Problem 67 - Calculus Single Variable

Question

Calculate the derivative of the given expression. $$ x \sin (x) \cos (x) $$

Step-by-Step Solution

Answer

The derivative is \( \sin(x) \cos(x) + x \cos^2(x) - x \sin^2(x) \).

Step 1: Recognize the Product Rule

To differentiate the expression \( x \sin(x) \cos(x) \), notice that it is a product of two functions: \( x \) and \( \sin(x) \cos(x) \). Thus, we will use the product rule for derivatives, which is \((u v)' = u'v + uv'\), where \( u = x \) and \( v = \sin(x) \cos(x) \).

Step 2: Differentiate \( x \)

Apply the derivative rule for \( x \), which is simply \( 1 \) because the derivative of \( x \) with respect to \( x \) is \( 1 \). Therefore, \( u' = 1 \).

Step 3: Differentiate \( \sin(x) \cos(x) \) Using the Product Rule

\( v = \sin(x) \cos(x) \) is also a product of two functions, so use the product rule again: \((ab)' = a'b + ab'\), where \( a = \sin(x) \) and \( b = \cos(x) \). The derivatives are: \( a' = \cos(x) \) and \( b' = -\sin(x) \). Therefore, \( v' = \cos(x) \cdot \cos(x) + \sin(x) (-\sin(x)) = \cos^2(x) - \sin^2(x) \).

Step 4: Substitute the Derivatives into the Product Rule

Now substitute \( u' = 1 \), \( v = \sin(x) \cos(x) \), and \( v' = \cos^2(x) - \sin^2(x) \) into the product rule formula: \( (uv)' = u'v + uv' = 1 \cdot \sin(x) \cos(x) + x(\cos^2(x) - \sin^2(x)) \).

Step 5: Simplify the Expression

Simplify the expression: \( \sin(x) \cos(x) + x(\cos^2(x) - \sin^2(x)) = \sin(x) \cos(x) + x \cos^2(x) - x \sin^2(x) \). This is the derivative of the original function.

Key Concepts

Understanding the Product Rule

The product rule is a fundamental technique in calculus used to find the derivative of a product of two functions. When two functions, say \( u(x) \) and \( v(x) \), are multiplied together, their derivative is not simply the product of their individual derivatives. Instead, the product rule must be used.

According to the product rule, if \( y = u(x) \cdot v(x) \), then the derivative \( y' \) is given by:

• \( y' = u'(x) \cdot v(x) + u(x) \cdot v'(x) \)

In the context of our example, we recognize our function \( x \sin(x) \cos(x) \) as a product of two separate functions, \( u(x) = x \) and \( v(x) = \sin(x) \cos(x) \). This means we need to apply the product rule carefully, first identifying each function and then calculating its derivative.

Identifying the correct functions and their derivatives is crucial in applying the product rule. Many common errors originate from not setting up these components correctly. Always verify the breakdown of functions before proceeding to differentiation.

Deep Dive into Trigonometric Functions

Trigonometric functions such as \( \sin(x) \) and \( \cos(x) \) are fundamental in calculus and have many important properties and identities. These functions are periodic with a specific pattern of derivatives. Understanding these basic derivatives is essential for solving problems effectively.

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

In our problem, we have \( \sin(x) \cos(x) \), a product of two trigonometric functions. To differentiate this, we use the product rule again, showing the recurrence and importance of this rule in differentiating complex or composite functions.Moreover, simplifying expressions involving trigonometric functions can sometimes rely on identities such as the Pythagorean identity \( \sin^2(x) + \cos^2(x) = 1 \). Knowing these identities provides a toolkit for both simplifying derivatives and checking the correctness of each computation stage.

Differentiation Techniques Demystified

Differentiation is the process of finding the derivative of a function, which gives the function's rate of change or the slope at any point. Mastery of different differentiation techniques is essential for solving a wide array of problems across calculus.The main differentiation techniques include:

• Basic Rules: These are the power rule, product rule, quotient rule, and chain rule. Each rule applies to specific kinds of function combinations.
• Understanding Function Types: Differentiate between linear, trigonometric, exponential, and logarithmic functions.
• Composite Functions: Use the chain rule for functions nested within each other, which is crucial when functions are not straightforward polynomials or trigonometric forms.

In our task, the product of a linear function \( x \) and a composite trigonometric function \( \sin(x) \cos(x) \) meant using the product rule not once, but twice. This iterated application demonstrates the interwoven nature of these techniques.

Practicing these techniques improves speed and accuracy in solving calculus problems, preparing students for more complex mathematical challenges.