differentiate these functions is crucial because they frequently appear in calculus problems. The basic derivatives you’ll encounter with trigonometric functions include:

• \( \frac{d}{dx}[\sin x] = \cos x \)
• \( \frac{d}{dx}[\cos x] = -\sin x \)
• \( \frac{d}{dx}[\tan x] = \sec^2 x \)

In the original exercise, differentiating \( \cos x \) and \( \sin x \) was necessary to find the derivative of terms like \(-2x \sin x \) and \(-2 \cos x \). When dealing with derivatives of trigonometric functions, always remember these basic rules. Moreover, understanding the derivatives of trigonometric functions is also stepping stone to tackling more complicated trigonometric identities and integrals.