Problem 11 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The derivative \( \frac{dy}{dx} = x^2 \cos x \).

Step 1: Identify the Function Components

We need to find the derivative \( \frac{dy}{dx} \) of the given function: \( y = x^2 \sin x + 2x \cos x - 2 \sin x \). The function has three components: \( x^2 \sin x \), \( 2x \cos x \), and \( -2 \sin x \). The derivative will be the sum of the derivatives of these components.

Step 2: Derivation Using the Product Rule

For \( x^2 \sin x \), we use the product rule: \((uv)' = u'v + uv'\). Here, \( u = x^2 \) and \( v = \sin x \). Compute the derivatives: \( u' = 2x \) and \( v' = \cos x \). Therefore, the derivative is \( 2x \sin x + x^2 \cos x \).

Step 3: Differentiate Second Term

For the term \( 2x \cos x \), we again use the product rule. With \( u = 2x \) and \( v = \cos x \), the derivatives are \( u' = 2 \) and \( v' = -\sin x \). Thus, the derivative is \( 2 \cos x - 2x \sin x \).

Step 4: Derivative of Third Term

For the term \(-2 \sin x\), the derivative is straightforward: the derivative of \( \sin x \) is \( \cos x \). Therefore, this part becomes \(-2 \cos x\).

Step 5: Combine the Derivatives

Now, sum the derivatives obtained in steps 2, 3, and 4 to find \( \frac{dy}{dx} \). This gives us: \[ \frac{dy}{dx} = (2x \sin x + x^2 \cos x) + (2 \cos x - 2x \sin x) - 2 \cos x \].

Step 6: Simplify the Expression

Simplify the expression by combining like terms. Notice \( 2x \sin x \) cancels out with \(-2x \sin x\) and \( 2 \cos x \) cancels out with \(-2 \cos x\). We are left with \( \frac{dy}{dx} = x^2 \cos x \).

Key Concepts

product rule

The product rule is a fundamental technique used in calculus to evaluate the derivative of a product of two functions. Let's break it down using a simple notation. Suppose you have two functions, say, \( u(x) \) and \( v(x) \). The product rule states that the derivative of their product \( (uv)' \) is given by:

• \( (uv)' = u'v + uv' \)

Here, \( u' \) represents the derivative of \( u \) with respect to \( x \), and \( v' \) represents the derivative of \( v \).
This rule comes in handy when you encounter expressions consisting of the product of two simpler functions, as is the case in our original exercise. In our example, we applied the product rule to the term \( x^2 \sin x \), considering \( x^2 \) and \( \sin x \) as our two functions. Each derivative is calculated separately and then combined according to the product rule. This technique is essential to mastering differentiation for more complex algebraic expressions.

trigonometric functions

Trigonometric functions like \( \sin x \) and \( \cos x \) play a significant role in calculus problems, including differentiation. These functions are periodic, with derivatives that cycle through a distinct pattern. Knowing their derivatives is crucial to solving problems involving trigonometric terms:

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

These properties are used when differentiating expressions that include trigonometric functions. In our exercise, these derivatives were necessary for each trigonometric part of the expression:
- For \( x^2 \sin x \), we used the sine function's derivative along with the product rule.
- For \( 2x \cos x \), the cosine derivative was applied similarly.
Being familiar with these trigonometric derivatives allows us to handle intricate expressions involving sinusoidal components effectively.

simplification

Simplification in calculus involves combining like terms and eliminating redundant components of an expression to reach a more concise form. During this step, we focus on making the expression easier to interpret and manage. In our example problem, simplification was vital after separately differentiating each term of the function.
Once the derivatives were combined, we noticed cancellations:- \( 2x \sin x \) and \( -2x \sin x \) canceled each other out.- Similarly, \( 2 \cos x \) and \( -2 \cos x \) neutralized.
Through simplification, we reduced the expression to \( x^2 \cos x \).
This process not only provides the final result in a more readable format but also reduces the complexity of dealing with the result algebraically. Simplification is a critical stage in derivative computations to ensure clarity and correctness.

differentiation steps

Differentiation steps are the sequential methods followed to find the derivative of a given function. It's important to approach each function systematically:

• First, identify the main components of the function. Break it down into parts that can be individually differentiated, as seen with the terms \( x^2 \sin x \), \( 2x \cos x \), and \( -2\sin x \).
• Next, apply the appropriate rules, such as the product rule and basic derivatives of trigonometric functions, for each component. This was demonstrated by using separate approaches for each term according to its structure.
• Finally, combine the resulting derivatives and simplify.

This structured method ensures that no step is overlooked and that the derived expression is accurate. Focusing on each differentiation step allows students to grasp the rationale behind the rules and see how they apply in varied contexts. With practice, these steps become second nature and significantly enhance problem-solving skills in calculus.