Problem 16 - University Calculus: Early Transcendentals

Question

Find \(d y / d x\). $$y=x^{2} \cos x-2 x \sin x-2 \cos x$$

Step-by-Step Solution

Answer

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

Step 1: Differentiate Each Term

We need to find the derivative of the function \( y = x^2 \cos x - 2x \sin x - 2 \cos x \). The function consists of three terms that we will differentiate separately: \(x^2 \cos x\), \(-2x \sin x\), and \(-2 \cos x\).

Step 2: Differentiate the First Term Using the Product Rule

The first term is \(x^2 \cos x\) which is a product of two functions: \(x^2\) and \(\cos x\). Use the product rule: \((uv)' = u'v + uv'\). Here, \(u = x^2\) and \(v = \cos x\). So, \(u' = 2x\) and \(v' = -\sin x\). The derivative of \(x^2 \cos x\) is \((2x \cdot \cos x) + (x^2 \cdot (-\sin x)) = 2x \cos x - x^2 \sin x\).

Step 3: Differentiate the Second Term Using the Product Rule

The second term is \(-2x \sin x\) which is also a product of two functions: \(-2x\) and \(\sin x\). Again using the product rule: \(u = -2x\), \(v = \sin x\), \(u' = -2\), \(v' = \cos x\). The derivative of \(-2x \sin x\) is \((-2 \cdot \sin x) + (-2x \cdot \cos x) = -2 \sin x - 2x \cos x\).

Step 4: Differentiate the Third Term

The third term is \(-2 \cos x\), which is a simple cosine term. The derivative of \(-2 \cos x\) is \(2 \sin x\) since the derivative of \(\cos x\) is \(-\sin x\).

Step 5: Combine All the Derivatives

Combine the derivatives of the individual terms: \[\frac{d}{dx}(x^2 \cos x) = 2x \cos x - x^2 \sin x\], \[\frac{d}{dx}(-2x \sin x) = -2 \sin x - 2x \cos x\], \[\frac{d}{dx}(-2 \cos x) = 2 \sin x\].Add them all together to get: \[\frac{dy}{dx} = (2x \cos x - x^2 \sin x) + (-2 \sin x - 2x \cos x) + 2 \sin x\].

Step 6: Simplify the Expression

Simplify the expression by combining like terms: \[2x \cos x - x^2 \sin x - 2 \sin x - 2x \cos x + 2 \sin x = -x^2 \sin x\].So, the derivative \( \frac{dy}{dx} \) simplifies to \( -x^2 \sin x \).

Key Concepts

Understanding the Product Rule

The product rule is a fundamental technique in calculus used to differentiate functions that are combined through multiplication. It's especially useful when you encounter terms like \(x^n \cdot f(x)\), where both parts are functions needing their own derivatives. The formula for the product rule is \((uv)' = u'v + uv'\), where \(u\) and \(v\) are functions of \(x\).

• First, identify \(u\) and \(v\). These are the two parts of your product that you want to differentiate.
• Differentiate each part: Find \(u'\) and \(v'\), the derivatives of \(u\) and \(v\) respectively.
• Apply the rule: Plug \(u\), \(v\), \(u'\), and \(v'\) into \(u'v + uv'\).

In the given exercise, to differentiate \(x^2 \cos x\), the product rule helped us break it into manageable derivatives. Here, \(u = x^2\) with \(u' = 2x\), and \(v = \cos x\) with \(v' = -\sin x\). This leads to the derivative: \(2x \cos x - x^2 \sin x\). Applying the product rule correctly ensures you account for both changing parts of the product.

Mastering Trigonometric Functions

Trigonometric functions, such as \(\sin x\) and \(\cos x\), are pivotal in calculus due to their periodic properties and derivatives. Understanding these functions and their derivatives is crucial.

• Basic Derivatives: For calculus, the three main functions you'll encounter are \(\sin x\), \(\cos x\), and \(\tan x\). The derivatives are \(\frac{d}{dx}\sin x = \cos x\), \(\frac{d}{dx}\cos x = -\sin x\), and \(\frac{d}{dx}\tan x = \sec^2 x\).
• Applying to Problem Solving: When differentiating expressions like \(\cos x\) or \(\sin x\), utilize these derivatives directly, like in \(-2 \cos x\) where the derivative becomes \(2 \sin x\). Trigonometric identities and derivatives simplify complex expressions.

Keeping these derivatives in mind will greatly aid in handling calculus problems, especially those involving more complex trigonometric equations. Each function translates natural phenomena into a mathematical framework, better comprehensible through calculus.

The Art of Simplification

Simplification is the final stage of solving a calculus problem, where you make the expression as concise as possible. While it might seem like a straightforward task, it requires careful attention to combining like terms.

• Identify Like Terms: Look for terms with identical variables and exponents. Here, terms like \(2x \cos x\) from different parts of the derivative can be combined and simplified.
• Check and Double-Check: Ensure that all terms are accounted for and simplified correctly. Avoid missing negative signs or terms that cancel each other out.
• Simplified Expression: The simplified derivative for this exercise was \(-x^2 \sin x\), derived by cancelling out components such as \(2 \sin x - 2 \sin x\) and \(2x \cos x - 2x \cos x\).

Proper simplification not only provides the correct derivative but also makes the expression much easier to understand and utilize in further applications. Accurate simplification is a signature step in solving calculus problems.