Problem 7 - Calculus

Question

Find the derivatives of the functions in \(1-26\). $$ x^{2} \cos x+2 x \sin x $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = 4x \cos x - x^2 \sin x + 2 \sin x \).

Step 1: Understand the Problem

We need to find the derivative of the function \( f(x) = x^2 \cos x + 2x \sin x \). This expression consists of two terms. We will apply derivative rules to each term separately.

Step 2: Apply the Derivative Sum Rule

Since the function is a sum of two terms, \( x^2 \cos x \) and \( 2x \sin x \), we apply the sum rule: the derivative of the sum is the sum of the derivatives.\[ f'(x) = (x^2 \cos x)' + (2x \sin x)' \]

Step 3: Differentiate the First Term using Product Rule

For the first term \( x^2 \cos x \), we need to use the product rule, which states: \((u v)' = u' v + u v'\). Let \( u = x^2 \) and \( v = \cos x \).- Differentiate \( u \): \( u' = 2x \).- Differentiate \( v \): \( v' = -\sin x \).Apply the product rule:\[ (x^2 \cos x)' = 2x \cos x - x^2 \sin x \]

Step 4: Differentiate the Second Term using Product Rule

For the second term \( 2x \sin x \), apply the product rule, where \( u = 2x \) and \( v = \sin x \).- Differentiate \( u \): \( u' = 2 \).- Differentiate \( v \): \( v' = \cos x \).Apply the product rule:\[ (2x \sin x)' = 2 \sin x + 2x \cos x \]

Step 5: Combine the Two Derivatives

Now, combine the derivatives we calculated for each term:\[ f'(x) = (x^2 \cos x)' + (2x \sin x)' = (2x \cos x - x^2 \sin x) + (2 \sin x + 2x \cos x) \]

Step 6: Simplify the Expression

Add the expressions to simplify:1. Combine \(2x \cos x\) terms: \(2x \cos x + 2x \cos x = 4x \cos x\).2. Then write down the remaining terms without any further simplification, as they do not combine anymore.Thus,\[ f'(x) = 4x \cos x - x^2 \sin x + 2 \sin x \]

Key Concepts

Understanding the Product Rule

The product rule is a fundamental concept in calculus, used especially when you need to differentiate expressions that consist of two multiplied functions. Let's break it down in simple steps.

When you have a function that looks like a product of two other functions, say \( u(x) \) and \( v(x) \), you must use the product rule to find its derivative. The rule is formulated as follows:

\( (uv)' = u'v + uv' \).

This equation tells us that the derivative of the product of two functions is the derivative of the first function multiplied by the second one, plus the first function multiplied by the derivative of the second function. This rule helps in handling complex expressions as you divide the task into smaller, easily manageable steps.

• First, identify each function in the product.
• Find the derivatives of the individual functions.
• Apply the product rule formula to obtain the final derivative.

In our example with \( x^2 \cos x \), we differentiate \( x^2 \) to get \( 2x \), and \( \cos x \) to get \( -\sin x \). Then, we follow the product rule to find the derivative of the term.

Remember, practice makes perfect when it comes to mastering the product rule! It becomes intuitive the more you work with such expressions.

Exploring the Sum Rule in Derivatives

The sum rule is another basic yet crucial concept when dealing with derivatives. It primarily makes the process of differentiation straightforward for functions that are sums of several terms.

If you have a function expressed as a sum, say \( f(x) = g(x) + h(x) \), the sum rule allows you to differentiate each term separately and then simply add those derivatives together. The mathematical representation of the sum rule is:

\( (f + g)' = f' + g' \).

Using this rule often simplifies complicated functions considerably:

• Identify individual terms in the sum.
• Find the derivative of each term separately.
• Add the derivatives together to arrive at the result.

In our original problem, the function \( f(x) = x^2 \cos x + 2x \sin x \) is a sum of two parts. By applying the sum rule, we differentiated \( x^2 \cos x \) and \( 2x \sin x \) and summed their derivatives. This simplification by focusing on smaller parts helps manage errors and understand each component's role in the final function.

Diving into Trigonometric Functions

Trigonometric functions like \( \sin x \) and \( \cos x \) appear often in calculus due to their periodic nature and roots in geometry. Differentiating these basic functions is an essential skill as they form building blocks for more complex problems.

Here are some fundamental derivatives related to these functions:

• The derivative of \( \sin x \) is \( \cos x \).
• The derivative of \( \cos x \) is \( -\sin x \).
• The derivative of \( \tan x \) is \( \sec^2 x \) (useful in further studies).

Each derivative represents how the function changes, which is crucial in understanding their behavior over periodic intervals. In our original function, both \( \sin x \) and \( \cos x \) were present, demanding a clear grasp of their derivatives to correctly apply in the product and sum rules when calculating \( f'(x) \).

The ability to quickly and correctly differentiate trig functions is not only important for exams but also lays the groundwork for more advanced topics in calculus and physics.